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IB Maths Applications & Interpretation · HL · Topic 1

Number and algebra

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1IBMI-030-04-r2-01Complex numbers and matrices Paper 1Easy6 marks

In an alternating-current circuit, impedances are modelled as complex numbers measured in ohms (Ω). Two components have impedances \(Z_1 = 3 + 4i\) and \(Z_2 = 2 - 5i\).
When components are connected in series, the total impedance is \(Z_S = Z_1 + Z_2\). When they are connected in parallel, the total impedance \(Z_P\) satisfies \(\frac{1}{Z_P} = \frac{1}{Z_1} + \frac{1}{Z_2}\).

(a)

Write down \(Z_S\) in the form \(a + bi\).

(1)
(b)

Show that \(Z_P = \frac{Z_1 Z_2}{Z_1 + Z_2}\) and hence find \(Z_P\) in the form \(a + bi\).

(3)
(c)

Find the magnitude of the parallel impedance, \(|Z_P|\).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
\(Z_S = 5 - i\) Ω✓ 1
Part (b)
Combines the fractions: \(\frac{1}{Z_P} = \frac{Z_1+Z_2}{Z_1Z_2}\), so \(Z_P = \frac{Z_1Z_2}{Z_1+Z_2}\)✓ 1
Substitutes: \(Z_P = \frac{26 - 7i}{5 - i}\)✓ 1
\(Z_P = 5.27 - 0.346i\) Ω✓ 1
Part (c)
Uses \(\sqrt{a^2+b^2}\), e.g. \(\sqrt{5.269^2 + 0.3462^2}\)✓ 1
\(|Z_P| = 5.28\) Ω✓ 1

Exam tip: Your GDC can divide complex numbers directly. Still write the unsimplified fraction first, because that working step earns a method mark even if you mistype a value.

2IBMI-030-04-r2-02Complex numbers and matrices Paper 1Easy6 marks

A café sells coffee, tea and cake. The numbers sold on Monday, Tuesday and Wednesday of one week are given by the matrix
\(Q = \begin{pmatrix} 12 & 8 & 15 \\ 10 & 14 & 9 \\ 20 & 11 & 13 \end{pmatrix}\)
where the rows are Monday, Tuesday and Wednesday, and the columns are coffee, tea and cake. In that week the prices, in dollars, are \(P = \begin{pmatrix} 2.80 \\ 2.20 \\ 3.50 \end{pmatrix}\).

(a)

Find \(QP\) and state what its entries represent.

(2)
(b)

The following week the café sells exactly the same quantities but changes its prices. The revenues on Monday, Tuesday and Wednesday are $116.00, $101.00 and $139.50 respectively. By forming a matrix equation, find the new price of a cake.

(4)
Show mark scheme
Marking pointMarkNotes
Part (a)
\(QP = \begin{pmatrix} 103.70 \\ 90.30 \\ 125.70 \end{pmatrix}\)✓ 1
These are the total revenues, in dollars, on Monday, Tuesday and Wednesday✓ 1
Part (b)
Writes \(Q\begin{pmatrix} c \\ t \\ k \end{pmatrix} = \begin{pmatrix} 116 \\ 101 \\ 139.5 \end{pmatrix}\)✓ 1
Uses the inverse: \(\begin{pmatrix} c \\ t \\ k \end{pmatrix} = Q^{-1}\begin{pmatrix} 116 \\ 101 \\ 139.5 \end{pmatrix}\)✓ 1
Obtains \(c = 3.00\), \(t = 2.50\)✓ 1
Cake costs $4.00✓ 1

Common error: Pre-multiply by the inverse: the answer is \(Q^{-1}R\), not \(RQ^{-1}\). The product \(RQ^{-1}\) is not even defined for a 3×1 column matrix \(R\).

3IBMI-030-04-r1-01Complex numbers and matrices Paper 1Easy5 marks

An engineer models two components in an alternating-current circuit using complex impedances, measured in ohms (Ω):
\(Z_1 = 4 + 3i\) and \(Z_2 = 2 - 6i\).
When the components are connected in series, the total impedance is \(Z_S = Z_1 + Z_2\). When they are connected in parallel, the total impedance \(Z_P\) satisfies \(\frac{1}{Z_P} = \frac{1}{Z_1} + \frac{1}{Z_2}\).

(a)

Find \(Z_S\), giving your answer in the form \(a + bi\).

(1)
(b)

Find \(Z_P\), giving your answer in the form \(a + bi\).

(3)
(c)

The magnitude of the impedance is \(|Z_P|\). Find \(|Z_P|\).

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(Z_S = 6 - 3i\) Ω✓ 1
Part (b)
M1 uses \(Z_P = \frac{Z_1 Z_2}{Z_1 + Z_2}\) or adds the reciprocals \(\frac{1}{Z_1}+\frac{1}{Z_2}\)✓ 1
A1 \(Z_1 Z_2 = 26 - 18i\) (or equivalent correct intermediate value)✓ 1
A1 \(Z_P = 4.67 - 0.667i\) Ω (exact: \(\frac{14}{3} - \frac{2}{3}i\))✓ 1
Part (c)
A1 \(|Z_P| = \sqrt{4.667^2 + 0.667^2} = 4.71\) Ω✓ 1

Common error: Writing \(Z_P = \frac{1}{Z_1} + \frac{1}{Z_2}\) and forgetting to take the reciprocal at the end. Using \(\frac{Z_1Z_2}{Z_1+Z_2}\) avoids this, and a GDC in complex mode handles the division directly.

4IBMI-030-04-r1-02Complex numbers and matrices Paper 1Easy5 marks

A graphic designer uses the transformation with matrix \(T = \begin{pmatrix} 2 & 1 \\ -1 & 3 \end{pmatrix}\) to distort a company logo drawn on a coordinate grid, where units are centimetres. A point with position vector \(\begin{pmatrix} x \\ y \end{pmatrix}\) is mapped to \(T\begin{pmatrix} x \\ y \end{pmatrix}\).

(a)

Find the image of the point \((3, -2)\).

(1)
(b)

A point on the original logo is mapped to \((5, 8)\). Find the coordinates of this original point.

(2)
(c)

The original logo has an area of 12 cm2. Find the area of the transformed logo.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \((4, -9)\)✓ 1
Part (b)
M1 uses \(T^{-1}\begin{pmatrix}5\\8\end{pmatrix}\) with \(T^{-1} = \frac{1}{7}\begin{pmatrix}3 & -1\\1 & 2\end{pmatrix}\), or solves \(2x + y = 5\), \(-x + 3y = 8\)✓ 1
A1 \((1, 3)\)✓ 1
Part (c)
M1 \(\det T = 2\times3 - 1\times(-1) = 7\)✓ 1
A1 area \(= 7 \times 12 = 84\) cm2✓ 1

Exam tip: The absolute value of the determinant is the area scale factor. If a determinant is negative, the orientation is reversed but the area is still multiplied by its absolute value.

5IBMI3H-1-02-01Complex numbers in Cartesian, polar and exponential form; matrices, inverses and solving systems; eigenvalues and eigenvectors in context Paper 1Easy4 marks

In an AC circuit, two components are connected in series. Their impedances are \(Z_1=3+4i\) ohms and \(Z_2=5-2i\) ohms. The total impedance is \(Z=Z_1+Z_2\).

(a)

Find \(Z\).

(1)
(b)

Find \(|Z|\), to 3 significant figures.

(1)
(c)

Find \(\arg Z\) in radians, to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(8+2i\) ohms✓ 1
Part (b)
A1 8.25 ohms✓ 1
Part (c)
M1 \(\arctan\left(\frac{2}{8}\right)\)✓ 1
A1 0.245 rad✓ 1

Exam tip: Check that your calculator is in radian mode when the question asks for an argument in radians.

6IBMI3H-1-02-02Complex numbers in Cartesian, polar and exponential form; matrices, inverses and solving systems; eigenvalues and eigenvectors in context Paper 1Easy5 marks

A voltage in an electronic circuit is represented by the complex number \(V_1=120e^{i\pi/3}\) volts. A second voltage is \(V_2=50-50i\) volts.

(a)

Write \(V_1\) in the form \(a+bi\), giving \(a\) and \(b\) to 3 significant figures.

(2)
(b)

Find \(|V_2|\), to 3 significant figures.

(1)
(c)

Find \(\arg V_2\) in radians, giving the principal argument to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(120\cos\frac{\pi}{3}\) and \(120\sin\frac{\pi}{3}\)✓ 1
A1 \(60.0+104i\)✓ 1
Part (b)
A1 70.7 volts✓ 1
Part (c)
M1 \(\arctan\left(\frac{-50}{50}\right)\) with \(V_2\) in the fourth quadrant✓ 1
A1 \(-0.785\) rad✓ 1

Common error: The principal argument lies in \((-\pi,\pi]\); for \(50-50i\) the angle is negative, not \(\frac{7\pi}{4}\).

7IBMI3H-1-02-03Complex numbers in Cartesian, polar and exponential form; matrices, inverses and solving systems; eigenvalues and eigenvectors in context Paper 1Easy4 marks

A bakery buys flour, butter and sugar at prices (in euros per kg) given by the matrix \(P=\begin{pmatrix}2.5&4.0&1.8\end{pmatrix}\). One batch of recipe A needs 3 kg flour, 1 kg butter and 4 kg sugar. One batch of recipe B needs 2 kg flour, 2 kg butter and 1 kg sugar. These are the columns of \(Q=\begin{pmatrix}3&2\\1&2\\4&1\end{pmatrix}\).

(a)

Find the matrix product \(PQ\).

(3)
(b)

Find how much more a batch of recipe A costs than a batch of recipe B.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt at matrix multiplication of a \(1\times3\) by a \(3\times2\) matrix✓ 1
A1 18.7✓ 1
A1 \(PQ=\begin{pmatrix}18.7&14.8\end{pmatrix}\)✓ 1
Part (b)
A1 3.90 euros✓ 1

Exam tip: A \(1\times3\) matrix times a \(3\times2\) matrix gives a \(1\times2\) matrix: check the dimensions before you multiply.

8IBMI3H-1-02-04Complex numbers in Cartesian, polar and exponential form; matrices, inverses and solving systems; eigenvalues and eigenvectors in context Paper 1Easy4 marks

A taxi company charges a fixed fee plus a constant rate per kilometre. A 6 km journey costs 14.40 euros and an 11 km journey costs 24.40 euros. Let the fixed fee be \(f\) euros and the rate be \(r\) euros per km.

(a)

Write down two equations connecting \(f\) and \(r\).

(2)
(b)

Find the rate per kilometre.

(1)
(c)

Find the fixed fee.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(f+6r=14.4\)✓ 1
A1 \(f+11r=24.4\)✓ 1
Part (b)
A1 \(r=2.00\) euros per km✓ 1
Part (c)
A1 \(f=2.40\) euros (ECF from (b))✓ 1

Exam tip: Subtracting the two equations eliminates the fixed fee and gives the rate directly.

9IBMI3H-1-02b-01Complex numbers in Cartesian, polar and exponential form; matrices, inverses and solving systems; eigenvalues and eigenvectors in context Paper 1Easy5 marks

Two components in an audio filter are connected in series. Their impedances, in ohms (Ω), are Z1 = 8 + 6i and Z2 = 3 − 4i. The total impedance is Z = Z1 + Z2.

(a)

Write Z in the form a + bi.

(1)
(b)

Find |Z|, giving your answer to 3 significant figures.

(2)
(c)

Find the argument of Z in radians, to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 Z = 11 + 2i✓ 1
Part (b)
M1 √(112 + 22) = √125✓ 1
A1 11.2 Ω✓ 1
Part (c)
M1 arg Z = tan−1(2 ÷ 11)✓ 1
A1 0.180 rad✓ 1

Common error: Use arctan(imaginary part ÷ real part), not the reciprocal, and check your calculator is in radian mode.

10IBMI3H-1-02b-02Complex numbers in Cartesian, polar and exponential form; matrices, inverses and solving systems; eigenvalues and eigenvectors in context Paper 1Easy5 marks

Two farms plant wheat, barley and maize. The areas planted, in hectares, are given by the matrix A, where each row is a farm (Farm 1, Farm 2) and the columns are wheat, barley, maize:
A = \(\begin{pmatrix} 12 & 8 & 5 \\ 9 & 14 & 6 \end{pmatrix}\)
The expected income per hectare, in euros, is wheat 1450, barley 1180 and maize 1620, written as the column matrix p = \(\begin{pmatrix} 1450 \\ 1180 \\ 1620 \end{pmatrix}\).

(a)

Calculate the matrix product Ap.

(3)
(b)

State which farm has the greater expected income and find the difference.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 multiplies a 2 × 3 matrix by a 3 × 1 matrix✓ 1
A1 Farm 1: 12 × 1450 + 8 × 1180 + 5 × 1620 = 34940✓ 1
A1 Farm 2: 39290, so Ap = \(\begin{pmatrix} 34940 \\ 39290 \end{pmatrix}\)✓ 1
Part (b)
A1 Farm 2 has the greater expected income✓ 1
A1 €4350✓ 1

Exam tip: Check that the number of columns of the first matrix equals the number of rows of the second before multiplying; the order matters.

11IBMI3H-1-02b-03Complex numbers in Cartesian, polar and exponential form; matrices, inverses and solving systems; eigenvalues and eigenvectors in context Paper 1Easy5 marks

A musical tone is represented by the phasor z = 7(cos(2π/5) + i sin(2π/5)), where the modulus is the amplitude.

(a)

Write z in the form a + bi, with a and b correct to 3 significant figures.

(2)
(b)

Use de Moivre's theorem to write z3 in the form reiθ.

(2)
(c)

Write down the argument of z3 in the interval −π < θ ≤ π.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 7cos(2π/5) and 7sin(2π/5)✓ 1
A1 z = 2.16 + 6.66i✓ 1
Part (b)
M1 modulus 73 and argument 3 × 2π/5✓ 1
A1 z3 = 343ei6π/5✓ 1
Part (c)
A1 6π/5 − 2π = −4π/5 = −2.51 rad✓ 1

Common error: The principal argument must lie in (−π, π]; an argument of 6π/5 is correct as a direction but is not the principal value.

12IBMI3H-1-02b-04Complex numbers in Cartesian, polar and exponential form; matrices, inverses and solving systems; eigenvalues and eigenvectors in context Paper 1Easy5 marks

A garden centre sells seed mixes X and Y at constant prices of €x and €y per packet. Three packets of X and two packets of Y cost €19. Five packets of X and four packets of Y cost €34. This is written as M\(\begin{pmatrix} x \\ y \end{pmatrix}\) = \(\begin{pmatrix} 19 \\ 34 \end{pmatrix}\), where M = \(\begin{pmatrix} 3 & 2 \\ 5 & 4 \end{pmatrix}\).

(a)

Find the determinant of M.

(1)
(b)

Find M−1.

(2)
(c)

Hence find the price of one packet of Y.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 3 × 4 − 2 × 5 = 2✓ 1
Part (b)
M1 swaps the diagonal entries and changes the signs of the other two✓ 1
A1 M−1 = ½ \(\begin{pmatrix} 4 & -2 \\ -5 & 3 \end{pmatrix}\)✓ 1
Part (c)
M1 \(\begin{pmatrix} x \\ y \end{pmatrix}\) = M−1\(\begin{pmatrix} 19 \\ 34 \end{pmatrix}\)✓ 1
A1 y = ½(−5 × 19 + 3 × 34) = €3.50 (and x = €4)✓ 1

Exam tip: Multiply by the inverse on the left: M⁻¹ must come before the column matrix of totals.

13IBMI3H-1-03r7-01Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy6 marks

A light aircraft flies in a region of constant wind. Using a complex-plane model, a velocity is written as \(x+yi\), where \(x\) is the eastward component and \(y\) the northward component, both in km h−1.
The velocity of the aircraft relative to the air is \(v_a = 210+120i\) and the velocity of the wind is \(w = -30+45i\).

(a)

Find the velocity of the aircraft relative to the ground, \(v_g = v_a + w\), in the form \(x+yi\).

(2)
(b)

Calculate the ground speed of the aircraft, correct to 3 significant figures.

(2)
(c)

Find the angle of the ground velocity above due east, in degrees, correct to 1 decimal place.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to add real parts and imaginary parts✓ 1
A1 \(v_g = 180+165i\)✓ 1
Part (b)
M1 attempt to find the modulus, \(\sqrt{180^2+165^2}\)✓ 1
A1 244 km h−1 (244.18…)✓ 1
Part (c)
M1 \(\arg v_g = \tan^{-1}\left(\frac{165}{180}\right)\) (in degree mode)✓ 1
A1 \(42.5^\circ\) (42.51…)✓ 1

Exam tip: Check that your calculator is in degree mode before finding an angle in degrees, and in radian mode for an argument in radians.

14IBMI3H-1-03r7-02Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy8 marks

In a digital synthesizer, the complex amplitude of a tone is \(A = 4e^{\frac{2\pi}{3}i}\). A filter multiplies the amplitude by \(0.5e^{-\frac{\pi}{4}i}\), producing the output amplitude \(B\).

(a)

Write \(B\) in the form \(re^{i\theta}\), giving \(r\) and the exact value of \(\theta\).

(3)
(b)

Write \(B\) in the form \(a+bi\), with \(a\) and \(b\) correct to 3 significant figures.

(2)
(c)

A second tone has complex amplitude \(C = 1.2-0.5i\). The two tones are added to give \(B+C\). Using unrounded values of \(B\), find the modulus of \(B+C\), correct to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to multiply moduli and add arguments✓ 1
A1 \(r=2\)✓ 1
A1 \(\theta=\frac{5\pi}{12}\)✓ 1
Part (b)
M1 \(a=2\cos\frac{5\pi}{12}\), \(b=2\sin\frac{5\pi}{12}\)✓ 1
A1 \(B=0.518+1.93i\)✓ 1
Part (c)
M1 attempt to add real and imaginary parts, \(B+C\approx1.7176+1.4319i\)✓ 1
M1 attempt to find the modulus✓ 1
A1 2.24✓ 1

Common error: Arguments are added when complex numbers are multiplied, not multiplied; moduli are the quantities that multiply.

15IBMI3H-1-03r7-06Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy6 marks

An audio effect sends a stereo signal through repeated echoes. The left and right amplitudes \((L_n, R_n)\) after \(n\) echoes satisfy
\(\begin{pmatrix}L_{n+1}\\R_{n+1}\end{pmatrix}=\mathbf{M}\begin{pmatrix}L_n\\R_n\end{pmatrix}\), where \(\mathbf{M}=\begin{pmatrix}0.6&0.3\\0.2&0.5\end{pmatrix}\).

(a)

Show that \(\begin{pmatrix}3\\2\end{pmatrix}\) is an eigenvector of \(\mathbf{M}\), and state the corresponding eigenvalue.

(2)
(b)

The initial amplitudes are \(L_0=1.5\) and \(R_0=1.0\). Find \(L_5\), correct to 3 significant figures.

(2)
(c)

Use the trace of \(\mathbf{M}\) to find the other eigenvalue of \(\mathbf{M}\).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(\mathbf{M}\begin{pmatrix}3\\2\end{pmatrix}=\begin{pmatrix}2.4\\1.6\end{pmatrix}\)✓ 1
A1 \(=0.8\begin{pmatrix}3\\2\end{pmatrix}\) so it is an eigenvector with eigenvalue 0.8✓ 1
Part (b)
M1 recognise that the vector is multiplied by \(0.8\) at each echo, so \(L_5=1.5\times0.8^5\)✓ 1
A1 \(L_5=0.492\)✓ 1
Part (c)
M1 sum of eigenvalues equals the trace, \(0.6+0.5=1.1\)✓ 1
A1 \(0.3\)✓ 1

Exam tip: The sum of the eigenvalues of a square matrix equals its trace, which gives a quick check on eigenvalues you have found.

16IBMI3H-1-03r7-13Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy6 marks

In equal temperament, the frequency of the note \(n\) semitones above the note A4 (440 Hz) is \(f=440\times2^{\frac{n}{12}}\) Hz.

(a)

Find the frequency of the note 7 semitones above A4, correct to the nearest hertz.

(2)
(b)

A string vibrates at 1000 Hz. Find the value of \(n\) for this frequency, correct to 1 decimal place, and state whether it matches a note on the equal-tempered scale.

(4)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(440\times2^{\frac{7}{12}}\)✓ 1
A1 659 Hz✓ 1
Part (b)
M1 \(1000=440\times2^{\frac{n}{12}}\) and attempt to isolate the power✓ 1
M1 take logarithms, \(n=12\log_2\frac{1000}{440}\)✓ 1
A1 \(n=14.2\) (14.21…)✓ 1
R1 no, \(n\) is not a whole number, so it lies between two notes✓ 1

Common error: \(440\times2^{n/12}\) is not \(880^{n/12}\); apply the exponent to the 2 only, then take logarithms of the isolated power.

17IBMI3H-1-04r10-01Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy7 marks

An archaeologist records positions in a trench as complex numbers, with the real axis pointing east, the imaginary axis pointing north and the datum peg at 0, distances in metres. A corner stone is found at z = 6 + 8i.

(a)

Find the modulus and argument of z. Give the argument in radians to 3 significant figures.

(2)
(b)

The trench plan is rotated 60° anticlockwise about the datum peg. Find the new position of the corner stone in the form a + bi, with a and b to 3 significant figures.

(3)
(c)

Find the distance between the original and the new positions of the corner stone.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
|z| = √(62 + 82) = 10✓ 1
arg z = arctan(8/6) = 0.927 radians✓ 1
Part (b)
multiplies by eiπ/3 = cos 60° + i sin 60°✓ 1
(6 + 8i)(½ + (√3/2)i) = (3 − 4√3) + (3√3 + 4)i✓ 1
−3.93 + 9.20i✓ 1
Part (c)
distance = |w − z| with w = −3.928... + 9.196...i, or equilateral triangle argument (sides 10, 10, angle 60°)✓ 1
distance = 10 m✓ 1

Exam tip: Multiplying by eiθ rotates a point anticlockwise by θ about the origin and leaves its distance from the origin unchanged.

18IBMI3H-1-04r10-04Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy8 marks

A cycling shop sells helmets at €h each and locks at €l each. A customer pays €89 for 3 helmets and 2 locks. Another pays €124 for 4 helmets and 3 locks.

(a)

Write this as a matrix equation Mp = c, where p is the column vector (h, l).

(2)
(b)

Find M−1.

(2)
(c)

Hence find h and l.

(2)
(d)

Find the cost of 5 helmets and 3 locks.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M = rows (3, 2), (4, 3)✓ 1
c = column (89, 124)✓ 1
Part (b)
det M = 9 − 8 = 1✓ 1
M−1 = rows (3, −2), (−4, 3)✓ 1
Part (c)
p = M−1c✓ 1
h = 19 and l = 16✓ 1
Part (d)
5(19) + 3(16)✓ 1
€143✓ 1

Common error: Multiply by the inverse on the left of both sides: p = M⁻¹c. Writing it on the right gives a product that is not defined.

19IBMI3H-1-04r10-11Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy5 marks

A cycling shop sells an electric bike for €3000 on a loan at a nominal annual interest rate of 6%, compounded monthly. The loan is repaid in 24 equal payments made at the end of each month.

(a)

Find the monthly payment, to the nearest cent.

(3)
(b)

Using the payment rounded to the nearest cent, find the total interest paid over the loan.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
TVM values: N = 24, I% = 6, PV = 3000, FV = 0, P/Y = C/Y = 12✓ 1
payments at end of period✓ 1
€132.96✓ 1
Part (b)
24 × 132.96 = 3191.04✓ 1
interest = 3191.04 − 3000 = €191.04✓ 1

Exam tip: In TVM mode enter the loan as a positive PV and the payments as negative, or the other way round, but never both with the same sign.

20IBMI3H-1-04r10-12Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy7 marks

The amount of data a data centre stores each day forms a geometric sequence. On day 1 it stores 40 TB and each day it stores 5% more than the day before.

(a)

Find the amount stored on day 10, to 3 significant figures.

(2)
(b)

Find the first day on which more than 100 TB is stored.

(3)
(c)

Find the total amount stored over the first 25 days, to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
u10 = 40 × 1.059✓ 1
62.1 TB✓ 1
Part (b)
40 × 1.05n−1 > 100 so (n − 1) ln 1.05 > ln 2.5✓ 1
n > 19.8✓ 1
day 20✓ 1
Part (c)
S25 = 40(1.0525 − 1)/(1.05 − 1)✓ 1
1910 TB✓ 1

Common error: For a first-term-a geometric sequence the n-th term uses rⁿ⁻¹, not rⁿ.

21IBMI3H-1-03r11-01Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy4 marks

In a treasure-hunting game, the map is modelled as the complex plane with the player at the origin, the real axis pointing east and the imaginary axis pointing north (units: metres). A treasure chest is at \(z = 4 - 7i\).

(a)

Find \(|z|\), giving your answer to 3 significant figures.

(2)
(b)

Find \(\arg z\) in radians, to 3 significant figures, with \(-\pi < \arg z \le \pi\).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to use \(\sqrt{4^2+(-7)^2}\)✓ 1
A1 \(|z| = \sqrt{65} = 8.06\) m✓ 1
Part (b)
M1 attempt to use \(\tan^{-1}\frac{7}{4}\) and recognise that \(z\) is in the fourth quadrant (negative argument)✓ 1
A1 \(\arg z = -1.05\)✓ 1

Common error: Giving +1.05 for the argument. Sketch the point first: with a negative imaginary part the principal argument must be negative.

22IBMI3H-1-03r11-02Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy6 marks

A sprite in a game is at the point \(z_0 = 2 + 3i\) in the complex plane (units: cm). A special effect moves every sprite to a new position by multiplying its coordinate by \(m = 1.5\,e^{0.7i}\).

(a)

Write \(m\) in the form \(a + bi\), giving \(a\) and \(b\) to 4 significant figures.

(2)
(b)

Find the new position \(w = m z_0\) in the form \(x + yi\), giving \(x\) and \(y\) to 3 significant figures. Use unrounded values of \(a\) and \(b\).

(3)
(c)

Find the distance of the new position from the origin, in cm, to 3 significant figures.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(m = 1.5(\cos 0.7 + i\sin 0.7)\) (radians)✓ 1
A1 \(a = 1.147\) and \(b = 0.9663\) i.e. \(m = 1.147 + 0.9663i\)✓ 1
Part (b)
M1 attempt to multiply: \(w = (a+bi)(2+3i)\) or \(1.5e^{0.7i}(2+3i)\)✓ 1
A1 real part \(2a-3b = -0.604\)✓ 1
A1 imaginary part \(3a+2b = 5.37\)✓ 1
Part (c)
A1 \(|w| = 1.5\sqrt{13} = 5.41\) cm✓ 1

Exam tip: Multiplying by \(re^{i\theta}\) scales distances from the origin by \(r\) and rotates by \(\theta\) radians, so \(|w| = r|z_0|\) gives a quick check.

23IBMI3H-1-03r11-07Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy5 marks

In a puzzle game, a pair of scores \((p, q)\) is encoded as the vector \(\mathbf{c} = \mathbf{A}\begin{pmatrix}p\\q\end{pmatrix}\) where \(\mathbf{A} = \begin{pmatrix}2&3\\1&k\end{pmatrix}\) and \(k\) is a constant.

(a)

Find the value of \(k\) for which the code cannot be decoded.

(2)
(b)

For \(k = 2\), the encoded vector is \(\begin{pmatrix}23\\14\end{pmatrix}\). Find the original scores \(p\) and \(q\).

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(\det\mathbf{A} = 2k - 3 = 0\)✓ 1
A1 \(k = 1.5\)✓ 1
Part (b)
A1 \(\mathbf{A}^{-1} = \begin{pmatrix}2&-3\\-1&2\end{pmatrix}\) (since \(\det\mathbf{A} = 1\))✓ 1
M1 \(\begin{pmatrix}p\\q\end{pmatrix} = \mathbf{A}^{-1}\begin{pmatrix}23\\14\end{pmatrix}\)✓ 1
A1 \(p = 4\) and \(q = 5\)✓ 1

Exam tip: A matrix has no inverse exactly when its determinant is zero, so set the determinant equal to 0 and solve for the unknown.

24IBMI3H-1-03r13-01Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy5 marks

A space probe's antenna signal is represented by the complex number z = 5 − 12i (volts, as a phasor).

(a)

Find |z|.

(2)
(b)

Find arg z in radians, giving your answer to 3 significant figures, with −π < arg z ≤ π.

(2)
(c)

Write z in the form reiθ, using the value of θ from part (b).

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to use √(52 + 122)✓ 1
A1 |z| = 13✓ 1
Part (b)
M1 attempt to use arctan(12/5) with the correct quadrant (fourth quadrant)✓ 1
A1 arg z = −1.18 rad (−1.1760…)✓ 1
Part (c)
A1 z = 13e−1.18i✓ 1

Common error: Using arctan(12/5) without checking the quadrant gives the right size but the wrong sign for the argument when the imaginary part is negative.

25IBMI3H-1-03r13-02Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy5 marks

The displacement of a bridge sensor's vibration is modelled by the complex number z = 7e2πi/5 (in mm).

(a)

Write z in the form a + bi, giving a and b to 3 significant figures.

(3)
(b)

Write z2 in the form reiθ.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to use a = 7cos(2π/5) and b = 7sin(2π/5)✓ 1
A1 a = 2.16 (2.1631…)✓ 1
A1 b = 6.66 (6.6574…)✓ 1
Part (b)
A1 modulus of z2 is 49 (= 72)✓ 1
A1 argument is 4π/5, so z2 = 49e4πi/5✓ 1

Exam tip: Convert between forms using rcosθ and rsinθ; make sure the calculator is in radian mode when the angle is given in radians.

26IBMI3H-1-03r13-10Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy5 marks

A town sets up a bridge-repair fund with €25 000 invested at a nominal annual interest rate of 3.2%, compounded quarterly.

(a)

Find the value of the fund after 8 years, to the nearest euro.

(2)
(b)

Find how many years it takes for the fund to reach €35 000, giving your answer to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to use 25000(1 + 0.032/4)32✓ 1
A1 €32 261 (€32 260.94)✓ 1
Part (b)
M1 set 25000(1.008)4t = 35000 (or use the finance solver)✓ 1
M1 take logarithms or use technology to solve✓ 1
A1 t = 10.6 years (about 42.2 quarters)✓ 1

Common error: With quarterly compounding the rate per period is 3.2% ÷ 4, and the number of periods is 4 times the number of years.

27IBMI3H-1-03r13-13Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy5 marks

A space probe is 2.37 × 108 km from Earth. A radio signal travels at 3.00 × 108 m s−1.

(a)

Calculate the time for a signal to reach Earth, in seconds, giving your answer in the form a × 10k where 1 ≤ a < 10 and k is an integer.

(2)
(b)

Write your answer to (a) in minutes, to 3 significant figures.

(1)
(c)

A technician estimates the time as 800 s. Find the percentage error in this estimate, to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 convert the distance to metres: 2.37 × 1011 m and divide by the speed✓ 1
A1 7.90 × 102 s✓ 1
Part (b)
A1 13.2 minutes (13.17)✓ 1
Part (c)
M1 attempt to use |estimate − exact|/exact × 100✓ 1
A1 1.27%✓ 1

Common error: Convert km to m before dividing by a speed in m s⁻¹; a factor of 1000 is the usual slip.

28IBMI3H-1-04r14-01Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy6 marks

In an aviation weather model, a wind velocity is represented by the complex number \(w=-9+12i\) knots, where the real axis points east and the imaginary axis points north.

(a)

Find the wind speed, \(|w|\).

(2)
(b)

Find \(\arg w\) in radians, with \(-\pi<\arg w\le\pi\), to 3 significant figures.

(2)
(c)

Hence find the bearing (measured clockwise from north) towards which the wind is blowing, to the nearest 0.1°.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(\sqrt{(-9)^2+12^2}\)✓ 1
A1 15 knots✓ 1
Part (b)
M1 recognises \(w\) is in the second quadrant: \(\pi-\arctan\frac{12}{9}\)✓ 1
A1 2.21 rad✓ 1
Part (c)
M1 \(\arg w=126.87^\circ\) and bearing \(=90^\circ-126.87^\circ\) (+360°)✓ 1
A1 323.1°✓ 1

Common error: \(\arctan\frac{12}{-9}\) on a calculator gives an angle in the fourth quadrant; sketch the point to choose the correct quadrant.

29IBMI3H-1-04r14-02Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy6 marks

Two pure tones of the same frequency reach a microphone. Their pressure amplitudes and phases (in millipascals and radians) are represented by the complex numbers \(z_1=2e^{i\pi/3}\) and \(z_2=3e^{i\pi/6}\). The combined tone is represented by \(z_1+z_2\).

(a)

Write \(z_1\) and \(z_2\) in the form \(a+bi\), with \(a\) and \(b\) to 3 significant figures.

(2)
(b)

Find the amplitude of the combined tone, to 3 significant figures.

(3)
(c)

Find the phase of the combined tone, to 3 significant figures.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(z_1=1.00+1.73i\)✓ 1
A1 \(z_2=2.60+1.50i\)✓ 1
Part (b)
M1 adds the real and imaginary parts✓ 1
A1 \(z_1+z_2=3.598\ldots+3.232\ldots i\)✓ 1
A1 amplitude \(=|z_1+z_2|=4.84\) mPa✓ 1
Part (c)
A1 \(\arg(z_1+z_2)=0.732\) rad✓ 1

Exam tip: Convert to Cartesian form before adding; polar forms cannot be added directly unless the arguments are equal.

30IBMI3H-1-04r14-03Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy5 marks

A farmer applies fertiliser to three fields. The matrix \(M\) gives the nutrient application (kg per hectare) for nitrogen, phosphorus and potassium in fields A, B and C:
\(M=\begin{pmatrix}120&40&60\\90&50&80\\150&30&40\end{pmatrix}\) (rows: fields A, B, C; columns: N, P, K).
The areas of the fields in hectares are given by \(H=\begin{pmatrix}12&8&5\end{pmatrix}\). The prices of N, P and K are €0.85, €1.90 and €0.70 per kg respectively, written as the column matrix \(p=\begin{pmatrix}0.85\\1.90\\0.70\end{pmatrix}\).

(a)

Write down the order of the matrix \(HM\).

(1)
(b)

Calculate \(HM\) and state what each element represents.

(2)
(c)

Calculate the total cost of the fertiliser, in euros.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(1\times3\)✓ 1
Part (b)
A1 \(HM=\begin{pmatrix}2910&1030&1560\end{pmatrix}\)✓ 1
A1 total mass in kg of N, P and K respectively applied over all three fields✓ 1
Part (c)
M1 \((HM)p\), i.e. \(2910\times0.85+1030\times1.90+1560\times0.70\)✓ 1
A1 €5522.50✓ 1

Common error: Matrix multiplication is not commutative: check that the number of columns of the first matrix equals the number of rows of the second.

31IBMI3H-1-04r14-13Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy6 marks

A plane flies 1260 km from airport P to airport Q along a straight route and back again, in a steady wind that blows along the route. The outward flight takes 2.1 hours and the return flight takes 2.4 hours. Let \(v\) km h−1 be the airspeed of the plane and \(w\) km h−1 the wind speed.

(a)

Find the ground speed of the plane on each leg.

(2)
(b)

Write down two equations in \(v\) and \(w\) and solve them.

(2)
(c)

In still air, find the time to fly 900 km at this airspeed. Give your answer in hours and minutes.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 speed \(=\frac{\text{distance}}{\text{time}}\)✓ 1
A1 600 km h−1 outward and 525 km h−1 return✓ 1
Part (b)
M1 \(v+w=600\) and \(v-w=525\)✓ 1
A1 \(v=562.5\) and \(w=37.5\)✓ 1
Part (c)
M1 \(\frac{900}{562.5}\) (ECF from (b))✓ 1
A1 1.6 hours = 1 hour 36 minutes✓ 1

Common error: The wind adds to the airspeed on one leg and subtracts on the other; check which leg is faster before writing the equations.

32IBMI3H-1-04r15-01Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy6 marks

The displacement of a warehouse robot from its charging point is modelled by complex numbers, in metres. Move one is \(z = 2 + 5i\) and move two is \(w = 3 - i\).

(a)

Find \(zw\) in the form \(a + bi\).

(2)
(b)

Find \(\dfrac{z}{w}\) in the form \(a + bi\).

(2)
(c)

Find \(|zw|\), giving your answer to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 expand \((2+5i)(3-i)\) and use \(i^2=-1\)✓ 1
A1 \(11 + 13i\)✓ 1
Part (b)
M1 multiply numerator and denominator by the conjugate \(3 + i\)✓ 1
A1 \(\dfrac{1+17i}{10} = 0.1 + 1.7i\)✓ 1
Part (c)
M1 \(\sqrt{11^2+13^2}\) (ECF from (a)) or \(|z||w| = \sqrt{29}\sqrt{10}\)✓ 1
A1 17.0✓ 1

Exam tip: The modulus of a product equals the product of the moduli, which gives a quick check on your Cartesian answer.

33IBMI3H-1-04r15-02Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy4 marks

A delivery robot's target is located at the point represented by the complex number \(z = -4 + 7i\) on a floor plan, with distances in metres.

(a)

Find \(|z|\) to 3 significant figures.

(1)
(b)

Find \(\arg z\) in radians to 3 significant figures, with \(-\pi < \arg z \leq \pi\).

(2)
(c)

Write z in the form \(re^{i\theta}\), using your answers to (a) and (b).

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 8.06✓ 1
Part (b)
M1 recognising that z lies in the second quadrant, e.g. \(\pi - \tan^{-1}\!\left(\tfrac{7}{4}\right)\)✓ 1
A1 2.09✓ 1
Part (c)
A1 \(8.06e^{2.09i}\) (ECF from (a) and (b))✓ 1

Common error: A calculator gives \(\tan^{-1}(7/(-4))\) in the fourth quadrant, so check the quadrant of the point before stating the argument.

34IBMI3H-1-04r15-06Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy5 marks

A small fashion label makes jackets, dresses and skirts. The fabric needed per garment, in metres, is shown in matrix F, with columns for wool and silk. In one week the label makes 12 jackets, 20 dresses and 30 skirts, written as the row matrix \(\mathbf{P} = (12\ \ 20\ \ 30)\).
\(\mathbf{F} = \begin{pmatrix}3.5&0.5\\2.0&1.5\\1.2&0.8\end{pmatrix}\)

(a)

Calculate \(\mathbf{P}\mathbf{F}\) and state what its two entries represent.

(3)
(b)

Wool costs £8.40 per metre and silk costs £22.50 per metre. Use matrix multiplication to find the total cost of the fabric for the week.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt at matrix multiplication✓ 1
A1 \((118\ \ 60)\)✓ 1
R1 total metres of wool (118) and silk (60) used in the week✓ 1
Part (b)
M1 \((118\ \ 60)\begin{pmatrix}8.40\\22.50\end{pmatrix}\) (ECF from (a))✓ 1
A1 £2341.20✓ 1

Exam tip: Check the dimensions before multiplying: a \(1\times3\) matrix times a \(3\times2\) matrix gives a \(1\times2\) matrix.

35IBMI3H-1-04r15-13Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy6 marks

The biomass of a fish stock in a protected bay is modelled by \(S(t) = 480 \times 0.92^t\) tonnes, where t is the number of years since monitoring began.

(a)

Find the biomass after 3 years, to 3 significant figures.

(2)
(b)

Interpret the value 0.92 in this model.

(1)
(c)

Find how long it takes for the biomass to fall below 300 tonnes. Give your answer in years to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(480 \times 0.92^3\)✓ 1
A1 374 tonnes✓ 1
Part (b)
R1 the biomass falls by 8% each year✓ 1
Part (c)
M1 \(480 \times 0.92^t = 300\)✓ 1
M1 \(t = \dfrac{\ln(300/480)}{\ln 0.92}\) (or use of GDC)✓ 1
A1 5.64 years✓ 1

Exam tip: When the unknown is an exponent, take logarithms of both sides or use the graphing or solver function on your calculator.

36IBMI3H-1-04r17-01Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy5 marks

Archaeologists date a wooden beam from a settlement using the amount of carbon-14 it contains. The fraction of carbon-14 remaining after \(t\) years is \(F=\left(\tfrac12\right)^{t/5730}\).

(a)

Find the fraction remaining after 2000 years. Give your answer as a percentage to 3 significant figures.

(2)
(b)

A sample from the beam contains 62% of its original carbon-14. Find the age of the beam, giving your answer to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(\left(\tfrac12\right)^{2000/5730}\)✓ 1
A1 \(78.5\%\)✓ 1
Part (b)
M1 sets \(\left(\tfrac12\right)^{t/5730}=0.62\)✓ 1
M1 takes logarithms: \(\frac{t}{5730}=\frac{\ln0.62}{\ln0.5}\)✓ 1
A1 \(t=3950\) years✓ 1

Exam tip: Half-life questions are solved by taking logarithms of both sides, not by trial and error; show the log step for the method mark.

37IBMI3H-1-04r17-02Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy7 marks

A cyclist's training plan has her ride 40 km in week 1, and each week she rides 6 km more than the week before.

(a)

Find the distance she rides in week 12.

(2)
(b)

Find the first week in which she rides more than 150 km.

(3)
(c)

Find the total distance she rides in the first 12 weeks.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(u_{12}=40+11\times6\)✓ 1
A1 \(106\) km✓ 1
Part (b)
M1 \(40+6(n-1)>150\)✓ 1
A1 \(n>19.3\ldots\)✓ 1
A1 week 20✓ 1
Part (c)
M1 \(S_{12}=\frac{12}{2}\left(2\times40+11\times6\right)\) (or \(\frac{12}{2}(40+106)\))✓ 1
A1 \(876\) km✓ 1

Common error: The \(n\)th term of an arithmetic sequence uses \((n-1)d\), not \(nd\); check by testing \(n=1\).

38IBMI3H-1-04r17-06Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy7 marks

Two components in an electronic cooling circuit have complex impedances \(Z_1=(12+5i)\) ohms and \(Z_2=(8-3i)\) ohms.

(a)

Find the total impedance \(Z_s=Z_1+Z_2\) when the components are connected in series.

(1)
(b)

When the components are connected in parallel the total impedance is \(Z_p=\dfrac{Z_1Z_2}{Z_1+Z_2}\). Write \(Z_p\) in the form \(a+bi\), giving \(a\) and \(b\) to 3 significant figures.

(4)
(c)

Find the modulus of \(Z_p\), giving your answer to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(Z_s=20+2i\)✓ 1
Part (b)
M1 \(Z_1Z_2=(12+5i)(8-3i)\)✓ 1
A1 \(=111+4i\)✓ 1
M1 multiplies numerator and denominator by the conjugate \(20-2i\) (or divides on a GDC)✓ 1
A1 \(Z_p=\frac{2228-142i}{404}=5.51-0.351i\)✓ 1
Part (c)
M1 \(\sqrt{5.5148\ldots^2+0.3514\ldots^2}\) (unrounded values)✓ 1
A1 \(5.53\) ohms✓ 1

Exam tip: On a GDC, store \(Z_1\) and \(Z_2\) as complex numbers so that the parallel formula is entered once; still show the product \(111+4i\).

39IBMI3H-1-04r17-08Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy7 marks

Six standing stones at an ancient site are located at the points \(z_k=5e^{ik\pi/3}\), \(k=0,1,2,3,4,5\), in the complex plane. Distances are in metres and the origin is the centre of the site.

(a)

Write \(z_2\) in the form \(a+bi\), with \(a\) and \(b\) to 3 significant figures.

(2)
(b)

Find \(z_5\times e^{i\pi/3}\) and state which stone it corresponds to.

(2)
(c)

Find the distance between the stones at \(z_0\) and \(z_2\), to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(5\left(\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}\right)\)✓ 1
A1 \(-2.50+4.33i\)✓ 1
Part (b)
M1 adds arguments: \(5e^{i(5\pi/3+\pi/3)}=5e^{2i\pi}\)✓ 1
A1 \(=5=z_0\) (the stone \(z_0\))✓ 1
Part (c)
M1 \(|z_2-z_0|=|-7.5+4.33i|\)✓ 1
M1 \(\sqrt{7.5^2+4.330^2}\)✓ 1
A1 \(8.66\) m✓ 1

Exam tip: Multiplying by \(e^{i\theta}\) rotates a point anticlockwise by \(\theta\) about the origin without changing its modulus.

40IBMI3H-1-04r18-01Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy5 marks

A beekeeper models a forager bee's flight in the complex plane, with the hive at the origin, the real axis pointing east and the imaginary axis pointing north. Distances are in metres. The bee's first leg is the displacement \(z=90+120i\). Its second leg is the displacement \(w=-60+80i\).

(a)

Find the modulus and the argument of \(z\). Give the argument in radians to 3 significant figures.

(2)
(b)

Find the modulus and argument of the total displacement \(z+w\) from the hive, giving both to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(|z|=\sqrt{90^2+120^2}=150\) m✓ 1
A1 \(\arg z=\arctan\frac{120}{90}=0.927\) rad✓ 1
Part (b)
A1 \(z+w=30+200i\)✓ 1
A1 \(|z+w|=202\) m✓ 1
A1 \(\arg(z+w)=\arctan\frac{200}{30}=1.42\) rad✓ 1

Common error: Using \(\arctan(\text{real}/\text{imaginary})\). The argument is measured from the positive real axis, so use \(\arctan(\text{imaginary}/\text{real})\) and check the quadrant.

41IBMI3H-1-04r18-05Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy5 marks

A hospital orders three items for each of three wards. Matrix A gives the number of boxes (gloves, masks, gel bottles) ordered by Wards 1, 2 and 3 (rows):
\(A=\begin{pmatrix}12&8&5\\20&15&10\\6&4&2\end{pmatrix}\)
Matrix B gives the price, in £, of one unit of each item (rows) from Supplier 1 and Supplier 2 (columns):
\(B=\begin{pmatrix}4.50&4.20\\6.00&6.40\\3.25&3.00\end{pmatrix}\)

(a)

State the order (dimensions) of the matrix \(AB\).

(1)
(b)

Calculate \(AB\).

(2)
(c)

Interpret the entry in row 2, column 1 of \(AB\).

(1)
(d)

Find how much the hospital saves in total by buying all three wards' orders from Supplier 2 rather than Supplier 1.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(3\times2\)✓ 1
Part (b)
M1 matrix multiplication of correct rows and columns, e.g. \(12(4.5)+8(6)+5(3.25)\)✓ 1
A1 \(AB=\begin{pmatrix}118.25&116.60\\212.50&210.00\\57.50&56.80\end{pmatrix}\)✓ 1
Part (c)
R1 the cost, in £, of Ward 2's whole order if bought from Supplier 1 (£212.50)✓ 1
Part (d)
A1 \((118.25+212.50+57.50)-(116.60+210.00+56.80)=£4.85\)✓ 1

Exam tip: Check the dimensions before multiplying: an \(m\times n\) matrix times an \(n\times p\) matrix gives an \(m\times p\) matrix, and the column index of the first must equal the row index of the second.

42IBMI3H-1-04r18-09Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy7 marks

In a hospital, patients are moved each day between three wards. The numbers of patients in the wards at the start of day n are given by \(\mathbf{p}_{n+1}=M\mathbf{p}_n\), where \(M=\begin{pmatrix}0.5&0.25&0.25\\0.25&0.5&0.25\\0.25&0.25&0.5\end{pmatrix}\) and the total number of patients is constant.

(a)

Show that \(\begin{pmatrix}1\\1\\1\end{pmatrix}\) is an eigenvector of \(M\) and state the eigenvalue.

(2)
(b)

Show that \(\begin{pmatrix}1\\-1\\0\end{pmatrix}\) is an eigenvector of \(M\) and state its eigenvalue.

(2)
(c)

Use the trace of \(M\) to find the third eigenvalue.

(1)
(d)

At day 0 all 90 patients are in Ward 1. Find the long-term number of patients in each ward and justify your answer.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(M\begin{pmatrix}1\\1\\1\end{pmatrix}=\begin{pmatrix}1\\1\\1\end{pmatrix}\)✓ 1
A1 eigenvalue 1✓ 1
Part (b)
M1 \(M\begin{pmatrix}1\\-1\\0\end{pmatrix}=\begin{pmatrix}0.25\\-0.25\\0\end{pmatrix}\)✓ 1
A1 eigenvalue \(0.25\)✓ 1
Part (c)
A1 \(1.5=1+0.25+\lambda\), so \(\lambda=0.25\)✓ 1
Part (d)
A1 30 patients in each ward✓ 1
R1 the other eigenvalues have modulus less than 1, so their contributions vanish; the component along \((1,1,1)^T\) stays constant at \((30,30,30)^T\)✓ 1

Exam tip: The trace of a matrix equals the sum of its eigenvalues, which gives the last eigenvalue quickly when the other two are known.

43IBMI3H-1-04r18-14Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy6 marks

During a Formula 1 stint, the lap time t seconds is modelled by \(t=ax^2+bx+c\), where x is the age of the tyres in laps. The model passes through the points (1, 90.2), (5, 88.6) and (9, 89.8).

(a)

Write down three equations for \(a\), \(b\) and \(c\).

(2)
(b)

Solve the equations to find \(a\), \(b\) and \(c\).

(2)
(c)

Find the minimum lap time predicted by the model, to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 substitutes the points into the model✓ 1
A1 \(a+b+c=90.2\), \(25a+5b+c=88.6\), \(81a+9b+c=89.8\)✓ 1
Part (b)
M1 solves the system✓ 1
A1 \(a=0.0875\), \(b=-0.925\), \(c=91.0375\)✓ 1
Part (c)
M1 minimum at \(x=-\frac{b}{2a}=5.29\) laps (or uses calculator minimum)✓ 1
A1 \(t=88.6\) s✓ 1

Common error: Rounding \(a\), \(b\) and \(c\) early. Keep the exact values (or stored calculator values) when finding the minimum, as small changes in \(a\) change the answer.

44IBMI3H-1-05r19-01Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy6 marks

A warehouse robot is located by a complex number \(z\), where the real part is the distance east and the imaginary part is the distance north of its docking station, in metres. The robot is at \(z = 4 + 3i\).

(a)

Find the modulus of \(z\) and the argument of \(z\), giving the argument in radians correct to 3 significant figures.

(3)
(b)

The controller turns the robot through 90° anticlockwise about the docking station. This is modelled by multiplying \(z\) by \(i\). Find the new position as a complex number.

(2)
(c)

Find the argument of the new position, in radians, correct to 3 significant figures.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 use of \(|z|=\sqrt{4^2+3^2}\) or \(\arg z=\tan^{-1}\frac{3}{4}\)✓ 1
A1 \(|z| = 5\)✓ 1
A1 \(\arg z = 0.644\) (0.6435…)✓ 1
Part (b)
M1 attempt to compute \(i(4+3i)\)✓ 1
A1 \(-3 + 4i\)✓ 1
Part (c)
A1 \(\arg(-3+4i) = \pi - \tan^{-1}\frac{4}{3} = 2.21\) (2.2143…), which is \(0.6435\ldots + \frac{\pi}{2}\)✓ 1

Exam tip: For a point in the second quadrant, a calculator's inverse tangent gives an angle that is too small by a multiple of π; sketch the point on an Argand diagram to choose the right angle.

45IBMI3H-1-05r19-02Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy6 marks

A fashion studio orders scarves and hats for three of its shops. The number of each item ordered is given by the matrix \(\mathbf{A}\), where the rows are Shop 1, Shop 2 and Shop 3 and the columns are scarves and hats.
\(\mathbf{A}=\begin{pmatrix}12&8\\20&5\\15&10\end{pmatrix}\)
A scarf costs €18.50 and a hat costs €12.00 to make. The costs are written as the column matrix \(\mathbf{c}=\begin{pmatrix}18.5\\12\end{pmatrix}\).

(a)

Find \(\mathbf{A}\mathbf{c}\) and state what each element of the result represents.

(3)
(b)

Find the total cost of the orders for all three shops.

(1)
(c)

The cost of making each item rises by 8%. Find the new total cost of the three orders, in €, to 2 decimal places.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt at matrix product \(\mathbf{A}\mathbf{c}\)✓ 1
A1 \(\begin{pmatrix}318\\430\\397.5\end{pmatrix}\) (at least two entries correct)✓ 1
A1 each element is the total cost (in €) of the order for the corresponding shop, all three entries correct✓ 1
Part (b)
A1 €1145.50✓ 1
Part (c)
M1 \(1145.5\times1.08\)✓ 1
A1 €1237.14✓ 1

Common error: Writing the product as \(\mathbf{c}\mathbf{A}\) gives a matrix product that is not defined here; check that the number of columns of the first matrix equals the number of rows of the second.

46IBMI3H-1-05r19-03Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy6 marks

Two fishing boats record their catches after a day at sea. Boat 1 used 3 nets of type X and 2 nets of type Y and landed 1260 kg of fish. Boat 2 used 5 nets of type X and 4 nets of type Y and landed 2300 kg. Let each net of type X catch \(x\) kg and each net of type Y catch \(y\) kg, and assume every net of the same type catches the same mass.

(a)

Write the information as a matrix equation of the form \(\mathbf{M}\begin{pmatrix}x\\y\end{pmatrix}=\mathbf{b}\).

(1)
(b)

Find \(\det\mathbf{M}\) and write down \(\mathbf{M}^{-1}\).

(2)
(c)

Use \(\mathbf{M}^{-1}\) to find \(x\) and \(y\).

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(\begin{pmatrix}3&2\\5&4\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}1260\\2300\end{pmatrix}\)✓ 1
Part (b)
A1 \(\det\mathbf{M}=3\times4-2\times5=2\)✓ 1
A1 \(\mathbf{M}^{-1}=\frac{1}{2}\begin{pmatrix}4&-2\\-5&3\end{pmatrix}\)✓ 1
Part (c)
M1 \(\begin{pmatrix}x\\y\end{pmatrix}=\mathbf{M}^{-1}\begin{pmatrix}1260\\2300\end{pmatrix}\)✓ 1
A1 \(x=220\)✓ 1
A1 \(y=300\) (kg)✓ 1

Exam tip: Pre-multiply both sides by the inverse: \(\mathbf{M}^{-1}\mathbf{M}\mathbf{x}=\mathbf{M}^{-1}\mathbf{b}\). Multiplying on the right is not valid because matrix multiplication is not commutative.

47IBMI3H-1-05r19-13Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy5 marks

A rectangular fish tank has internal dimensions 1.85 m by 0.62 m by 0.48 m.

(a)

Calculate the volume of the tank, in m³, exactly as the calculator displays it.

(2)
(b)

Write the volume from (a) correct to 2 significant figures.

(1)
(c)

Find the percentage error in using the answer to (b) as the volume. Give your answer to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(1.85\times0.62\times0.48\)✓ 1
A1 0.55056 m³✓ 1
Part (b)
A1 0.55 m³✓ 1
Part (c)
M1 \(\frac{|0.55-0.55056|}{0.55056}\times100\)✓ 1
A1 0.102% (0.10171…)✓ 1

Exam tip: Percentage error is the size of the error divided by the exact value, times 100; use the exact value from (a), not the rounded one, in the denominator.

48IBMI3H-1-05r20-01Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy4 marks

Calculator

In a strategy game, positions on the map are stored as complex numbers, with the base at the origin. A scout camp is located at \(z=-6+8i\), where the real part is the distance east (in map units) and the imaginary part is the distance north.

(a)

Find the modulus of \(z\).

(2)
(b)

Find the argument of \(z\) in radians, giving your answer to 3 significant figures, with \(-\pi<\arg z\le\pi\).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 use of \(\sqrt{(-6)^2+8^2}\)✓ 1
A1 \(|z|=10\)✓ 1
Part (b)
M1 recognising that \(z\) is in the second quadrant, e.g. \(\pi-\arctan\left(\frac{8}{6}\right)\)✓ 1
A1 \(\arg z=2.21\)✓ 1

Common error: Always sketch the point first: a negative real part and positive imaginary part puts the argument in the second quadrant, so a calculator value of \(\arctan(8/-6)\) needs adjusting.

49IBMI3H-1-05r20-02Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy5 marks

Calculator

A hospital pharmacy supplies three medicines to two wards. The matrix \(\mathbf{N}\) shows the number of boxes ordered per week, with rows for Ward A and Ward B and columns for medicines P, Q and R:
\(\mathbf{N}=\begin{pmatrix}12&8&5\\ 9&15&10\end{pmatrix}\)
The cost per box, in euros, is €4.50 for P, €7.20 for Q and €12.00 for R.

(a)

Write down a column matrix \(\mathbf{C}\) for the cost per box of each medicine.

(1)
(b)

Calculate the matrix product \(\mathbf{N}\mathbf{C}\).

(2)
(c)

Hence find the total weekly cost for the two wards together, and state what each entry of \(\mathbf{N}\mathbf{C}\) represents.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(\mathbf{C}=\begin{pmatrix}4.50\\7.20\\12.00\end{pmatrix}\)✓ 1
Part (b)
M1 attempt to multiply, e.g. \(12(4.5)+8(7.2)+5(12)\)✓ 1
A1 \(\begin{pmatrix}171.6\\ 268.5\end{pmatrix}\), accept \(\begin{pmatrix}171.60\\ 268.50\end{pmatrix}\)✓ 1
Part (c)
A1 total \(=171.6+268.5=\) €440.10✓ 1
R1 each entry is the weekly medicine cost (in €) for that ward (row 1 Ward A, row 2 Ward B)✓ 1

Exam tip: When multiplying matrices, check that the number of columns of the first equals the number of rows of the second, then say in words what each entry of the product represents.

50IBMI3H-1-05r20-03Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy5 marks

Calculator

A puzzle game encodes each secret position \((x,y)\) into a pair of display numbers \((p,q)\) using the matrix
\(\mathbf{M}=\begin{pmatrix}4&3\\2&2\end{pmatrix},\qquad \begin{pmatrix}p\\q\end{pmatrix}=\mathbf{M}\begin{pmatrix}x\\y\end{pmatrix}.\)

(a)

Find \(\det\mathbf{M}\).

(1)
(b)

Find \(\mathbf{M}^{-1}\).

(2)
(c)

The display shows \((p,q)=(31,18)\). Find the secret position \((x,y)\).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(\det\mathbf{M}=4(2)-3(2)=2\)✓ 1
Part (b)
M1 swap leading-diagonal entries and negate the others, with factor \(\frac{1}{\det\mathbf{M}}\)✓ 1
A1 \(\mathbf{M}^{-1}=\frac{1}{2}\begin{pmatrix}2&-3\\-2&4\end{pmatrix}=\begin{pmatrix}1&-1.5\\-1&2\end{pmatrix}\)✓ 1
Part (c)
M1 \(\begin{pmatrix}x\\y\end{pmatrix}=\mathbf{M}^{-1}\begin{pmatrix}31\\18\end{pmatrix}\)✓ 1
A1 \((x,y)=(4,5)\)✓ 1

Exam tip: To undo a matrix transformation, multiply by the inverse on the left of the column vector. A quick check is to multiply \(\mathbf{M}\) by your answer and see that you recover the display numbers.

51IBMI3H-1-05r20-04Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy5 marks

Calculator

The voltage of a pump motor controller is modelled by the complex number \(v=6e^{i\pi/6}\) (in volts). A second component has voltage \(w=2e^{i\pi/3}\) (in volts).

(a)

Write \(v\) in the form \(a+bi\), giving \(a\) and \(b\) to 3 significant figures.

(2)
(b)

Find \(vw\), giving your answer in the form \(re^{i\theta}\).

(2)
(c)

State the value of \(vw\) in the form \(a+bi\).

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(a=6\cos\frac{\pi}{6}\), \(b=6\sin\frac{\pi}{6}\)✓ 1
A1 \(v=5.20+3.00i\)✓ 1
Part (b)
M1 multiply moduli and add arguments✓ 1
A1 \(vw=12e^{i\pi/2}\)✓ 1
Part (c)
A1 \(vw=12i\)✓ 1

Exam tip: In exponential form, multiplying complex numbers multiplies the moduli and adds the arguments, which is much quicker than expanding Cartesian brackets.

52IBMI3H-1-04r21-01Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy6 marks

In a strategy game, a scout stands at the point represented by the complex number z = 6 + 8i, where distances are in metres from the base at the origin.

(a)

Find the modulus of z and its argument in radians, to 3 significant figures.

(3)
(b)

The scout is moved by a rotation of 60° anticlockwise about the base. Find the scout's new position as a complex number in the form a + bi, with a and b given to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to use √(62 + 82) or tan−1(8/6)✓ 1
A1 modulus = 10✓ 1
A1 argument = 0.927✓ 1
Part (b)
M1 multiply by eiπ/3, or add π/3 to the argument and keep the modulus 10✓ 1
A1 real part −3.93✓ 1
A1 imaginary part 9.20 (unrounded 9.196…)✓ 1

Common error: Rotating through 60° but adding 60 to an argument that is in radians. Convert the angle to π/3 first, or keep the calculator in degree mode throughout.

53IBMI3H-1-04r21-02Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy4 marks

In a puzzle game, a magic portal sits at the point z = 4e2πi/3 in the complex plane.

(a)

Write z in the form a + bi, giving a and b to 3 significant figures.

(2)
(b)

A second portal is at z2. Find the exact argument of z2, in the interval −π < θ ≤ π.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 a = 4cos(2π/3), b = 4sin(2π/3)✓ 1
A1 −2.00 + 3.46i✓ 1
Part (b)
M1 argument of z2 is 2 × 2π/3 = 4π/3, outside the interval, so subtract 2π✓ 1
A1 -2π/3 (= -2.09 rad)✓ 1

Exam tip: When you square a complex number the modulus is squared and the argument is doubled. Always check whether the new argument lies in the required interval.

54IBMI3H-1-04r21-13Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy5 marks

Chlorine is added to a water storage tank. The concentration of chlorine is modelled by C = 2.4 × 0.85t mg per litre, where t is the time in hours after the chlorine is added.

(a)

Find the concentration after 3 hours, to 3 significant figures.

(2)
(b)

The water needs to be treated again when the concentration falls below 0.5 mg per litre. Find the time at which this happens, in hours to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 substitute t = 3, 2.4 × 0.853✓ 1
A1 1.47 mg per litre✓ 1
Part (b)
M1 set 2.4 × 0.85t = 0.5✓ 1
M1 0.85t = 0.5/2.4 and take logarithms of both sides✓ 1
A1 t = 9.65 hours✓ 1

Exam tip: To solve an equation where the unknown is an exponent, isolate the power first and then take logarithms. You can check the result by substituting it back in.

55IBMI3H-1-04r21-14Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy7 marks

In a role-playing game, the experience points (XP) needed to move from level n to level n + 1 form an arithmetic sequence. Moving from level 1 to level 2 needs 400 XP, and each further level-up needs 75 XP more than the one before.

(a)

Find the XP needed to move from level 12 to level 13.

(2)
(b)

A player has earned a total of 52000 XP since level 1. Find the greatest number of complete level-ups the player has made.

(3)
(c)

Find how many XP the player has earned towards the next level-up.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 u12 = 400 + 11 × 75✓ 1
A1 1225 XP✓ 1
Part (b)
M1 total for n level-ups = (n/2)(2 × 400 + (n − 1) × 75) ≤ 52000✓ 1
M1 evidence of solving, such as comparing the totals 50000 for n = 32 and 52800 for n = 33✓ 1
A1 32 level-ups✓ 1
Part (c)
M1 52000 − 50000✓ 1
A1 2000 XP (ECF from (b))✓ 1

Exam tip: A question about a whole number of steps needs a whole-number answer. After solving the inequality, check the values either side of the boundary.

56IBMI3H-1-04r22-01Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy7 marks

Two components of a railway signalling circuit are connected in series. Their impedances are \(Z_1 = (12 + 5i)\) Ω and \(Z_2 = (3 - 9i)\) Ω. The circuit is supplied with a voltage of 24 V, which is represented by the real number 24.

(a)

Find the total impedance \(Z = Z_1 + Z_2\) in the form a + bi.

(1)
(b)

Find the modulus and the argument of Z. Give the argument in radians to 3 significant figures.

(2)
(c)

The current is \(I = \dfrac{24}{Z}\). Write I in the form a + bi with a and b given to 3 significant figures.

(2)
(d)

Find the magnitude of the current, in amperes, to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(Z = 15 - 4i\) Ω✓ 1
Part (b)
A1 \(|Z| = \sqrt{15^2 + 4^2} = \sqrt{241} = 15.52\) Ω (15.5)✓ 1
A1 \(\arg Z = -\arctan(4/15) = -0.261\) rad (fourth quadrant)✓ 1
Part (c)
M1 multiplies numerator and denominator by the conjugate \(15 + 4i\)✓ 1
A1 \(I = \dfrac{24(15 + 4i)}{241} = 1.49 + 0.398i\) A✓ 1
Part (d)
M1 \(|I| = \dfrac{24}{|Z|}\) (or modulus of the answer to (c))✓ 1
A1 1.55 A✓ 1

Exam tip: Dividing by a complex number: multiply top and bottom by its conjugate so the denominator becomes the real number \(a^2 + b^2\).

57IBMI3H-1-04r22-02Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy6 marks

Two radio signals from a telescope array are represented by the complex numbers \(z_1 = 7e^{i\pi/5}\) and \(z_2 = 2e^{-i\pi/3}\).

(a)

Write the product \(z_1z_2\) in the form \(re^{i\theta}\), where \(r > 0\) and \(-\pi < \theta \le \pi\).

(2)
(b)

Write the quotient \(\dfrac{z_1}{z_2}\) in the form \(re^{i\theta}\), where \(r > 0\) and \(-\pi < \theta \le \pi\).

(2)
(c)

Write \(z_1z_2\) in the form a + bi, with a and b given to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 modulus \(7 \times 2 = 14\)✓ 1
A1 argument \(\dfrac{\pi}{5} - \dfrac{\pi}{3} = -\dfrac{2\pi}{15}\), so \(z_1z_2 = 14e^{-2\pi i/15}\)✓ 1
Part (b)
A1 modulus \(7 \div 2 = 3.5\)✓ 1
A1 argument \(\dfrac{\pi}{5} + \dfrac{\pi}{3} = \dfrac{8\pi}{15}\), so \(\dfrac{z_1}{z_2} = 3.5e^{8\pi i/15}\)✓ 1
Part (c)
M1 \(14\cos\left(\dfrac{2\pi}{15}\right) - 14i\sin\left(\dfrac{2\pi}{15}\right)\)✓ 1
A1 \(a = 12.8\), \(b = -5.69\), so \(z_1z_2 = 12.8 - 5.69i\)✓ 1

Common error: When multiplying numbers in Euler form, the moduli multiply and the arguments add; when dividing, the moduli divide and the arguments subtract.

58IBMI3H-1-04r22-10Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy6 marks

An observatory borrows €60 000 to renew a telescope mirror. The loan has a nominal annual interest rate of 5.4%, compounded monthly (that is 0.45% per month). The observatory makes a payment of €1100 at the end of each month.

(a)

Find the number of months, to 3 significant figures, for the loan to be repaid. Assume the formula for a non-integer number of payments.

(3)
(b)

The loan therefore needs 63 payments, and the last payment is smaller than €1100. Find the amount of the last payment, to the nearest cent.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 uses a finance solver with I% = 5.4, PV = 60000, PMT = −1100, FV = 0, P/Y = C/Y = 12 and solves for N (or the annuity formula)✓ 1
M1 e.g. \(60000 = 1100\,\dfrac{1 - 1.0045^{-n}}{0.0045}\) leading to \(1.0045^{-n} = 1 - \dfrac{60000 \times 0.0045}{1100}\)✓ 1
A1 62.7 months✓ 1
Part (b)
M1 balance after 62 payments = 60000 × 1.004562 − 1100 × (1.004562 − 1)/0.0045 = €796.96✓ 1
M1 adds one month of interest: balance × 1.0045✓ 1
A1 €800.55✓ 1

Common error: When the number of payments is not a whole number, round up to complete the loan and recalculate the final payment; do not assume the last payment equals the others.

59IBMI3H-1-04r22-14Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy5 marks

A fund manager invests €8000 in an account that pays compound interest with the interest added each month. After 7 years the account contains €10 450.

(a)

Find the nominal annual interest rate, as a percentage to 3 significant figures.

(3)
(b)

Find the annual percentage growth rate that would give the same value after 7 years if interest were added once a year. Give your answer to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 equation \(8000\left(1 + \dfrac{r}{12}\right)^{84} = 10450\) (or finance solver with N = 84, PV = −8000, FV = 10450, P/Y = C/Y = 12)✓ 1
M1 \(1 + \dfrac{r}{12} = \left(\dfrac{10450}{8000}\right)^{1/84}\)✓ 1
A1 3.82%✓ 1
Part (b)
M1 \(\left(\dfrac{10450}{8000}\right)^{1/7} - 1\)✓ 1
A1 3.89%✓ 1

Common error: A nominal rate compounded monthly gives a slightly smaller annual growth rate than the same figure compounded yearly; the annual growth is always a little higher than the nominal rate.

60IBMI3H-1-05r22-01Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy5 marks

Two radio signals arrive at a receiving antenna. Their amplitudes and phases, in millivolts, are represented by the complex numbers s1 = 6 + 8i and s2 = −2 + 3i. The combined signal is s = s1 + s2.

(a)

Find s in the form a + bi.

(1)
(b)

Find the modulus of s, giving your answer to 3 significant figures.

(2)
(c)

Find the argument of s in radians, to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 s = 4 + 11i✓ 1
Part (b)
M1 attempt to use \(\sqrt{4^2+11^2}\)✓ 1
A1 11.7 mV✓ 1
Part (c)
M1 attempt to use \(\tan^{-1}\left(\frac{11}{4}\right)\)✓ 1
A1 1.22 rad✓ 1

Common error: Check that the calculator is in radian mode when an argument is requested in radians.

61IBMI3H-1-05r22-05Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy5 marks

On a trek, a climber ascends 600 m on the first day. On each later day the ascent is 80 % of the previous day's ascent.

(a)

Find the ascent on day 4.

(2)
(b)

Find the total ascent over the first 6 days, to the nearest metre.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(600(0.8)^3\)✓ 1
A1 307.2 m✓ 1
Part (b)
M1 use of the sum formula for a geometric series✓ 1
A1 \(\frac{600(1-0.8^6)}{1-0.8}\)✓ 1
A1 2214 m✓ 1

Common error: The sum formula uses n terms: check that you have used 6, not 5 or 7.

62IBMI3H-1-05r22-07Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy6 marks

To protect a data link, pairs of numbers \(\begin{pmatrix}x\\y\end{pmatrix}\) are encoded by multiplying by the matrix \(E=\begin{pmatrix}3&2\\4&3\end{pmatrix}\).

(a)

Encode the pair \(\begin{pmatrix}6\\9\end{pmatrix}\).

(2)
(b)

Find \(E^{-1}\).

(2)
(c)

The receiver gets the encoded pair \(\begin{pmatrix}47\\64\end{pmatrix}\). Find the original pair.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(E\begin{pmatrix}6\\9\end{pmatrix}\)✓ 1
A1 \(\begin{pmatrix}36\\51\end{pmatrix}\)✓ 1
Part (b)
A1 det E = 1✓ 1
A1 \(E^{-1}=\begin{pmatrix}3&-2\\-4&3\end{pmatrix}\)✓ 1
Part (c)
M1 \(E^{-1}\begin{pmatrix}47\\64\end{pmatrix}\)✓ 1
A1 \(\begin{pmatrix}13\\4\end{pmatrix}\)✓ 1

Exam tip: Check a decoded answer by re-encoding it: multiplying by E should return the received pair.

63IBMI3H-1-05r22-11Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy5 marks

The altitudes of the camps on a mountain form an arithmetic sequence. Camp 2 is at 4550 m and camp 6 is at 6070 m.

(a)

Find the common difference and the altitude of camp 1.

(3)
(b)

Find the altitude of camp 9.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 two equations, e.g. \(u_1+d=4550\) and \(u_1+5d=6070\)✓ 1
A1 d = 380 m✓ 1
A1 u1 = 4170 m✓ 1
Part (b)
M1 \(4170+8(380)\)✓ 1
A1 7210 m (ECF from (a))✓ 1

Exam tip: Subtracting the two equations eliminates u₁ and gives 4d directly.

64IBMI3H-1-04r24-01Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy5 marks

Two sonar signals received by a research vessel are represented by the complex numbers \(z_1=2+3i\) and \(z_2=4-i\).

(a)

Find \(z_1z_2\) in the form \(a+bi\).

(2)
(b)

Find \(\dfrac{z_1z_2}{1+i}\) in the form \(a+bi\).

(2)
(c)

Find the modulus of \(\dfrac{z_1z_2}{1+i}\) to 3 significant figures.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 expand \((2+3i)(4-i)\) using \(i^2=-1\)✓ 1
A1 \(11+10i\)✓ 1
Part (b)
M1 multiply numerator and denominator by \(1-i\) (ECF from (a))✓ 1
A1 \(10.5-0.5i\)✓ 1
Part (c)
A1 \(\sqrt{10.5^2+0.5^2}=10.5\)✓ 1

Common error: Remember that \(i^2=-1\) when expanding; a sign slip here changes both the real and imaginary parts.

65IBMI3H-1-04r24-02Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy7 marks

In the plan of a rotating façade, a marker on the panel is at the point represented by \(z=-2+2\sqrt{3}\,i\), where lengths are in metres.

(a)

Find the modulus and argument of \(z\), giving the argument in radians as an exact value.

(3)
(b)

Use de Moivre’s theorem to find \(z^2\) in the form \(a+bi\), giving exact values.

(2)
(c)

Find the smallest positive integer \(n\) for which \(z^n\) is a positive real number.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(|z|=\sqrt{4+12}=4\)✓ 1
M1 identify that \(z\) is in the second quadrant✓ 1
A1 \(\arg z=\dfrac{2\pi}{3}\)✓ 1
Part (b)
M1 \(z^2=16\left(\cos\dfrac{4\pi}{3}+i\sin\dfrac{4\pi}{3}\right)\)✓ 1
A1 \(-8-8\sqrt{3}\,i\)✓ 1
Part (c)
M1 \(\arg z^n=\dfrac{2\pi n}{3}\) must be a multiple of \(2\pi\)✓ 1
A1 \(n=3\)✓ 1

Exam tip: Sketch the point on an Argand diagram before finding the argument, so that you choose the correct quadrant.

66IBMI3H-1-04r24-06Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy5 marks

A logistics firm uses small vans carrying \(x\) crates each and large vans carrying \(y\) crates each. A load of 2 small vans and 1 large van carries 140 crates, and a load of 5 small vans and 3 large vans carries 380 crates.

(a)

Write the information as a matrix equation of the form \(\mathbf{A}\binom{x}{y}=\mathbf{b}\).

(1)
(b)

Find \(\mathbf{A}^{-1}\).

(2)
(c)

Hence find the number of crates a large van carries.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(\begin{pmatrix}2&1\\5&3\end{pmatrix}\binom{x}{y}=\binom{140}{380}\)✓ 1
Part (b)
A1 \(\det\mathbf{A}=1\)✓ 1
A1 \(\mathbf{A}^{-1}=\begin{pmatrix}3&-1\\-5&2\end{pmatrix}\)✓ 1
Part (c)
M1 \(\binom{x}{y}=\mathbf{A}^{-1}\binom{140}{380}\)✓ 1
A1 \(y=60\) crates (and \(x=40\))✓ 1

Exam tip: Multiply by the inverse on the left of both sides; matrix multiplication is not commutative.

67IBMI3H-1-04r24-09Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy6 marks

A logistics company has two warehouses, A and B. Each week 20% of the stock in A is moved to B and 30% of the stock in B is moved to A. The proportions in (A, B) change according to \(\binom{a_{n+1}}{b_{n+1}}=\mathbf{T}\binom{a_n}{b_n}\), where \(\mathbf{T}=\begin{pmatrix}0.8&0.3\\0.2&0.7\end{pmatrix}\).

(a)

Show that 1 is an eigenvalue of \(\mathbf{T}\).

(2)
(b)

Find the other eigenvalue of \(\mathbf{T}\).

(2)
(c)

Find the long-term percentage of stock held in warehouse A.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(\det(\mathbf{T}-\mathbf{I})=(-0.2)(-0.3)-(0.3)(0.2)\)✓ 1
A1 \(=0\), so 1 is an eigenvalue (AG)✓ 1
Part (b)
M1 sum of eigenvalues equals the trace \(1.5\) (or solve the characteristic equation)✓ 1
A1 \(0.5\)✓ 1
Part (c)
M1 solve \((\mathbf{T}-\mathbf{I})\mathbf{v}=\mathbf{0}\), giving \(-0.2a+0.3b=0\)✓ 1
A1 \(a:b=3:2\), so 60%✓ 1

Exam tip: The long-term state of a transition matrix is the eigenvector for eigenvalue 1, scaled so its entries sum to 1 (or 100%).

68IBMI3H-1-04r25-01Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy6 marks

The cooling fan in a data centre has a complex impedance Z1 = 6 + 8i ohms. It is connected in series with a second component of impedance Z2 = 4 − 3i ohms.

(a)

Find the modulus of Z1.

(1)
(b)

Find the argument of Z1, in radians, to 3 significant figures.

(2)
(c)

The total impedance of the series circuit is Z1 + Z2. Find its modulus to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 |Z1| = √(36 + 64) = 10 ohms✓ 1
Part (b)
M1 arctan(8/6) used (first quadrant)✓ 1
A1 0.927✓ 1
Part (c)
A1 Z1 + Z2 = 10 + 5i✓ 1
M1 √(102 + 52)✓ 1
A1 11.2 ohms✓ 1

Exam tip: For the argument, sketch the point on an Argand diagram first so that you choose the correct quadrant.

69IBMI3H-1-04r25-02Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy7 marks

On an archaeological site, the position of a post-hole relative to a datum point is represented by the complex number z = 6eiπ/3 metres, where the real axis points east and the imaginary axis points north. A second feature is at w = 2eiπ/6 metres from the same datum.

(a)

Write z in the form a + bi, giving exact values.

(2)
(b)

Find zw in the form reiθ, and state its value in Cartesian form.

(2)
(c)

Find the distance between the two features. Give your answer to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 6(cos(π/3) + i sin(π/3))✓ 1
A1 z = 3 + 3√3 i✓ 1
Part (b)
A1 zw = 12eiπ/2 (multiply moduli, add arguments)✓ 1
A1 zw = 12i✓ 1
Part (c)
A1 w = √3 + i (or 1.73 + i)✓ 1
M1 |z − w| = √((3 − √3)2 + (3√3 − 1)2)✓ 1
A1 4.38 m✓ 1

Common error: When multiplying numbers in polar form, add the arguments but multiply the moduli; do not multiply the arguments.

70IBMI3H-1-04r25-10Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy5 marks

In a cycle computer circuit, the voltage is V = (7 − i) volts and the impedance is Z = (2 + 3i) ohms. The current is I = V/Z.

(a)

Find I in the form a + bi, giving exact values.

(3)
(b)

Find |I|, the magnitude of the current, in amperes, to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 multiply numerator and denominator by 2 − 3i✓ 1
A1 numerator (7 − i)(2 − 3i) = 11 − 23i and denominator 13✓ 1
A1 I = 11/13 − (23/13)i✓ 1
Part (b)
M1 √((11/13)2 + (23/13)2) (or |V|/|Z|)✓ 1
A1 1.96 A✓ 1

Exam tip: |V/Z| = |V|/|Z|, so the magnitude can be found without first dividing in Cartesian form.

71IBMI3H-1-04r25-12Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy7 marks

A data centre adds storage capacity every month. In month 1 it adds 40 TB and in each later month it adds 8% more than in the previous month, so the amounts added form a geometric sequence.

(a)

Find the capacity added in month 12, to 3 significant figures.

(2)
(b)

Find the total capacity added during the first 12 months, to 3 significant figures.

(2)
(c)

The data centre needs a total of more than 2000 TB. Find the smallest number of months, n, after which this is achieved.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 u12 = 40 × 1.0811✓ 1
A1 93.3 TB✓ 1
Part (b)
M1 S12 = 40(1.0812 − 1)/(1.08 − 1)✓ 1
A1 759 TB (759.1 before rounding)✓ 1
Part (c)
M1 40(1.08n − 1)/0.08 > 2000, so 1.08n > 5✓ 1
M1 take logarithms: n > ln 5 / ln 1.08 = 20.9✓ 1
A1 n = 21✓ 1

Common error: After solving an inequality with logarithms, round up to the next whole number of months, not to the nearest.

72IBMI3H-1-08r27-01Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy7 marks

The tip of a robot arm is at position z = 0.45 + 0.60i metres in the complex plane, where the base of the arm is at the origin, the real axis is horizontal and the imaginary axis is vertical.

(a)

Find the modulus of z and the argument of z in radians. Give the argument to 3 significant figures.

(3)
(b)

The arm rotates anticlockwise through 25° about the base, keeping its length. Write the new position in the form reiθ, with θ in radians to 3 significant figures.

(2)
(c)

Find the new height of the tip above the horizontal line through the base, in metres, to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 modulus = √(0.452 + 0.602)✓ 1
A1 |z| = 0.75 m✓ 1
A1 arg z = tan−1(0.60 ÷ 0.45) = 0.927 rad✓ 1
Part (b)
A1 r = 0.75 (length unchanged)✓ 1
A1 θ = 0.92730 + 0.4363 = 1.36 (using the unrounded argument)✓ 1
Part (c)
M1 imaginary part = 0.75 sin(1.3636…) (using the unrounded angle)✓ 1
A1 0.734 m✓ 1

Common error: Convert the rotation angle to radians (25° = 0.436 rad) before adding it to the argument, and keep the calculator in radian mode for the sine.

73IBMI3H-1-08r27-03Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy5 marks

A boutique sells scarves and belts. A customer pays €112 for 3 scarves and 2 belts. Another customer pays €94 for 1 scarf and 4 belts. Let a scarf cost €s and a belt cost €b. The situation is modelled by
A(s, b)T = (112, 94)T, where A = [[3, 2], [1, 4]] (rows listed in turn).

(a)

Find the determinant of A and hence write down A−1.

(2)
(b)

Use A−1 to find the price of one belt.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 det A = 3 × 4 − 2 × 1 = 10✓ 1
A1 A−1 = (1/10)[[4, −2], [−1, 3]]✓ 1
Part (b)
M1 (s, b)T = A−1(112, 94)T✓ 1
A1 second row: (−1 × 112 + 3 × 94) ÷ 10✓ 1
A1 €17.00 (and scarf €26.00)✓ 1

Common error: Matrix multiplication is not commutative: the inverse multiplies from the left, so the solution is A⁻¹B, not BA⁻¹.

74IBMI3H-1-08r27-09Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy8 marks

A camera calibration for a robot maps each point (x, y) of the robot's frame, in centimetres, to image coordinates using the matrix M = [[3, −1], [2, 4]] (rows listed in turn), so that the image is M(x, y)T.

(a)

Find det M. A triangular region of the frame has area 36 cm2. Find the area of its image.

(3)
(b)

Find M−1.

(2)
(c)

A point in the robot's frame is imaged at (9, 20). Find its coordinates in the frame.

(2)
(d)

Find the distance of this point from the origin of the frame, in centimetres.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 det M = 3 × 4 − (−1) × 2✓ 1
A1 14✓ 1
A1 area of image = 14 × 36 = 504 cm2✓ 1
Part (b)
M1 use of 1/det × adjusted matrix (swap leading diagonal, negate the other entries)✓ 1
A1 M−1 = (1/14)[[4, 1], [−2, 3]]✓ 1
Part (c)
M1 M−1(9, 20)T✓ 1
A1 (4, 3)✓ 1
Part (d)
A1 √(42 + 32) = 5 cm (ECF from (c))✓ 1

Common error: The area scale factor is the absolute value of the determinant, not the determinant squared; a negative determinant would only mean the orientation is reversed.

75IBMI3H-1-08r27-14Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy7 marks

The capacity of the battery in a robot vacuum cleaner is measured when fully charged. At the first charge it is 4500 mAh, and each later full charge gives a capacity that is 98.5% of the previous one. So the capacities form a geometric sequence u1 = 4500, u2 = 4500 × 0.985, …

(a)

Find the capacity at the 80th charge, to 3 significant figures.

(2)
(b)

The battery is replaced when its capacity first falls below 3000 mAh. Find the charge number at which this first happens.

(3)
(c)

Find the total capacity delivered over the first 40 charges, to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 u80 = 4500 × 0.98579✓ 1
A1 1360 mAh (1.36 × 103)✓ 1
Part (b)
M1 4500 × 0.985n−1 < 3000✓ 1
M1 n − 1 > ln(2/3) ÷ ln(0.985) = 26.83 (reverse inequality, or check n = 27, 28 by substitution)✓ 1
A1 28th charge✓ 1
Part (c)
M1 sum of a geometric series 4500(1 − 0.98540) ÷ (1 − 0.985)✓ 1
A1 136 000 mAh (1.36 × 105)✓ 1

Common error: In a geometric sequence the nth term is u₁rⁿ⁻¹, not u₁rⁿ; the 80th charge uses the power 79.

76IBMI3H-1-04r30-01Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy5 marks

A ground station tracks a space probe. In a plane through the station, the probe's position is represented by the complex number \(z=4.2+3.1i\), where the real and imaginary parts are in thousands of kilometres.

(a)

Calculate \(|z|\), the distance of the probe from the station in thousands of kilometres. Give your answer to 3 significant figures.

(2)
(b)

Find \(\arg z\) in radians, to 3 significant figures, and hence write \(z\) in the form \(re^{i\theta}\) with \(r\) and \(\theta\) to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(\sqrt{4.2^2+3.1^2}\)✓ 1
A1 5.22✓ 1
Part (b)
M1 \(\arctan\left(\frac{3.1}{4.2}\right)\)✓ 1
A1 0.636✓ 1
A1 \(z=5.22e^{0.636i}\)✓ 1

Exam tip: Check that your calculator is in radian mode when the question asks for an argument in radians, and that the point is in the first quadrant before you accept the arctan value.

77IBMI3H-1-04r30-02Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy4 marks

In a motion-capture study, a marker on an athlete's knee is represented by the complex number \(z_1=6e^{i\pi/6}\) (in cm). When the athlete's leg moves, the marker's new position is \(z_2=wz_1\), where \(w=0.8e^{i\pi/4}\).

(a)

Write \(z_2\) in the form \(re^{i\theta}\), giving \(r\) and \(\theta\) exactly.

(2)
(b)

Write \(z_2\) in the form \(a+bi\), with \(a\) and \(b\) correct to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 multiply moduli and add arguments✓ 1
A1 \(z_2=4.8e^{i\frac{5\pi}{12}}\)✓ 1
Part (b)
M1 \(a=4.8\cos\frac{5\pi}{12}\) and \(b=4.8\sin\frac{5\pi}{12}\)✓ 1
A1 \(1.24+4.64i\)✓ 1

Common error: When multiplying complex numbers in exponential form, the moduli multiply but the arguments add; do not multiply the arguments.

78IBMI3H-1-04r30-03Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy5 marks

A rail ticket machine records its sales. On Monday it sold 3 single tickets and 2 day-return tickets for a total of €21. On Tuesday it sold 5 single tickets and 4 day-return tickets for a total of €37. Let \(s\) and \(r\) be the prices in euros of a single and a day-return ticket. The system can be written as \(A\begin{pmatrix}s\\r\end{pmatrix}=\begin{pmatrix}21\\37\end{pmatrix}\) where \(A=\begin{pmatrix}3&2\\5&4\end{pmatrix}\).

(a)

Find \(\det A\) and write down \(A^{-1}\).

(3)
(b)

Hence find the price of a single ticket and the price of a day-return ticket.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(\det A=2\)✓ 1
M1 attempt to use \(\frac{1}{\det A}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}\)✓ 1
A1 \(A^{-1}=\frac12\begin{pmatrix}4&-2\\-5&3\end{pmatrix}\)✓ 1
Part (b)
M1 \(\begin{pmatrix}s\\r\end{pmatrix}=A^{-1}\begin{pmatrix}21\\37\end{pmatrix}\)✓ 1
A1 single €5, day-return €3✓ 1

Exam tip: Multiply the inverse on the left of the column vector; \(A^{-1}b\) and \(bA^{-1}\) are not the same, and the second is not even defined here.

79IBMI3H-1-04r30-04Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy5 marks

A sports club buys 12 training bibs and 8 cones for €62. A second order, of 5 bibs and 9 cones, costs €40. Let a bib cost €\(x\) and a cone cost €\(y\).

(a)

Write down two equations in \(x\) and \(y\).

(2)
(b)

Solve the equations to find \(x\) and \(y\).

(2)
(c)

Find the cost of 7 bibs and 6 cones.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(12x+8y=62\)✓ 1
A1 \(5x+9y=40\)✓ 1
Part (b)
M1 use of simultaneous-equation solver or elimination✓ 1
A1 \(x=3.5\), \(y=2.5\)✓ 1
Part (c)
A1 €39.50✓ 1

Exam tip: Check your solution by substituting it back into the second equation as well as the first.

80IBMI3H-1-06r30-01Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy6 marks

Two antennas on a space probe produce signals represented by the complex numbers z1 = 3 + 4i and z2 = 1 − 2i, where i2 = −1.

(a)

Find z1z2 in the form a + bi.

(2)
(b)

Find z1 / z2 in the form a + bi.

(2)
(c)

Find |z1z2|, correct to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 expands (3 + 4i)(1 − 2i) and uses i2 = −1✓ 1
A1 11 − 2i✓ 1
Part (b)
M1 multiplies numerator and denominator by 1 + 2i, the conjugate of the denominator✓ 1
A1 (−5 + 10i)/5 = −1 + 2i✓ 1
Part (c)
M1 √(112 + (−2)2) = √125✓ 1
A1 11.2✓ 1

Common error: When dividing, multiply top and bottom by the conjugate of the denominator, not by its negative.

81IBMI3H-1-06r30-02Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy4 marks

A sensor on a bridge deck records a vibration as the complex number z = −5 + 12i.

(a)

Find the modulus of z.

(2)
(b)

Find the argument of z in radians, with −π < arg z ≤ π, correct to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 √((−5)2 + 122)✓ 1
A1 13✓ 1
Part (b)
M1 recognises that z is in the second quadrant, so arg = π − arctan(12/5)✓ 1
A1 1.97 radians✓ 1

Common error: Arctan(12/5) alone gives an angle in the first quadrant; check the quadrant from the signs of the real and imaginary parts.

82IBMI3H-1-06r30-03Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy5 marks

A coffee roaster buys two kinds of green bean, Arabica and Robusta. One mixed batch uses 2 kg of Arabica and 3 kg of Robusta and costs €39.50. A second batch uses 4 kg of Arabica and 1 kg of Robusta and costs €47.50. Let a and r be the prices in € per kg of Arabica and Robusta.

(a)

Write the information as the matrix equation A\(\begin{pmatrix}a\\r\end{pmatrix}\) = b, stating A and b.

(1)
(b)

Find A−1.

(2)
(c)

Hence find the price per kg of Arabica beans.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 A = \(\begin{pmatrix}2&3\\4&1\end{pmatrix}\) and b = \(\begin{pmatrix}39.5\\47.5\end{pmatrix}\)✓ 1
Part (b)
M1 determinant = 2×1 − 3×4 = −10✓ 1
A1 A−1 = −(1/10)\(\begin{pmatrix}1&-3\\-4&2\end{pmatrix}\)✓ 1
Part (c)
M1 multiplies A−1 by b✓ 1
A1 a = (39.5 − 142.5)/(−10) = €10.30 per kg✓ 1

Exam tip: Check a matrix solution by substituting back into both original equations.

83IBMI3H-1-06r30-10Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy5 marks

A coffee shop sells bags of light roast and dark roast. On Friday it sells 40 bags of light roast and 25 of dark roast. On Saturday it sells 55 bags of light roast and 30 of dark roast. A bag of light roast costs €6.50 and a bag of dark roast costs €8.20.

(a)

Write down a 2×2 matrix S of sales (rows are Friday and Saturday) and a column matrix p of prices.

(1)
(b)

Calculate Sp and state what its entries represent.

(3)
(c)

Find the total income over the two days.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 S = \(\begin{pmatrix}40&25\\55&30\end{pmatrix}\) and p = \(\begin{pmatrix}6.50\\8.20\end{pmatrix}\)✓ 1
Part (b)
M1 multiplies the matrices✓ 1
A1 \(\begin{pmatrix}465.00\\603.50\end{pmatrix}\)✓ 1
A1 the income in € on Friday and on Saturday respectively✓ 1
Part (c)
A1 465 + 603.50 = €1068.50✓ 1

Exam tip: Check the order of multiplication: a 2×2 matrix times a 2×1 matrix is valid and gives a 2×1 matrix.

84IBMI3H-1-05r32-01Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy6 marks

In a telecommunications link, two signal components are represented by the complex numbers \(z = 5 + 2i\) and \(w = 3 - 4i\).

(a)

Find \(zw\), giving your answer in the form \(a + bi\).

(2)
(b)

Calculate \(|zw|\), giving your answer to 3 significant figures.

(2)
(c)

Find \(\arg(zw)\) in radians, to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 expands \((5+2i)(3-4i)\) and uses \(i^2=-1\)✓ 1
A1 \(zw = 23 - 14i\)✓ 1
Part (b)
M1 \(\sqrt{23^2+(-14)^2}=\sqrt{725}\) (ECF from (a))✓ 1
A1 \(26.9\)✓ 1
Part (c)
M1 \(\arctan\left(\frac{14}{23}\right)\) with the 4th quadrant recognised (ECF from (a))✓ 1
A1 \(-0.547\) rad✓ 1

Common error: \(zw\) lies in the fourth quadrant (positive real part, negative imaginary part), so its argument is negative; a calculator's \(\tan^{-1}(14/23)\) alone gives the wrong sign.

85IBMI3H-1-05r32-07Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy5 marks

A mountain is estimated to be 6800 m high. Its true height is 6749 m.

(a)

Calculate the percentage error in the estimate, to 3 significant figures.

(2)
(b)

Write the true height in standard form, correct to 2 significant figures.

(1)
(c)

The air temperature falls by 6.5 °C for every 1000 m climbed. At sea level it is 15 °C. Predict the temperature at the summit, to 3 significant figures, using the true height.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(\frac{|6800-6749|}{6749}\times100\)✓ 1
A1 \(0.756\%\)✓ 1
Part (b)
A1 \(6.7\times10^{3}\) m✓ 1
Part (c)
M1 \(15-6.5\times6.749\)✓ 1
A1 \(-28.9\) °C✓ 1

Common error: Percentage error is divided by the true (exact) value, not by the estimate.

86IBMI3H-1-05r32-10Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy4 marks

The acidity of a food sample is measured by its pH, where \(\text{pH}=-\log_{10}[\text{H}^+]\) and \([\text{H}^+]\) is the hydrogen ion concentration in mol dm\(^{-3}\).

(a)

A yoghurt has pH 4.20. Find \([\text{H}^+]\), to 3 significant figures.

(2)
(b)

Fresh milk has pH 6.70. How many times greater is \([\text{H}^+]\) in the yoghurt than in the milk? Give your answer to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \([\text{H}^+]=10^{-4.20}\)✓ 1
A1 \(6.31\times10^{-5}\) mol dm\(^{-3}\)✓ 1
Part (b)
M1 \(10^{6.70-4.20}\) or \(\frac{10^{-4.20}}{10^{-6.70}}\)✓ 1
A1 \(316\)✓ 1

Common error: A lower pH means a higher concentration: a difference of 1 pH unit is a factor of 10, not an increase of 10%.

87IBMI3H-1-05r32-12Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy5 marks

The zeros of the transfer function of a telecoms filter are the solutions of \(z^2-8z+41=0\).

(a)

Solve the equation, giving your answers in the form \(a\pm bi\).

(2)
(b)

Write down the exact modulus of the solution with positive imaginary part.

(1)
(c)

Find the argument of this solution in radians, to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 uses the quadratic formula with \(\sqrt{-100}=10i\)✓ 1
A1 \(z=4\pm5i\)✓ 1
Part (b)
A1 \(\sqrt{41}\)✓ 1
Part (c)
M1 \(\arctan\left(\frac{5}{4}\right)\)✓ 1
A1 \(0.896\)✓ 1

Exam tip: Complex roots of a real quadratic come in conjugate pairs, so the two arguments are equal in size and opposite in sign.

88IBMI3H-1-08r33-01Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy5 marks

A seismometer near a volcano records a tremor signal represented by the complex number z = −4 + 7i (units: µm/s).

(a)

Find the modulus of z and its argument, giving the argument in radians to 3 significant figures, with −π < arg z ≤ π.

(3)
(b)

A processing step delays the phase of the signal by π/4, giving the new signal w = ze−iπ/4. Find the real part of w to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 |z| = √65 = 8.06✓ 1
M1 attempt at argument, recognising z is in the second quadrant, e.g. π − arctan(7/4)✓ 1
A1 arg z = 2.09✓ 1
Part (b)
M1 w = (−4 + 7i)(cos(π/4) − i sin(π/4)) = (−4 + 7i)(1 − i) ÷ √2, or modulus × cos(2.0899 − π/4)✓ 1
A1 real part = 3/√2 = 2.12 (the imaginary part is 11/√2 = 7.78)✓ 1

Common error: A negative real part and positive imaginary part puts z in the second quadrant, so arctan(7/−4) alone gives the wrong argument; add π to the calculator value.

89IBMI3H-1-08r33-06Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy4 marks

A theatre sells three ticket types (stalls, circle, balcony) for each of three performances. The numbers of tickets sold are given by the matrix N with rows for the performances and columns for the ticket types: row 1 is (120, 80, 40), row 2 is (150, 90, 60), row 3 is (100, 70, 50). The ticket prices in euros are (45, 30, 18) for stalls, circle and balcony respectively.

(a)

Write down a matrix product that gives the revenue for each performance, and evaluate it.

(2)
(b)

Find the total revenue from the three performances.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 N multiplied by the column of prices (45, 30, 18)✓ 1
A1 revenues €8520, €10530 and €7500✓ 1
Part (b)
M1 8520 + 10 530 + 7500✓ 1
A1 €26550✓ 1

Exam tip: In a product such as N p, the number of columns of N must equal the number of rows of p; set the prices as a column vector.

90IBMI3H-1-08r33-08Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy5 marks

Two cargo ships are loaded with crates of type P and type Q. Ship 1 carries 3 crates of type P and k crates of type Q. Ship 2 carries 2 crates of type P and 5 crates of type Q. Let a type P crate have mass x tonnes and a type Q crate have mass y tonnes. The matrix of crate numbers is M with rows (3, k) and (2, 5).

(a)

Find the value of k for which the masses x and y cannot be found uniquely from the total masses of the two ships.

(2)
(b)

Now k = 4. The total masses are 26 t for ship 1 and 29 t for ship 2. Find the inverse of M, and hence find y, the mass of a type Q crate.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 det M = 15 − 2k = 0✓ 1
A1 k = 7.5✓ 1
Part (b)
A1 M−1 = (1/7) × rows (5, −4) and (−2, 3)✓ 1
M1 (x, y)T = M−1(26, 29)T✓ 1
A1 y = 5 t (and x = 2 t)✓ 1

Exam tip: For a 2×2 matrix with rows (a, b) and (c, d), the inverse is (1/(ad − bc)) × rows (d, −b) and (−c, a); check by multiplying back to get the identity.

91IBMI3H-1-08r33-14Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy5 marks

After an eruption, the thickness of the ash layer, T cm, at a distance d km from the vent is modelled by T = Ae−kd. The layer is 40 cm thick at 2 km and 10 cm thick at 6 km.

(a)

Find the values of A and k. Give k to 3 significant figures.

(3)
(b)

Find the distance from the vent at which the layer is 2 cm thick, to 3 significant figures. Use the unrounded value of k.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 two equations 40 = Ae−2k and 10 = Ae−6k, and dividing: e4k = 4✓ 1
A1 k = (ln 4) ÷ 4 = 0.347✓ 1
A1 A = 80✓ 1
Part (b)
M1 2 = 80e−kd, so d = ln(40) ÷ k✓ 1
A1 10.6 km✓ 1

Exam tip: Divide the two equations to eliminate A, then take natural logarithms to bring the exponent down.

92IBMI3H-1-08r34-01Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy6 marks

A theatre turntable is centred at the origin of an Argand diagram whose axes are measured in metres. A prop sits at the point represented by \(z = 3 + i\). The turntable rotates anticlockwise through \(60°\) about the origin, which is equivalent to multiplying z by \(e^{i\pi/3}\).

(a)

Write \(e^{i\pi/3}\) in the form \(a + bi\).

(1)
(b)

Find the new position of the prop, giving the real and imaginary parts to three significant figures.

(3)
(c)

Find the distance of the prop from the centre of the turntable after the rotation, and explain why it does not change.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3} = 0.5 + 0.866i\) (or \(\frac12 + \frac{\sqrt3}{2}i\))✓ 1
Part (b)
M1 \((3+i)\left(\frac12 + \frac{\sqrt3}{2}i\right)\) expanded✓ 1
A1 real part \(\frac32 - \frac{\sqrt3}{2} = 0.634\)✓ 1
A1 imaginary part \(\frac{3\sqrt3}{2} + \frac12 = 3.10\) (3.098…)✓ 1
Part (c)
R1 \(|e^{i\pi/3}| = 1\), so multiplying by it leaves the modulus unchanged✓ 1
A1 \(\sqrt{10} = 3.16\) m✓ 1

Exam tip: Multiplying by e^{iθ} rotates a point by θ about the origin and never changes its modulus.

93IBMI3H-1-08r34-02Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy7 marks

A port charges x euros per standard container and y euros per refrigerated container. A ship unloading 4 standard and 3 refrigerated containers is charged €1200 in total. A second ship unloading 2 standard and 5 refrigerated containers is charged €1300 in total.

(a)

Write the information as a matrix equation and find the determinant of the coefficient matrix.

(2)
(b)

Use the inverse matrix to find x and y.

(3)
(c)

A third ship unloads 10 standard and 8 refrigerated containers. Find its total charge.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(\begin{pmatrix}4&3\\2&5\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}1200\\1300\end{pmatrix}\)✓ 1
A1 determinant \(4\times5 - 3\times2 = 14\)✓ 1
Part (b)
M1 \(\begin{pmatrix}x\\y\end{pmatrix}=\frac{1}{14}\begin{pmatrix}5&-3\\-2&4\end{pmatrix}\begin{pmatrix}1200\\1300\end{pmatrix}\)✓ 1
A1 \(x = 150\)✓ 1
A1 \(y = 200\)✓ 1
Part (c)
M1 \(10\times150 + 8\times200\) (ECF from (b))✓ 1
A1 €3100✓ 1

Exam tip: Check an inverse-matrix solution by substituting back into both original equations.

94IBMI3H-1-08r34-09Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy5 marks

A volcano monitoring station uses a filter whose characteristic equation is \(z^3 - 5z^2 + 17z - 13 = 0\). One solution is \(z = 2 + 3i\).

(a)

Write down a second complex solution, giving a reason.

(1)
(b)

Find a quadratic factor of the polynomial with real coefficients.

(2)
(c)

Find the remaining solution.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(2 - 3i\), because the coefficients are real so complex roots occur in conjugate pairs✓ 1
Part (b)
M1 \((z - (2+3i))(z - (2-3i))\) or sum \(4\) and product \(13\) of the roots✓ 1
A1 \(z^2 - 4z + 13\)✓ 1
Part (c)
M1 divide the cubic by \(z^2 - 4z + 13\), or use the product of roots \(13 \div (2^2+3^2)\)✓ 1
A1 \(z = 1\)✓ 1

Exam tip: For a polynomial with real coefficients, a non-real root always brings its conjugate with it.

95IBMI3H-1-08r34-14Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy5 marks

Marine growth on a section of a ship's hull has mass \(m = 4.2e^{0.18t}\) kg, where t is the time in days since the hull was cleaned.

(a)

Find the mass of marine growth after 10 days, to three significant figures.

(2)
(b)

Find the time taken for the mass to reach 50 kg, giving your answer in days to three significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(4.2e^{0.18\times10}\)✓ 1
A1 25.4 kg (25.409…)✓ 1
Part (b)
M1 \(4.2e^{0.18t} = 50\)✓ 1
M1 take natural logarithms: \(0.18t = \ln\frac{50}{4.2}\)✓ 1
A1 13.8 days (13.761…)✓ 1

Exam tip: Divide by the constant 4.2 before taking logarithms so that the exponent stands alone.

96IBMI3H-1-06r35-01Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy6 marks

The centre of a revolving theatre stage is the origin of an Argand diagram, with distances in metres. A prop is at the point represented by z = 3 + 4i. The stage then turns anticlockwise through 40° about its centre, which moves the prop to the point represented by w = zeiθ, where θ is the angle of turn in radians.

(a)

Find the modulus of z and the argument of z in radians, to 3 significant figures.

(2)
(b)

Write w in the form reiφ, giving φ to 3 significant figures.

(2)
(c)

Using unrounded values, write w in the form a + bi, giving a and b to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 |z| = 5.00 m✓ 1
A1 arg z = tan−1(4 ÷ 3) = 0.927✓ 1
Part (b)
A1 r = 5.00 (the modulus is unchanged by the turn)✓ 1
A1 φ = 0.9273 + 40π ÷ 180 = 1.63✓ 1
Part (c)
M1 a = 5 cos(1.6254…) and b = 5 sin(1.6254…)✓ 1
A1 w = −0.273 + 4.99i✓ 1

Exam tip: Multiplying by eiθ turns a point through θ without changing its distance from the origin: add the arguments, keep the modulus.

97IBMI3H-1-06r35-04Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy7 marks

A theatre sells stalls tickets at €s each and balcony tickets at €b each. At the first show, 120 stalls and 80 balcony tickets were sold, taking €5280. At the second show, 90 stalls and 150 balcony tickets were sold, taking €5580.

(a)

Write the information as a matrix equation and find the determinant of the 2 × 2 matrix.

(2)
(b)

Use the inverse of the matrix to find s and b.

(3)
(c)

A third show sells 200 stalls and 140 balcony tickets. Find the total taken.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 [[120, 80], [90, 150]] × (s, b)T = (5280, 5580)T✓ 1
A1 determinant = 120 × 150 − 80 × 90 = 10 800✓ 1
Part (b)
M1 inverse = (1 ÷ 10 800) × [[150, −80], [−90, 120]]✓ 1
M1 multiply the inverse by (5280, 5580)T✓ 1
A1 s = 32 and b = 18✓ 1
Part (c)
M1 200 × 32 + 140 × 18 (ECF from (b))✓ 1
A1 €8920✓ 1

Common error: When using an inverse, place it on the left of the right-hand side vector: matrix multiplication is not commutative.

98IBMI3H-1-06r35-14Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy6 marks

The cargo hold of a ship is a cuboid measured as 38.6 m by 17.4 m by 9.2 m.

(a)

Calculate the volume of the hold, to the nearest m3.

(2)
(b)

A planner estimates the volume by rounding each dimension to 2 significant figures before multiplying. Find the planner's estimate.

(2)
(c)

Find the percentage error in the planner's estimate, to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 38.6 × 17.4 × 9.2✓ 1
A1 6179 m3✓ 1
Part (b)
M1 39 × 17 × 9✓ 1
A1 5967 m3✓ 1
Part (c)
M1 |5967 − 6179| ÷ 6179 × 100✓ 1
A1 3.43%✓ 1

Exam tip: Percentage error is measured against the exact value, not the estimate.

99IBMI3H-1-07r36-01Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy5 marks

An archaeological team lays out a grid with a fixed datum peg at the origin. The east axis is the real axis and the north axis is the imaginary axis, with distances in metres. A marker P is at the point represented by \(z = 7 + 4i\).

(a)

Marker Q is found by rotating P through 90° anticlockwise about the datum peg. Find the complex number that represents Q.

(2)
(b)

Find the distance of P from the datum peg, correct to 3 significant figures.

(1)
(c)

Find \(\arg z\) in radians, correct to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 multiplies \(z\) by \(i\), or rotates the point \((7,4)\) to \((-4,7)\)✓ 1
A1 \(-4+7i\)✓ 1
Part (b)
A1 \(|z|=\sqrt{7^2+4^2}=\sqrt{65}=8.06\) m✓ 1
Part (c)
M1 \(\tan^{-1}\left(\frac{4}{7}\right)\) or equivalent✓ 1
A1 \(0.519\)✓ 1

Common error: Calculate the argument with the calculator in radian mode; an answer of 29.7 means the calculator was left in degrees.

100IBMI3H-1-07r36-02Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy5 marks

A cycling team buys energy bars and energy gels. Let b be the price of one bar and g the price of one gel, both in euros. An order of 6 bars and 4 gels costs €17.40. An order of 3 bars and 5 gels costs €13.65.

(a)

Write this information as a matrix equation of the form \(\mathbf{M}\begin{pmatrix} b \\ g \end{pmatrix} = \mathbf{c}\).

(2)
(b)

Write down \(\mathbf{M}^{-1}\), giving the entries as fractions.

(1)
(c)

Hence find the price of one gel.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(\mathbf{M}=\begin{pmatrix}6&4\\3&5\end{pmatrix}\)✓ 1
A1 \(\mathbf{c}=\begin{pmatrix}17.40\\13.65\end{pmatrix}\)✓ 1
Part (b)
A1 \(\mathbf{M}^{-1}=\frac{1}{18}\begin{pmatrix}5&-4\\-3&6\end{pmatrix}\)✓ 1
Part (c)
M1 attempts \(\mathbf{M}^{-1}\mathbf{c}\), for example \(g=\frac{-3(17.40)+6(13.65)}{18}\)✓ 1
A1 €1.65✓ 1

Exam tip: Check a matrix solution by substituting the prices back into both original orders: 6(1.80) + 4(1.65) = 17.40 and 3(1.80) + 5(1.65) = 13.65.

101IBMI3H-1-07r36-03Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy6 marks

In a data centre, a cable multiplies a signal by a complex gain \(g\). The input signal is represented by \(2 - i\) and the output signal by \(7 + 4i\), so that output = \(g \times\) input.

(a)

Find \(g\) in the form \(a + bi\) where \(a, b \in \mathbb{R}\).

(3)
(b)

Write down \(|g|\) correct to 3 significant figures.

(1)
(c)

Find \(\arg g\) in radians, correct to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(g=\frac{7+4i}{2-i}\)✓ 1
M1 multiplies numerator and denominator by \(2+i\)✓ 1
A1 \(g=\frac{10+15i}{5}=2+3i\)✓ 1
Part (b)
A1 \(\sqrt{13}=3.61\) (ECF from (a))✓ 1
Part (c)
M1 \(\tan^{-1}\left(\frac{3}{2}\right)\)✓ 1
A1 \(0.983\) (ECF from (a))✓ 1

Exam tip: To divide complex numbers, multiply top and bottom by the conjugate of the denominator so the denominator becomes real.

102IBMI3H-1-07r36-12Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy5 marks

A cycling club needs £10 000 in 6 years to replace its track bikes. It invests a lump sum in an account paying a nominal annual interest rate of 3.0%, compounded quarterly.

(a)

Find the amount that must be invested now. Give your answer to the nearest pound.

(3)
(b)

Find the effective annual interest rate, as a percentage correct to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 uses \(10000=P\left(1+\frac{3.0}{400}\right)^{24}\) or a financial app with \(N=24\), \(I\%=3\), \(FV=10000\), \(C/Y=4\)✓ 1
A1 \(P=\frac{10000}{1.0075^{24}}\)✓ 1
A1 £8358✓ 1
Part (b)
M1 \(1.0075^4-1\)✓ 1
A1 \(3.03\%\)✓ 1

Common error: Divide the nominal rate by the number of compounding periods per year (4), and raise to the power of the total number of periods (24), not the number of years.

103IBMI3H-1-06r37-01Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy7 marks

Calculator

On an architect's site plan, two displacement vectors are represented by the complex numbers a = 4 + 3i and b = 2 − i (units: metres).

(a)

Find ab, giving your answer in the form x + yi.

(2)
(b)

Find a/b, giving your answer in the form x + yi.

(3)
(c)

Find |a/b|, to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 expands (4 + 3i)(2 − i) using i2 = −1✓ 1
A1 11 + 2i✓ 1
Part (b)
M1 multiplies numerator and denominator by 2 + i✓ 1
A1 numerator 5 + 10i and denominator 5✓ 1
A1 1 + 2i✓ 1
Part (c)
M1 √(12 + 22) (ECF from (b))✓ 1
A1 2.24✓ 1

Exam tip: To divide complex numbers, multiply top and bottom by the conjugate of the denominator so the denominator becomes real.

104IBMI3H-1-06r37-05Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy6 marks

Calculator

A freight depot loads two types of crate. A type A crate has mass x kg and a type B crate has mass y kg. Load 1 consists of 3 crates of type A and 2 of type B, with total mass 1450 kg. Load 2 consists of 4 crates of type A and 5 of type B, with total mass 2750 kg.

(a)

Write the information as a matrix equation of the form M(x, y)T = c, and find the determinant of M.

(2)
(b)

Write down M−1.

(2)
(c)

Hence find the mass of one crate of each type.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 M = [[3, 2], [4, 5]] and c = (1450, 2750)T✓ 1
A1 det M = 15 − 8 = 7✓ 1
Part (b)
M1 swaps leading diagonal, changes signs of the other entries, divides by 7✓ 1
A1 M−1 = (1/7)[[5, −2], [−4, 3]]✓ 1
Part (c)
M1 (x, y)T = M−1c (ECF from (b))✓ 1
A1 type A 250 kg, type B 350 kg✓ 1

Exam tip: Order matters in matrix multiplication: solve Mx = c by x = M⁻¹c, with the inverse on the left.

105IBMI3H-1-06r37-14Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy5 marks

Calculator

Light intensity under water decreases with depth x metres according to I = I0e−kx, where I0 is the intensity at the surface. At a depth of 10 m, the intensity is 25% of the surface intensity.

(a)

Find the value of k, to 3 significant figures.

(2)
(b)

Marine biologists need to know the depth at which the intensity falls to 1% of the surface intensity. Using the unrounded value of k, find this depth to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 e−10k = 0.25, so k = ln 4/10✓ 1
A1 0.139✓ 1
Part (b)
M1 e−kx = 0.01✓ 1
M1 x = ln 100/k = 10 ln 100/ln 4✓ 1
A1 33.2 m✓ 1

Common error: Rounding k to 0.139 before the second calculation gives 33.1 m; keep the unrounded value to protect the third significant figure.

106IBMI3H-1-06r39-01Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy5 marks

Two turbines in a wind farm produce alternating voltages, in volts, represented by the complex numbers \(z_1=4+3i\) and \(z_2=2-5i\).

(a)

Find \(z_1+z_2\).

(1)
(b)

Find \(z_1z_2\), giving your answer in the form \(a+bi\).

(2)
(c)

Find \(|z_1z_2|\) to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(6-2i\)✓ 1
Part (b)
M1 expands \((4+3i)(2-5i)\) using \(i^2=-1\)✓ 1
A1 \(23-14i\)✓ 1
Part (c)
M1 \(\sqrt{23^2+(-14)^2}\) (or \(|z_1||z_2|=5\sqrt{29}\))✓ 1
A1 \(26.9\) V✓ 1

Exam tip: The modulus of a product equals the product of the moduli, which gives a quick check: \(5\times\sqrt{29}\approx26.9\).

107IBMI3H-1-06r39-02Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy4 marks

The current phasor in a wind farm cable is \(z=-5+5\sqrt3\,i\) amps.

(a)

Find \(|z|\).

(1)
(b)

Find \(\arg z\) in radians, giving your answer to 3 significant figures.

(2)
(c)

Hence write \(z\) in the form \(re^{i\theta}\).

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(\sqrt{25+75}=10\)✓ 1
Part (b)
M1 recognises \(z\) lies in the second quadrant (or uses \(\pi-\arctan\sqrt3\))✓ 1
A1 \(\dfrac{2\pi}{3}=2.09\) rad✓ 1
Part (c)
A1 \(10e^{2\pi i/3}\) (ECF from (a) and (b))✓ 1

Common error: Using \(\arctan(5\sqrt3/-5)\) on a calculator gives a fourth-quadrant angle; sketch the point to see that the argument is in the second quadrant.

108IBMI3H-1-06r39-11Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy6 marks

A community wind co-operative invests €250 000 at a nominal annual interest rate of 4.2%, compounded monthly.

(a)

Find the value of the investment after 8 years, to the nearest euro.

(3)
(b)

Find the time, in years to 3 significant figures, taken for the investment to reach €400 000.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 uses the compound interest formula with monthly rate \(\dfrac{0.042}{12}\)✓ 1
M1 uses \(n=12\times8=96\) compounding periods✓ 1
A1 €349 630✓ 1
Part (b)
M1 sets up \(250\,000(1.0035)^n=400\,000\) and solves (graphically or with logarithms)✓ 1
A1 \(n=134.5\ldots\) months✓ 1
A1 \(11.2\) years✓ 1

Exam tip: Check that the compounding period matches the time unit: 8 years of monthly compounding means 96 periods, with a monthly rate of 0.042 ÷ 12.

109IBMI3H-1-06r39-12Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy7 marks

A swimmer builds up training distance over several weeks. In week 1 she swims 1.2 km, and the distance swum increases by 7% each week, so the weekly distances form a geometric sequence.

(a)

Find the distance she swims in week 12, to 3 significant figures.

(2)
(b)

Find the first week in which she swims more than 4 km.

(3)
(c)

Find the total distance she swims in the first 20 weeks, to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(u_{12}=1.2\times1.07^{11}\)✓ 1
A1 \(2.53\) km✓ 1
Part (b)
M1 solves \(1.2\times1.07^{n-1}>4\) (using logarithms or a table)✓ 1
A1 \(n-1>17.8\ldots\)✓ 1
A1 week 19✓ 1
Part (c)
M1 \(S_{20}=\dfrac{1.2(1.07^{20}-1)}{0.07}\)✓ 1
A1 \(49.2\) km✓ 1

Common error: The common ratio for a 7% increase is 1.07, not 0.07, and the n-th term uses the power \(n-1\).

110IBMI3H-1-04r40-01Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy6 marks

Two hydrophones in a marine reserve record signals that are modelled by the complex numbers z = 4 − 3i and w = 1 + 2i.

(a)

Find, in the form a + bi, (i) z + w, (ii) zw, (iii) z ÷ w.

(4)
(b)

Find the modulus of z, and the argument of z in radians correct to 3 significant figures, with −π < arg ≤ π.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 z + w = 5 − i✓ 1
A1 zw = 10 + 5i✓ 1
M1 multiply numerator and denominator by the conjugate 1 − 2i, or use technology✓ 1
A1 z ÷ w = −0.4 − 2.2i (accept −2/5 − 11i/5)✓ 1
Part (b)
A1 |z| = 5✓ 1
A1 arg z = −0.644 (= arctan(−3/4), fourth quadrant)✓ 1

Common error: The argument of 4 − 3i lies in the fourth quadrant, so it is negative; check the sign against the position of the point.

111IBMI3H-1-04r40-02Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy7 marks

A roof joint carries a force that is represented in an Argand diagram by the complex number z = −6 + 8i, where the real part is the horizontal component and the imaginary part the vertical component, both in kN.

(a)

Find |z|.

(1)
(b)

Find the argument of z in radians, correct to 3 significant figures, with −π < arg ≤ π.

(2)
(c)

Write z3 in the form reiθ, with r > 0. Give θ to 3 significant figures; use the unrounded argument of z.

(2)
(d)

Write z3 in the form a + bi.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 10 kN✓ 1
Part (b)
M1 π − arctan(8/6) (second quadrant; accept use of technology)✓ 1
A1 2.21✓ 1
Part (c)
M1 r = 103 and θ = 3 × 2.2143…✓ 1
A1 1000e6.64i (accept 1000e0.360i)✓ 1
Part (d)
M1 expand (−6 + 8i)3 or convert 1000ei × 6.6429… using unrounded values✓ 1
A1 936 + 352i✓ 1

Exam tip: Cubing a complex number cubes the modulus and triples the argument; an argument outside (−π, π] can be reduced by subtracting 2π.

112IBMI3H-1-04r40-06Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy6 marks

A logistics firm hires x trucks of type X and y trucks of type Y to move a delivery. The numbers of pallets and the masses carried give the equations
2x + ky = 18
3x + 6y = 30
where k is the number of pallets one type Y truck can carry.

(a)

Find the value of k for which the system does not have a unique solution.

(2)
(b)

For this value of k, show that the system has no solution.

(2)
(c)

Given instead that k = 3, find the number of trucks of each type.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 determinant 2 × 6 − 3k = 0✓ 1
A1 k = 4✓ 1
Part (b)
M1 divide to obtain x + 2y = 9 from the first equation and x + 2y = 10 from the second✓ 1
R1 the two equations contradict each other, so there is no solution (parallel lines)✓ 1
Part (c)
M1 solve the system, e.g. by A−1B with A = [[2, 3], [3, 6]], or by technology✓ 1
A1 6 trucks of type X and 2 trucks of type Y✓ 1

Common error: A zero determinant does not always mean no solution: it means no unique solution, so check whether the equations are consistent (infinitely many solutions) or contradictory (none).

113IBMI3H-1-04r40-14Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy5 marks

A research vessel uses an echo sounder to measure the depth of the sea. A pulse travels to the seabed and back in 0.84 s. The depth d m is given by d = vt ÷ 2, where v m s−1 is the speed of sound in the water and t s is the time for the round trip. The crew assumes v = 1.50 × 103.

(a)

Calculate the depth that the crew finds.

(2)
(b)

The true speed of sound in the water is 1482 m s−1. Calculate the percentage error in the depth found in (a), correct to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 1500 × 0.84 ÷ 2✓ 1
A1 630 m✓ 1
Part (b)
M1 true depth = 1482 × 0.84 ÷ 2 = 622.44 m✓ 1
M1 |630 − 622.44| ÷ 622.44 × 100✓ 1
A1 1.21%✓ 1

Common error: Percentage error is divided by the true value (here the depth from the true speed), not by the value the crew found.

114IBMI3H-1-06r40-01Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy6 marks

A sonar receiver represents two echo signals by the complex numbers \(z = 3 + 4i\) and \(w = 1 - 2i\), where each number gives the amplitude and phase of a signal in a marine-biology survey.

(a)

Find \(zw\) in the form \(a + bi\).

(2)
(b)

Find \(\dfrac{z}{w}\) in the form \(a + bi\).

(2)
(c)

Find the exact value of \(\left|\dfrac{z}{w}\right|\).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 expand \((3+4i)(1-2i) = 3 - 6i + 4i - 8i^2\)✓ 1
A1 \(zw = 11 - 2i\)✓ 1
Part (b)
M1 multiply numerator and denominator by the conjugate \(1+2i\)✓ 1
A1 \(\dfrac{z}{w} = \dfrac{-5+10i}{5} = -1 + 2i\)✓ 1
Part (c)
M1 \(\sqrt{(-1)^2 + 2^2}\) ECF from (b)✓ 1
A1 \(\sqrt{5}\)✓ 1

Exam tip: To divide complex numbers, multiply top and bottom by the conjugate of the denominator so the denominator becomes real.

115IBMI3H-1-06r40-02Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy5 marks

On the plan of a circular atrium, a decorative panel is located at \(z = -2 + 2\sqrt{3}\,i\), where distances are in metres from the centre and the real axis points east.

(a)

Find the modulus of z.

(1)
(b)

Find the argument of z, in radians, giving an exact answer.

(2)
(c)

The panel is moved by multiplying z by \(e^{i\pi/3}\). Find the new position as a complex number in Cartesian form.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(|z| = \sqrt{4 + 12} = 4\) m✓ 1
Part (b)
M1 attempt to use \(\tan^{-1}\left(\frac{2\sqrt3}{-2}\right)\) with z in the second quadrant✓ 1
A1 \(\arg z = \frac{2\pi}{3}\)✓ 1
Part (c)
M1 \(4e^{i(2\pi/3 + \pi/3)} = 4e^{i\pi}\)✓ 1
A1 \(-4\) (that is, \(-4 + 0i\), 4 m west of the centre)✓ 1

Common error: Sketch the point first: \(\tan^{-1}\) alone on a calculator gives an angle in the wrong quadrant when the real part is negative.

116IBMI3H-1-06r40-10Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy6 marks

A delivery company borrows €45 000 to buy vans. The loan has a nominal annual interest rate of 6%, compounded monthly, and is repaid by equal monthly payments at the end of each month for 5 years.

(a)

Find the monthly payment, to the nearest cent.

(3)
(b)

Using the rounded payment from part (a), find the total amount repaid.

(2)
(c)

Find the total interest paid.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 use of a finance solver with \(N=60\), \(I=6\), \(PV=45000\), \(P/Y=12\), \(C/Y=12\)✓ 1
M1 solve for PMT✓ 1
A1 €869.98✓ 1
Part (b)
M1 \(869.98 \times 60\)✓ 1
A1 €52 198.80✓ 1
Part (c)
A1 \(52198.80 - 45000 = €7198.80\), ECF from (b)✓ 1

Exam tip: Enter money paid out as negative and money received as positive in the finance solver, or the answer will carry the wrong sign.

117IBMI3H-1-06r40-13Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy5 marks

A rectangular concrete floor slab is measured as 12.4 m by 8.6 m, each length given to the nearest 0.1 m. The estimated area is the product of these two measurements.

(a)

Find the lower and upper bounds for the area of the slab, to 2 decimal places.

(3)
(b)

The true dimensions are later found to be 12.38 m and 8.57 m. Find the percentage error in the estimated area, to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 use of 12.35, 8.55 and 12.45, 8.65✓ 1
A1 lower bound 105.59 m2✓ 1
A1 upper bound 107.69 m2✓ 1
Part (b)
M1 \(\dfrac{|106.64 - 12.38 \times 8.57|}{12.38 \times 8.57} \times 100\)✓ 1
A1 0.512%✓ 1

Common error: Percentage error divides by the exact value, not by the estimate.

118IBMI3H-1-07r40-06Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy5 marks

In a model of a stage-lighting circuit, impedance, current and voltage are treated as complex numbers. A component has impedance \(Z=3+4\mathrm{i}\) Ω and carries a current \(I=1-2\mathrm{i}\) A. The voltage across it is \(V=ZI\).

(a)

Find \(V\) in the form \(a+b\mathrm{i}\).

(2)
(b)

A second component has the same voltage \(V\) but carries a current of \(2+\mathrm{i}\) A. Find its impedance \(Z_2=\frac{V}{2+\mathrm{i}}\) in the form \(a+b\mathrm{i}\).

(2)
(c)

Find \(|Z_2|\), in ohms.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 expand \((3+4\mathrm{i})(1-2\mathrm{i})\) with \(\mathrm{i}^2=-1\)✓ 1
A1 \(V=11-2\mathrm{i}\)✓ 1
Part (b)
M1 multiply numerator and denominator by the conjugate \(2-\mathrm{i}\)✓ 1
A1 \(Z_2=4-3\mathrm{i}\)✓ 1
Part (c)
A1 \(|Z_2|=\sqrt{4^2+3^2}=5\) Ω (ECF from (b))✓ 1

Exam tip: To divide complex numbers, multiply top and bottom by the conjugate of the denominator so the denominator becomes real.

119IBMI3H-1-07r40-07Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy6 marks

A ship leaves port at the origin of a map. Its position is modelled by the complex number \(z=-6+8\mathrm{i}\) km, where the real axis points east and the imaginary axis points north.

(a)

Find the modulus and the argument of \(z\), giving the argument in radians to 3 significant figures.

(3)
(b)

Write \(z\) in the form \(re^{\mathrm{i}\theta}\).

(1)
(c)

Find the ship's bearing from the port, in degrees, to the nearest degree.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(|z|=10\) km✓ 1
M1 argument in the second quadrant, e.g. \(\pi-\arctan\frac{8}{6}\) or calculator Arg✓ 1
A1 \(\arg z=2.21\)✓ 1
Part (b)
A1 \(z=10e^{2.21\mathrm{i}}\) (ECF from (a))✓ 1
Part (c)
M1 angle from north \(=\arctan\frac{6}{8}=36.87^\circ\) west of north, or \(90^\circ-126.87^\circ\)✓ 1
A1 \(323^\circ\)✓ 1

Common error: The calculator gives \(\arctan(-\tfrac{4}{3})\) in the fourth quadrant; check which quadrant the point is in before stating the argument.

120IBMI3H-1-07r40-11Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy4 marks

Two types of crate are loaded onto a small barge. A type A crate has mass 3 tonnes and volume 4 m3. A type B crate has mass 1 tonne and volume 2 m3. The barge carries \(x\) type A crates and \(y\) type B crates, with total mass 26 tonnes and total volume 40 m3. This is written as \(\mathbf{M}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}26\\40\end{pmatrix}\), where \(\mathbf{M}=\begin{pmatrix}3&1\\4&2\end{pmatrix}\).

(a)

Find \(\mathbf{M}^{-1}\).

(2)
(b)

Hence find the number of type B crates on the barge.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(\det\mathbf{M}=3\times2-1\times4=2\)✓ 1
A1 \(\mathbf{M}^{-1}=\frac{1}{2}\begin{pmatrix}2&-1\\-4&3\end{pmatrix}\)✓ 1
Part (b)
M1 \(\begin{pmatrix}x\\y\end{pmatrix}=\mathbf{M}^{-1}\begin{pmatrix}26\\40\end{pmatrix}\)✓ 1
A1 \(y=8\) (and \(x=6\))✓ 1

Common error: Multiplying in the wrong order: the solution is \(\mathbf{M}^{-1}\mathbf{b}\), not \(\mathbf{b}\mathbf{M}^{-1}\).

121IBMI3H-1-04r41-01Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy5 marks

Calculator

A water pump motor has impedance Z = 12 + 16i ohms (Ω) in an alternating current circuit. The supply has rms voltage 240 V. The magnitude of the impedance is |Z| and its argument is the phase angle.

(a)

Find |Z|.

(2)
(b)

Find the phase angle, arg Z, in radians, correct to 3 significant figures.

(2)
(c)

The rms current is the rms voltage divided by |Z|. Find the rms current.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 √(122 + 162)✓ 1
A1 20 Ω✓ 1
Part (b)
M1 arctan(16/12)✓ 1
A1 0.927 radians✓ 1
Part (c)
A1 240 ÷ 20 = 12 A✓ 1

Exam tip: Set your calculator to radians before finding an argument unless the question asks for degrees.

122IBMI3H-1-04r41-05Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy5 marks

Calculator

A pharmacist mixes a 10% solution and a 25% stock solution of the same drug to make 150 mL of a 16% solution. Let x mL be the volume of the 10% solution and y mL the volume of the 25% stock solution.

(a)

Write the two conditions as a matrix equation of the form A(x, y)T = b, where the first row of A is for the total volume.

(1)
(b)

Find \(\mathbf{A}^{-1}\).

(2)
(c)

Hence find the volume of the 25% stock solution needed.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(\begin{pmatrix}1&1\\0.10&0.25\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}150\\24\end{pmatrix}\)✓ 1
Part (b)
M1 determinant 0.25 − 0.10 = 0.15 used in the inverse✓ 1
A1 inverse with entries 5/3, −20/3, −2/3, 20/3 (accept decimals 1.67, −6.67, −0.667, 6.67 or the matrix written with factor 1/0.15)✓ 1
Part (c)
M1 \(\mathbf{A}^{-1}\begin{pmatrix}150\\24\end{pmatrix}\) (or solves the system with technology)✓ 1
A1 y = 60 mL (and x = 90 mL)✓ 1

Exam tip: Write the amount of pure drug (16% of 150 mL = 24 mL) as the second equation: the matrix equation then follows directly.

123IBMI3H-1-04r41-09Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy5 marks

Calculator

A game encodes a pair of player IDs (p, q) as the pair (3p + 2q, 4p + 3q), which can be written as multiplication by the matrix A = \(\begin{pmatrix}3&2\\4&3\end{pmatrix}\).

(a)

Find det \(\mathbf{A}\).

(1)
(b)

Write down \(\mathbf{A}^{-1}\).

(2)
(c)

Decode the pairs (13, 18) and (29, 41) to find the original ID pairs.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 9 − 8 = 1✓ 1
Part (b)
M1 swaps the leading-diagonal entries and changes the signs of the others✓ 1
A1 \(\begin{pmatrix}3&-2\\-4&3\end{pmatrix}\)✓ 1
Part (c)
M1 multiplies each coded pair by A−1✓ 1
A1 (3, 2) and (5, 7)✓ 1

Common error: To decode, multiply by the inverse matrix on the left of the coded column vector; multiplying in the other order gives a different result.

124IBMI3H-1-04r41-13Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy5 marks

Calculator

A patient is prescribed a course of a drug. The dose on day 1 is 400 mg and each day the dose is 10% smaller than on the previous day.

(a)

Find the dose on day 8. Give your answer to the nearest mg.

(2)
(b)

Find the total amount of the drug taken over the first 8 days. Give your answer correct to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 400 × 0.97✓ 1
A1 191 mg✓ 1
Part (b)
M1 400(1 − 0.98)/(1 − 0.9)✓ 1
A1 2278.1…✓ 1
A1 2280 mg✓ 1

Common error: A 10% decrease is a common ratio of 0.9, not 0.1.

125IBMI3H-1-04r43-01Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy5 marks

In a game, a character's position on a map is represented by the complex number z = 6 + 2i, with distances in metres from the origin. A spell rotates the character about the origin through an angle of π⁄3 radians anticlockwise.

(a)

Find the modulus and argument of z, giving each to 3 significant figures.

(2)
(b)

Find the position of the character after the rotation, in the form a + bi with a and b given to 3 significant figures.

(2)
(c)

Find the angle, in radians to 3 significant figures, between the line from the origin to the new position and the positive real axis.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 |z| = √40 = 6.32✓ 1
A1 arg z = arctan(1⁄3) = 0.322 radians✓ 1
Part (b)
M1 multiplies z by cos(π⁄3) + i sin(π⁄3) (or adds π⁄3 to the argument and keeps the modulus)✓ 1
A1 1.27 + 6.20i✓ 1
Part (c)
A1 0.322 + 1.05 = 1.37 radians (arctan(1⁄3) + π⁄3)✓ 1

Exam tip: Multiplying by a complex number of modulus 1 is a pure rotation: the modulus stays the same and the arguments add.

126IBMI3H-1-04r43-06Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy5 marks

A water company runs two pumping stations, P and Q, for x and y hours respectively each day. District A needs 34 ML of water per day: station P supplies 3 ML each hour it runs and station Q supplies 2 ML each hour it runs. District B needs 62 ML per day: station P supplies 5 ML each hour and station Q supplies 4 ML each hour.

(a)

Write the information as a matrix equation of the form Av = b, where v is the column vector of x and y.

(1)
(b)

Find the determinant and the inverse of \(\begin{pmatrix}3&2\\5&4\end{pmatrix}\).

(2)
(c)

Hence find the number of hours each station runs, and the total number of pumping hours per day.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(\begin{pmatrix}3&2\\5&4\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}34\\62\end{pmatrix}\)✓ 1
Part (b)
A1 determinant = 3 × 4 − 2 × 5 = 2✓ 1
A1 inverse = \(\frac{1}{2}\begin{pmatrix}4&-2\\-5&3\end{pmatrix}\)✓ 1
Part (c)
M1 multiplies the inverse by the vector (34, 62), giving x = 6 and y = 8✓ 1
A1 total = 14 hours✓ 1

Exam tip: Multiply by the inverse on the left of both sides: v = A⁻¹b; multiplying on the right is not defined for these shapes.

127IBMI3H-1-04r43-11Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy5 marks

A patient takes a 250 mg dose of a drug every 6 hours. Just before each new dose, 35% of the drug from the previous amount remains in the body.

(a)

Show that the amount of the drug in the body immediately after the second dose is 337.5 mg.

(1)
(b)

Find the amount of the drug in the body immediately after the third dose. Give your answer to 3 significant figures.

(2)
(c)

Find the amount in the body immediately after a dose in the long term, to the nearest mg, and hence show that the amount never reaches 400 mg.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 250 × 0.35 + 250 = 87.5 + 250 = 337.5✓ 1
Part (b)
M1 250 + 250 × 0.35 + 250 × 0.352 (or 250 + 0.35 × 337.5)✓ 1
A1 368 mg✓ 1
Part (c)
M1 sum to infinity 250⁄1 − 0.35✓ 1
A1 385 mg (384.6…); the amounts increase towards this limit, which is below 400, so 400 is never reached✓ 1

Exam tip: The sum to infinity of a geometric series only exists when the common ratio satisfies |r| < 1; here r = 0.35, so the series converges.

128IBMI3H-1-04r43-12Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy5 marks

In an online game, completing level 1 needs 100 experience points (XP), and each later level needs 50 XP more than the previous level.

(a)

Find the XP needed to complete level 20.

(1)
(b)

Find the total XP needed to complete the first 20 levels.

(2)
(c)

A player has earned a total of 30 000 XP. Find the greatest number of complete levels this player has finished.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 100 + 19 × 50 = 1050 XP✓ 1
Part (b)
M1 uses the arithmetic series sum 20⁄2(100 + 1050)✓ 1
A1 11 500 XP✓ 1
Part (c)
M1 solves n⁄2(200 + 50(n − 1)) ≤ 30 000, i.e. 25n2 + 75n ≤ 30 000 (n ≤ 33.17…)✓ 1
A1 33 levels (the 34th level would need a total of 31 450 XP)✓ 1

Common error: A solution of 33.17… to an inequality about complete levels must be rounded down, not to the nearest whole number.

129IBMI3H-1-05r43-01Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy7 marks

A robot arm has its tip at the point represented by the complex number z = 2.4 + 1.8i, where the real and imaginary axes are measured in metres from the base of the arm.

(a)

Find the modulus and the argument of z. Give the argument in radians to 3 significant figures.

(3)
(b)

The arm is rotated 90° anticlockwise about its base. The new tip position is represented by w. Write w in the form a + bi.

(2)
(c)

Find arg w in radians to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to use modulus = √(2.42 + 1.82)✓ 1
A1 |z| = 3.00 m✓ 1
A1 arg z = tan−1(1.8/2.4) = 0.644 rad (0.6435…)✓ 1
Part (b)
M1 attempt to multiply z by i (or add π/2 to the argument)✓ 1
A1 w = −1.8 + 2.4i✓ 1
Part (c)
M1 attempt to add π/2 to the argument of z (or to use the second-quadrant angle for w)✓ 1
A1 2.21 rad (2.2143…); ECF from (a)✓ 1

Common error: Rotating by 90° anticlockwise means multiplying by i, not by −i; check the answer lies in the quadrant you expect.

130IBMI3H-1-05r43-07Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy5 marks

A pattern generator for printed fabric moves the point represented by z to the point represented by w = (1 + i)z + (2 − i).

(a)

Find w when z = 3 + 2i.

(2)
(b)

Find |w|.

(1)
(c)

The generator is applied again to the point 3 + 4i. Find the argument of the new point, in radians to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to expand (1 + i)(3 + 2i), using i2 = −1✓ 1
A1 w = 3 + 4i✓ 1
Part (b)
A1 5✓ 1
Part (c)
M1 attempt to apply the map: (1 + i)(3 + 4i) + 2 − i = 1 + 6i✓ 1
A1 arg = tan−16 = 1.41 rad✓ 1

Common error: When multiplying complex numbers, replace i² by −1 and collect the real and imaginary parts separately.

131IBMI3H-1-05r43-08Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy8 marks

The fishing authority sets a catch limit for a sea bass fishery. The limit in year 1 is 1200 tonnes, and each later year the limit is 5% lower than in the previous year.

(a)

Find the catch limit in year 6, to the nearest tonne.

(2)
(b)

Find the first year in which the catch limit is below 500 tonnes.

(3)
(c)

Find the total of the catch limits over the first 10 years, to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to use 1200 × 0.955✓ 1
A1 929 tonnes✓ 1
Part (b)
M1 set up 1200 × 0.95n−1 < 500✓ 1
M1 attempt to solve using logarithms or a table, giving n − 1 > 17.07…✓ 1
A1 year 19✓ 1
Part (c)
M1 use of the geometric series sum formula✓ 1
A1 1200(1 − 0.9510)/(1 − 0.95)✓ 1
A1 9630 tonnes✓ 1

Common error: A 5% decrease means multiplying by 0.95; the 6th term uses the power 5, not 6.

132IBMI3H-1-05r43-14Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy6 marks

The population of a fish species in a protected reserve is modelled by P = a × bt, where t is the number of years since monitoring began. The population was 5100 when t = 1 and 3132 when t = 4.

(a)

Find b and a, each to 3 significant figures.

(3)
(b)

Interpret the value of b in context.

(1)
(c)

Using P = 6000 × 0.85t, find the value of t when the population is 1500.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to divide: b3 = 3132/5100✓ 1
A1 b = 0.850✓ 1
A1 a = 6000 (6.00 × 103)✓ 1
Part (b)
A1 the population decreases by 15% each year✓ 1
Part (c)
M1 attempt to solve 6000 × 0.85t = 1500 using logarithms✓ 1
A1 t = 8.53✓ 1

Exam tip: To find an unknown exponent, take logarithms of both sides after isolating the power.

133IBMI3H-1-07r44-01Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy5 marks

Two hydrophones record the same whale call. Each signal is represented by a complex number whose modulus is the amplitude (in mPa) and whose argument is the phase (in radians). The signals are z1 = 4 + 3i and z2 = 2 − 5i. The combined signal is w = z1 + z2.

(a)

Write down w in the form a + bi.

(1)
(b)

Find the amplitude of the combined signal, giving your answer to 3 significant figures.

(2)
(c)

Find the phase of the combined signal in radians, giving your answer to 3 significant figures, with the argument taken in the interval −π < θ ≤ π.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 w = 6 − 2i✓ 1
Part (b)
M1 attempt to find |w| using √(6² + (−2)²)✓ 1
A1 6.32 mPa (√40)✓ 1
Part (c)
M1 attempt to use arctan(2/6) or tan−1(−2/6), with the correct quadrant identified (fourth quadrant, Re > 0)✓ 1
A1 −0.322 radians (ECF from (a))✓ 1

Exam tip: Check the quadrant before quoting an argument: a positive real part and a negative imaginary part give a negative angle between −π/2 and 0.

134IBMI3H-1-07r44-05Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy5 marks

The behaviour of a sensor on a research buoy is modelled by the quadratic equation z2 + bz + c = 0, where b and c are real numbers. One of the solutions is z = 2 − 3i.

(a)

Write down the other solution and give a reason for your answer.

(2)
(b)

Find the value of b and the value of c.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
R1 the coefficients are real, so non-real solutions occur as complex conjugate pairs✓ 1
A1 2 + 3i✓ 1
Part (b)
M1 attempt to use sum of roots = −b or product of roots = c (or expand (z − (2 − 3i))(z − (2 + 3i)))✓ 1
A1 b = −4✓ 1
A1 c = 13✓ 1

Exam tip: For a quadratic with real coefficients, the product of the two roots is c and their sum is −b; for conjugate roots the product is the square of the modulus.

135IBMI3H-1-07r44-13Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy6 marks

The light intensity I at a depth of z metres below the surface of the sea is modelled by I = I0e−kz, where I0 is the intensity at the surface and k is a positive constant. At a depth of 5 m, the intensity is 40% of the surface intensity.

(a)

Find the value of k, giving your answer to 3 significant figures.

(2)
(b)

Photosynthesis is not possible where the intensity is less than 1% of the surface intensity. Find the depth, in metres, at which the intensity falls to 1% of the surface intensity. Use the unrounded value of k and give your answer to 3 significant figures.

(2)
(c)

Find the percentage of the surface intensity that remains at a depth of 10 m.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 0.4 = e−5k or equivalent, then taking natural logarithms✓ 1
A1 k = 0.183✓ 1
Part (b)
M1 0.01 = e−kz and attempt to solve, e.g. z = ln 100 / k✓ 1
A1 25.1 m✓ 1
Part (c)
R1 10 m is two lots of 5 m, so intensity is 0.4² of the surface value (or 100e−10k)✓ 1
A1 16%✓ 1

Exam tip: Keep unrounded values in your calculator memory between parts; rounding k to 3 significant figures before using it again can change the third significant figure of the answer.

136IBMI3H-1-07r44-14Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy6 marks

A cylindrical storage tank at a port has a measured diameter of 2.4 m and a measured height of 3.5 m, each given to the nearest 0.1 m.

(a)

Write down the lower and upper bounds for the diameter and for the height of the tank.

(2)
(b)

Calculate the upper bound for the volume of the tank, in m³, to 3 significant figures.

(2)
(c)

A site manager estimates the volume of the tank as 16 m³. Taking the volume calculated from the measured values (2.4 m and 3.5 m) as the exact value, find the percentage error in the manager's estimate, to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 diameter: 2.35 m ≤ d < 2.45 m✓ 1
A1 height: 3.45 m ≤ h < 3.55 m✓ 1
Part (b)
M1 attempt to use V = π(d/2)²h with both upper bounds, i.e. π × 1.225² × 3.55✓ 1
A1 16.7 m³✓ 1
Part (c)
M1 attempt to use |16 − V| / V × 100 where V = π × 1.2² × 3.5 = 15.83…✓ 1
A1 1.05%✓ 1

Common error: The upper bound of a volume uses the upper bound of every measurement; the radius is half the diameter, so halve the bound, not the measured value.

137IBMI3H-1-04r46-01Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy5 marks

Two hydrophone signals are represented by the complex numbers \(z_1 = 4 + 3i\) and \(z_2 = 1 - 2i\).

(a)

Find \(z_1 z_2\), giving your answer in the form \(a + bi\).

(2)
(b)

Find \(\dfrac{z_1}{z_2}\), giving your answer in the form \(a + bi\).

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 expands \((4+3i)(1-2i)\) using \(i^2 = -1\)✓ 1
A1 \(10 - 5i\)✓ 1
Part (b)
M1 multiplies numerator and denominator by \(1 + 2i\) (the conjugate of \(z_2\))✓ 1
A1 numerator \(-2 + 11i\) and denominator 5✓ 1
A1 \(-0.4 + 2.2i\)✓ 1

Common error: Remember that \(i^2 = -1\); a sign slip when expanding the \(i \times i\) term is the usual cause of a wrong real part.

138IBMI3H-1-04r46-02Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy4 marks

A marine research buoy is located at the point represented by the complex number \(z = 5 - 12i\) on a map, where one unit represents 1 km and the origin is the harbour.

(a)

Find \(|z|\).

(1)
(b)

Find \(\arg z\) in radians, with \(-\pi < \arg z \le \pi\). Give your answer to 3 significant figures.

(2)
(c)

Hence write \(z\) in the form \(re^{i\theta}\).

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 13✓ 1
Part (b)
M1 \(\tan^{-1}(12/5)\) with the point in the fourth quadrant✓ 1
A1 −1.18✓ 1
Part (c)
A1 \(13e^{-1.18i}\)✓ 1

Exam tip: Sketch the point first; knowing the quadrant tells you whether the principal argument is positive or negative.

139IBMI3H-1-04r46-04Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy6 marks

A courier firm uses small vans and large vans. A small van carries \(x\) pallets per trip and a large van carries \(y\) pallets per trip. On Monday, 3 small vans and 2 large vans carried a total of 46 pallets. On Tuesday, 5 small vans and 4 large vans carried a total of 82 pallets. All vans were full on each trip.

(a)

Write the information as a matrix equation \(A\mathbf{v} = \mathbf{b}\) and find \(\det A\).

(2)
(b)

Find \(A^{-1}\).

(2)
(c)

Hence find the number of pallets a large van carries.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(\begin{pmatrix}3&2\\5&4\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}46\\82\end{pmatrix}\)✓ 1
A1 \(\det A = 12 - 10 = 2\)✓ 1
Part (b)
M1 swaps the leading diagonal entries and changes signs of the others, dividing by the determinant✓ 1
A1 \(A^{-1} = \dfrac12\begin{pmatrix}4&-2\\-5&3\end{pmatrix}\)✓ 1
Part (c)
M1 \(\mathbf{v} = A^{-1}\mathbf{b}\)✓ 1
A1 \(x = 10\) and \(y = 8\), so a large van carries 8 pallets✓ 1

Exam tip: Check your solution by substituting back into the original equations: \(3(10) + 2(8) = 46\).

140IBMI3H-1-04r46-12Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy5 marks

The mass \(B\) grams of an algal bloom in a marine tank is modelled by \(B = 12e^{0.18t}\), where \(t\) is the time in days after the bloom is first measured.

(a)

Find the mass of the bloom after 5 days, to 3 significant figures.

(2)
(b)

Find the time taken for the mass to reach 100 g, giving your answer in days to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(12e^{0.18 \times 5}\)✓ 1
A1 29.5 g✓ 1
Part (b)
M1 sets \(12e^{0.18t} = 100\)✓ 1
M1 takes natural logs: \(t = \ln(100/12) \div 0.18\)✓ 1
A1 11.8 days✓ 1

Exam tip: Isolate the exponential term first, then take \(\ln\) of both sides; do not take logs before dividing by 12.

141IBMI3H-1-05r46-01Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy5 marks

Calculator

An archaeologist walks two straight legs across a dig site. Taking east as the real axis and north as the imaginary axis, with distances in metres, the legs are the displacements \(z_1 = 14 + 8i\) and \(z_2 = -6 + 11i\).

(a)

Find the total displacement \(w = z_1 + z_2\).

(1)
(b)

Find the distance of the archaeologist from the starting point, correct to 3 significant figures.

(2)
(c)

Find the angle of the final position measured anticlockwise from east, in degrees, correct to 1 decimal place.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(w = 8 + 19i\)✓ 1
Part (b)
M1 use of \(|w| = \sqrt{8^2 + 19^2}\)✓ 1
A1 20.6 m✓ 1
Part (c)
M1 \(\arctan\left(\frac{19}{8}\right)\) (or equivalent argument, in degree mode)✓ 1
A1 67.2° (ECF from the real and imaginary parts of the total displacement)✓ 1

Exam tip: In degree questions check your calculator is in degree mode; the argument of a point in the first quadrant is just arctan(imaginary part ÷ real part).

142IBMI3H-1-05r46-02Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy6 marks

Calculator

A cycling club buys energy gels and energy bars. Order 1 is 3 gels and 2 bars and costs €14.10. Order 2 is 1 gel and 4 bars and costs €18.70. Let a gel cost €g and a bar cost €b.

(a)

Write the information as a matrix equation of the form \(\mathbf{A}\begin{pmatrix} g \\ b \end{pmatrix} = \mathbf{c}\).

(2)
(b)

Find \(\mathbf{A}^{-1}\).

(2)
(c)

Hence find the price of one bar.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 matrix \(\begin{pmatrix} 3 & 2 \\ 1 & 4 \end{pmatrix}\) multiplied by the column of prices✓ 1
A1 right-hand side \(\begin{pmatrix} 14.10 \\ 18.70 \end{pmatrix}\)✓ 1
Part (b)
M1 determinant \(3 \times 4 - 2 \times 1 = 10\)✓ 1
A1 \(\mathbf{A}^{-1} = \begin{pmatrix} 0.4 & -0.2 \\ -0.1 & 0.3 \end{pmatrix}\)✓ 1
Part (c)
M1 \(\begin{pmatrix} g \\ b \end{pmatrix} = \mathbf{A}^{-1}\begin{pmatrix} 14.10 \\ 18.70 \end{pmatrix}\) (ECF from (b))✓ 1
A1 €4.20✓ 1

Exam tip: Multiply by the inverse on the left of both sides: the order of matrix multiplication matters.

143IBMI3H-1-05r46-08Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy6 marks

Calculator

Two components in parallel in a server's power supply have impedances \(Z_1 = 4 + 3i\) ohms and \(Z_2 = 2 - i\) ohms. The combined impedance \(Z\) satisfies \(\frac{1}{Z} = \frac{1}{Z_1} + \frac{1}{Z_2}\).

(a)

Show that \(Z = \frac{Z_1 Z_2}{Z_1 + Z_2}\) and find \(Z\) in the form \(a + bi\).

(4)
(b)

Find the magnitude of \(Z\), correct to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 combine the fractions to get \(Z = \frac{Z_1Z_2}{Z_1+Z_2}\)✓ 1
M1 \(Z_1Z_2 = (4+3i)(2-i) = 11 + 2i\) and \(Z_1 + Z_2 = 6 + 2i\)✓ 1
M1 multiply numerator and denominator by the conjugate \(6 - 2i\)✓ 1
A1 \(Z = \frac{70 - 10i}{40} = 1.75 - 0.25i\)✓ 1
Part (b)
M1 \(\sqrt{1.75^2 + 0.25^2}\)✓ 1
A1 1.77 ohms✓ 1

Exam tip: To divide complex numbers, multiply top and bottom by the conjugate of the denominator so the denominator becomes real.

144IBMI3H-1-05r46-12Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy5 marks

Calculator

A stone column found at a dig is a cylinder of radius 0.45 m and height 3.2 m.

(a)

Calculate the volume of the column, correct to 3 significant figures.

(2)
(b)

A student estimates the volume by rounding the radius and height to 1 significant figure and using \(\pi \approx 3\). Find the student's estimate.

(1)
(c)

Calculate the percentage error in the student's estimate, using the unrounded volume from (a) as the exact value, and give your answer to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(\pi \times 0.45^2 \times 3.2\)✓ 1
A1 2.04 m3✓ 1
Part (b)
A1 \(3 \times 0.5^2 \times 3 = 2.25\) m3✓ 1
Part (c)
M1 \(\frac{|2.25 - 2.03575\ldots|}{2.03575\ldots} \times 100\)✓ 1
A1 10.5%✓ 1

Exam tip: Percentage error is the size of the error divided by the exact value; use the unrounded exact value to avoid a rounding error in your answer.

145IBMI3H-1-05r47-01Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy5 marks

A volcano monitoring station models a seismic ground displacement as the complex number z = −5 + 12i, measured in millimetres.

(a)

Find the modulus of z.

(2)
(b)

Find the argument of z in radians, giving your answer to 3 significant figures, with \(-\pi < \arg z \le \pi\).

(2)
(c)

Write z in the form \(re^{i\theta}\), using your answers to parts (a) and (b).

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to use \(\sqrt{(-5)^2+12^2}\)✓ 1
A1 \(|z| = 13\)✓ 1
Part (b)
M1 correct use of \(\pi - \arctan\left(\frac{12}{5}\right)\) or equivalent, recognising the second quadrant✓ 1
A1 \(\arg z = 1.97\)✓ 1
Part (c)
A1 \(z = 13e^{1.97i}\) (ECF from (a) and (b))✓ 1

Common error: Using \(\arctan\left(\frac{12}{-5}\right)\) directly on a calculator gives an angle in the fourth quadrant. Sketch the point first so you place the argument in the correct quadrant.

146IBMI3H-1-05r47-02Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy7 marks

Two lighting control signals in a theatre are represented by the complex numbers z1 = 3 + 2i and z2 = 2 − i.

(a)

Find z1z2, giving your answer in the form a + bi.

(2)
(b)

Find \(\dfrac{z_1}{z_2}\), giving your answer in the form a + bi.

(3)
(c)

Find \(\left|\dfrac{z_1}{z_2}\right|\), giving your answer to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 expand \((3+2i)(2-i)\) using \(i^2=-1\)✓ 1
A1 \(8 + i\)✓ 1
Part (b)
M1 multiply numerator and denominator by the conjugate \(2+i\)✓ 1
A1 numerator \(4 + 7i\) and denominator 5✓ 1
A1 \(0.8 + 1.4i\)✓ 1
Part (c)
M1 attempt to use \(\sqrt{0.8^2+1.4^2}\) (ECF from (b)) or \(\frac{|z_1|}{|z_2|}\)✓ 1
A1 \(1.61\)✓ 1

Exam tip: To divide complex numbers, multiply top and bottom by the conjugate of the denominator so that the denominator becomes a real number.

147IBMI3H-1-05r47-03Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy7 marks

A shipping company unloads 120 containers from a vessel in week 1 of a project. In each following week the number unloaded is 5% greater than in the previous week. Treat the weekly numbers as a geometric sequence, and give answers to the nearest whole container where appropriate.

(a)

Find the number of containers unloaded in week 10.

(2)
(b)

Find the total number of containers unloaded during the first 10 weeks.

(2)
(c)

Find the smallest number of weeks needed for the total number unloaded to exceed 2000.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 use \(u_n = 120(1.05)^{n-1}\) with \(n=10\)✓ 1
A1 186✓ 1
Part (b)
M1 use \(S_{10} = \frac{120(1.05^{10}-1)}{0.05}\)✓ 1
A1 1509✓ 1
Part (c)
M1 set up \(\frac{120(1.05^{n}-1)}{0.05} > 2000\)✓ 1
M1 correct method to solve for n (logarithms or table of values)✓ 1
A1 \(n = 13\) (the total is 1910 after 12 weeks and 2125.56 after 13 weeks)✓ 1

Common error: Week 10 is the tenth term, so the exponent is 9, not 10. The sum formula uses \(n\) in the exponent.

148IBMI3H-1-05r47-04Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy7 marks

A theatre sells standard tickets at €s each and premium tickets at €p each. One group buys 3 standard and 2 premium tickets for €102. Another group buys 1 standard and 4 premium tickets for €114.

(a)

Write this information as a matrix equation of the form \(\mathbf{M}\begin{pmatrix}s\\p\end{pmatrix}=\mathbf{c}\).

(2)
(b)

Find \(\mathbf{M}^{-1}\).

(2)
(c)

Hence find the price of a standard ticket and the price of a premium ticket.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(\mathbf{M}=\begin{pmatrix}3&2\\1&4\end{pmatrix}\)✓ 1
A1 \(\mathbf{c}=\begin{pmatrix}102\\114\end{pmatrix}\)✓ 1
Part (b)
M1 determinant \(=3\times4-2\times1=10\)✓ 1
A1 \(\mathbf{M}^{-1}=\frac{1}{10}\begin{pmatrix}4&-2\\-1&3\end{pmatrix}\)✓ 1
Part (c)
M1 attempt to calculate \(\mathbf{M}^{-1}\mathbf{c}\)✓ 1
A1 \(s=18\)✓ 1
A1 \(p=24\)✓ 1

Exam tip: Order matters in matrix multiplication: to solve \(\mathbf{M}\mathbf{x}=\mathbf{c}\) you multiply on the left by \(\mathbf{M}^{-1}\), giving \(\mathbf{x}=\mathbf{M}^{-1}\mathbf{c}\).

149IBMI3H-1-05r48-01Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy5 marks

Calculator

In a solar-farm inverter, the impedance of a filter circuit is modelled by the complex number \(Z = 6 + 8i\), measured in ohms (Ω).

(a)

Find the modulus of \(Z\).

(2)
(b)

Find the argument of \(Z\), in radians, correct to 3 significant figures.

(2)
(c)

Write \(Z\) in the form \(re^{i\theta}\), using your answers to (a) and (b).

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to use \(\sqrt{6^2+8^2}\)✓ 1
A1 \(|Z| = 10\) Ω✓ 1
Part (b)
M1 attempt to use \(\tan^{-1}\left(\frac{8}{6}\right)\) with the calculator in radian mode✓ 1
A1 \(0.927\) rad✓ 1
Part (c)
A1 \(Z = 10e^{0.927i}\) (ECF from (a) and (b))✓ 1

Common error: Set the calculator to radian mode before finding an argument that will be used in \(re^{i\theta}\) form; a degree value cannot go in the exponent.

150IBMI3H-1-05r48-02Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy6 marks

Calculator

Two alternating voltages in a tidal-turbine controller are represented by the complex numbers \(z_1 = 2 + 3i\) and \(z_2 = 1 - 4i\), in volts.

(a)

Find \(z_1 z_2\), giving your answer in the form \(a + bi\).

(3)
(b)

Find the real part of \(\frac{z_1}{z_2}\), correct to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to expand \((2+3i)(1-4i)\)✓ 1
M1 use of \(i^2 = -1\)✓ 1
A1 \(14 - 5i\)✓ 1
Part (b)
M1 multiply numerator and denominator by the conjugate \(1+4i\)✓ 1
A1 \(\frac{-10+11i}{17}\)✓ 1
A1 real part \(= -0.588\)✓ 1

Exam tip: To divide complex numbers, multiply the top and bottom by the conjugate of the denominator so that the denominator becomes real.

151IBMI3H-1-05r48-06Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy5 marks

Calculator

An energy company has three types of turbine, A, B and C, which generate \(x\), \(y\) and \(z\) megawatt-hours (MWh) respectively in one hour. One turbine of each type together generate 12 MWh. Two of A, three of B and one of C generate 22 MWh. One of A, four of B and two of C generate 26 MWh.

(a)

Write this information as a matrix equation of the form \(M\mathbf{v} = \mathbf{c}\), where \(\mathbf{v} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}\).

(2)
(b)

Solve the matrix equation to find \(x\), \(y\) and \(z\).

(2)
(c)

Find the hourly output of a group of two turbines of type A, one of type B and four of type C.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(M = \begin{pmatrix} 1&1&1 \\ 2&3&1 \\ 1&4&2 \end{pmatrix}\)✓ 1
A1 \(\mathbf{c} = \begin{pmatrix} 12 \\ 22 \\ 26 \end{pmatrix}\)✓ 1
Part (b)
M1 use of \(M^{-1}\mathbf{c}\) or a simultaneous-equation solver✓ 1
A1 \(x = 4\), \(y = 3\), \(z = 5\)✓ 1
Part (c)
A1 \(2(4) + 3 + 4(5) = 31\) MWh (ECF from (b))✓ 1

Exam tip: Check a solved system by substituting back into one equation that you did not use to solve it.

152IBMI3H-1-05r48-09Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Easy6 marks

Calculator

A bakery has two outlets, Harbour and Station. The matrix \(Q = \begin{pmatrix} 10 & 5 & 8 \\ 6 & 12 & 4 \end{pmatrix}\) gives the numbers of loaves, packs of rolls and cakes sold at Harbour (row 1) and Station (row 2) one morning. The prices are £2.50 for a loaf, £4.00 for a pack of rolls and £3.20 for a cake, written as the column matrix \(P = \begin{pmatrix} 2.5 \\ 4 \\ 3.2 \end{pmatrix}\).

(a)

Calculate \(QP\).

(3)
(b)

State what the entries of \(QP\) represent.

(1)
(c)

All prices are reduced by 15% in a sale. Find the takings at each outlet if the same items are sold. Give your answers to the nearest penny.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to multiply \(Q\) by \(P\) (a \(2\times1\) result)✓ 1
A1 \(70.6\) in row 1✓ 1
A1 \(75.8\) in row 2✓ 1
Part (b)
A1 the morning takings, in £, at Harbour and at Station respectively✓ 1
Part (c)
M1 attempt to evaluate \(0.85\,QP\)✓ 1
A1 Harbour £60.01 and Station £64.43✓ 1

Exam tip: Check that the number of columns of the first matrix equals the number of rows of the second before multiplying.

153IBMI-030-04-r2-06Complex numbers and matrices Paper 1Medium7 marks

A population of birds is split into juveniles \(J\) and adults \(D\), counted each spring. The model is
\(\begin{pmatrix} J_{n+1} \\ D_{n+1} \end{pmatrix} = M\begin{pmatrix} J_n \\ D_n \end{pmatrix}\), where \(M = \begin{pmatrix} 0 & 1.5 \\ 0.6 & 0.4 \end{pmatrix}\).
Initially there are 100 juveniles and 50 adults.

(a)

Find the total population after 5 years, to the nearest bird.

(2)
(b)

Find the eigenvalues of \(M\).

(3)
(c)

Hence state the long-term annual percentage growth rate of the population, and interpret this in context.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Uses \(M^5\begin{pmatrix} 100 \\ 50 \end{pmatrix} = \begin{pmatrix} 165.63 \\ 136.69 \end{pmatrix}\)✓ 1
Total = 302 birds✓ 1
Part (b)
\(\det(M - \lambda I) = -\lambda(0.4 - \lambda) - 0.9\)✓ 1
\(\lambda^2 - 0.4\lambda - 0.9 = 0\)✓ 1
\(\lambda = 1.17\) or \(\lambda = -0.770\)✓ 1
Part (c)
The dominant eigenvalue is 1.1695, so the growth rate is 17.0% per year✓ 1
In the long term the population increases by about 17.0% each year, assuming the survival and birth rates stay constant✓ 1

Exam tip: The long-term growth factor is the eigenvalue with the largest absolute value. The negative eigenvalue only causes oscillations, and these die away because \(|-0.770| < 1.17\).

154IBMI-030-04-r1-04Complex numbers and matrices Paper 1Medium6 marks

A bakery sells croissants at x euros each, muffins at y euros each and scones at z euros each. Three office orders were placed in one morning.
• Order 1: 3 croissants, 2 muffins and 4 scones cost €19.70.
• Order 2: 2 croissants, 5 muffins and 1 scone cost €16.25.
• Order 3: 4 croissants, 1 muffin and 3 scones cost €18.05.

(a)

Write this information as a matrix equation of the form \(A\mathbf{x} = \mathbf{b}\).

(2)
(b)

Hence find the price of each item.

(2)
(c)

The bakery offers a 10% discount on a box containing 6 of each item. Find the cost of this box.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 a 3×3 coefficient matrix multiplied by \(\begin{pmatrix}x\\y\\z\end{pmatrix}\)✓ 1
A1 \(\begin{pmatrix}3&2&4\\2&5&1\\4&1&3\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}19.70\\16.25\\18.05\end{pmatrix}\)✓ 1
Part (b)
M1 uses \(\mathbf{x} = A^{-1}\mathbf{b}\) (for example, with a GDC)✓ 1
A1 croissant €2.40, muffin €1.85, scone €2.20✓ 1
Part (c)
M1 \(0.9 \times 6 \times (2.40 + 1.85 + 2.20)\)✓ 1
A1 €34.83✓ 1

Exam tip: Write money answers to exactly 2 decimal places, for example €34.83 and not €34.8.

155IBMI-030-08-r2-04Exponents and logarithms Paper 1Medium7 marks

Two bacterial cultures are grown in a laboratory. The number of bacteria in each culture t hours after the start of an experiment is modelled by
Culture A: \(N_A = 800 \times 2^{t/5}\)
Culture B: \(N_B = 50 \times 3^{t/4}\)

(a)

Use the laws of logarithms to show that the two cultures contain the same number of bacteria when
\(t = \dfrac{\ln 16}{\frac{1}{4}\ln 3 - \frac{1}{5}\ln 2}\).

(4)
(b)

Find the value of t.

(1)
(c)

Find the number of bacteria in each culture at this time.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1: sets \(N_A = N_B\) and rearranges, \(16 \times 2^{t/5} = 3^{t/4}\)✓ 1
M1: takes natural logarithms of both sides✓ 1
M1: uses the product and power laws, \(\ln 16 + \frac{t}{5}\ln 2 = \frac{t}{4}\ln 3\)✓ 1
A1: collects terms in t, \(\ln 16 = t\left(\frac{1}{4}\ln 3 - \frac{1}{5}\ln 2\right)\), leading to the given answer✓ 1
Part (b)
A1: \(t = 20.4\) hours (20.383…)✓ 1
Part (c)
M1: substitutes the unrounded value of t into either model, \(800 \times 2^{20.383.../5}\)✓ 1
A1: 13 500 bacteria✓ 1

Common error: In a 'show that' question the logarithm laws must be written out step by step. Jumping straight to the given expression, or only checking it with the GDC, does not earn the method marks.

156IBMI-030-03-r2-05Exponents and logarithms Paper 1Medium7 marks

Astronomers describe the apparent brightness of a star using its magnitude \(m\). This is defined by
\(m = -2.5\log_{10}\left(\dfrac{F}{F_0}\right)\),
where \(F\) is the light flux received from the star and \(F_0\) is a fixed reference flux. A brighter star has a smaller magnitude.

(a)

Two stars have magnitudes \(m_1\) and \(m_2\) and fluxes \(F_1\) and \(F_2\). Use the laws of logarithms to show that \(m_1 - m_2 = 2.5\log_{10}\left(\dfrac{F_2}{F_1}\right)\).

(2)
(b)

Star A has magnitude 1.25 and star B has magnitude 4.85. Find \(\dfrac{F_A}{F_B}\).

(2)
(c)

A binary system consists of two stars that appear as a single point of light. Each star on its own has magnitude 3.40. Find the magnitude of the combined light from the system.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1: \(m_1 - m_2 = -2.5\left[\log_{10}F_1 - \log_{10}F_0 - \log_{10}F_2 + \log_{10}F_0\right]\), using \(\log(a/b) = \log a - \log b\)✓ 1
A1: \(= -2.5\log_{10}\left(\dfrac{F_1}{F_2}\right) = 2.5\log_{10}\left(\dfrac{F_2}{F_1}\right)\), using \(-\log x = \log(1/x)\)✓ 1
Part (b)
M1: \(4.85 - 1.25 = 2.5\log_{10}\left(\dfrac{F_A}{F_B}\right)\), so \(\dfrac{F_A}{F_B} = 10^{3.6/2.5}\)✓ 1
A1: 27.5✓ 1
Part (c)
M1: the combined flux is \(2F\), where \(F\) is the flux of one star✓ 1
M1: \(m_{\text{comb}} = 3.40 - 2.5\log_{10}2\)✓ 1
A1: 2.65✓ 1

Common error: The magnitudes of the two stars do not add, so the answer is not 6.80. It is the fluxes that add. A larger flux gives a smaller magnitude, so the combined magnitude must be less than 3.40.

157IBMI-030-08-r1-05Exponents and logarithms Paper 1Medium8 marks

A biologist believes that the resting metabolic rate, \(P\) watts, of a mammal is related to its mass, \(m\) kg, by a model of the form \(P = a m^b\), where \(a\) and \(b\) are constants.

(a)

Show that if \(P = a m^b\), then \(\ln P = b \ln m + \ln a\).

(2)
(b)

Using data from 12 species, the biologist finds that the regression line of \(\ln P\) on \(\ln m\) is \(\ln P = 0.742 \ln m + 1.25\). Find the value of \(a\) and the value of \(b\).

(3)
(c)

Use the model to estimate the resting metabolic rate of a 60 kg mammal.

(2)
(d)

The heaviest species in the data set has a mass of 450 kg. Comment on the reliability of using the model to estimate the metabolic rate of an elephant of mass 6000 kg.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1: takes ln of both sides: ln P = ln(a mb)✓ 1
A1: uses ln(xy) = ln x + ln y and ln(mb) = b ln m to obtain ln P = b ln m + ln a✓ 1
Part (b)
A1: b = 0.742✓ 1
M1: ln a = 1.25, so a = e1.25✓ 1
A1: a = 3.49✓ 1
Part (c)
M1: substitutes m = 60, e.g. 3.49×600.742 or ln P = 0.742 ln 60 + 1.25✓ 1
A1: P = 72.8 W✓ 1
Part (d)
R1: unreliable, because 6000 kg is well outside the range of the data, so this is extrapolation✓ 1

Common error: The intercept of the ln–ln line is ln a, not a. You must exponentiate it to find a.

158IBMI-030-03-r1-05Exponents and logarithms Paper 1Medium8 marks

A biologist collects data on the body mass, m kg, and resting metabolic rate, R watts, of mammals with masses between 0.02 kg and 100 kg. A log–log graph of the data is approximately linear, and the regression line of ln R on ln m is
\(\ln R = 0.74\ln m + 1.25\).

(a)

Show that the relationship can be written in the form \(R = am^b\), and find the values of a and b.

(3)
(b)

Estimate the resting metabolic rate of a mammal of mass 65 kg.

(2)
(c)

Find the factor by which body mass must be multiplied for the resting metabolic rate to double.

(2)
(d)

Comment on the reliability of using the model to estimate the metabolic rate of an elephant of mass 5000 kg.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
R = e0.74 ln m + 1.25 = e1.25 × m0.74 M1✓ 1
a = 3.49 (3.4903…) A1✓ 1
b = 0.74 A1✓ 1
Part (b)
R = 3.4903 × 650.74 (or ln R = 0.74 ln 65 + 1.25 = 4.339) M1✓ 1
R = 76.6 W A1✓ 1
Part (c)
2R = a(km)0.74 ⇒ k0.74 = 2 M1✓ 1
k = 21/0.74 = 2.55 A1✓ 1
Part (d)
Unreliable, since 5000 kg is far outside the data range of 0.02–100 kg, so this is extrapolation R1✓ 1

Exam tip: In a log–log model the gradient of the line becomes the power b and the intercept c becomes the coefficient a = ec (or 10c if base-10 logs are used).

159IBMI-030-04-r1-05Complex numbers and matrices Paper 1Medium7 marks

Two alternating voltages in a sound-system circuit are given by
\(V_1 = 12\sin(100\pi t)\) and \(V_2 = 8\sin\left(100\pi t + \frac{\pi}{3}\right)\),
where \(V_1\) and \(V_2\) are in volts and \(t\) is the time in seconds. All angles are in radians. Each voltage can be represented by a complex number (phasor) \(Ae^{i\phi}\), where \(A\) is the amplitude and \(\phi\) is the phase. So \(V_1\) is represented by \(12\) and \(V_2\) by \(8e^{i\pi/3}\).

(a)

Write \(8e^{i\pi/3}\) in the form \(a + bi\).

(2)
(b)

The total voltage is \(V = V_1 + V_2\). By adding the phasors, write \(V\) in the form \(R\sin(100\pi t + \alpha)\), where \(R > 0\) and \(0 < \alpha < \frac{\pi}{2}\).

(3)
(c)

Find the total voltage when \(t = 0.002\) s.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(8\cos\frac{\pi}{3} + 8i\sin\frac{\pi}{3}\)✓ 1
A1 \(4 + 4\sqrt{3}\,i\) (≈ \(4 + 6.93i\))✓ 1
Part (b)
M1 adds the phasors: \(12 + 4 + 4\sqrt{3}\,i = 16 + 4\sqrt{3}\,i\)✓ 1
A1 \(R = \sqrt{16^2 + 48} = \sqrt{304} = 4\sqrt{19} \approx 17.4\) V✓ 1
A1 \(\alpha = \arctan\frac{4\sqrt{3}}{16} \approx 0.409\) rad, so \(V = 17.4\sin(100\pi t + 0.409)\)✓ 1
Part (c)
M1 substitutes \(t = 0.002\) into the expression for \(V\) found in the previous part, or into \(V_1 + V_2\)✓ 1
A1 \(V = 15.0\) V✓ 1

Common error: Adding the amplitudes (12 + 8 = 20 V) ignores the phase difference. Add the phasors as complex numbers, and keep your GDC in radian mode.

160IBMI3H-1-02-05Complex numbers in Cartesian, polar and exponential form; matrices, inverses and solving systems; eigenvalues and eigenvectors in context Paper 1Medium6 marks

A cinema sells adult tickets at \(a\) euros and child tickets at \(c\) euros. Group 1 buys 3 adult and 2 child tickets for 34 euros. Group 2 buys 5 adult and 4 child tickets for 61 euros. Let \(M=\begin{pmatrix}3&2\\5&4\end{pmatrix}\).

(a)

Find \(\det M\) and write down \(M^{-1}\).

(3)
(b)

Write the information as a matrix equation and hence find the price of an adult ticket and of a child ticket.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(\det M=2\)✓ 1
M1 attempt at inverse using \(\frac{1}{\det M}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}\)✓ 1
A1 \(M^{-1}=\frac12\begin{pmatrix}4&-2\\-5&3\end{pmatrix}\)✓ 1
Part (b)
A1 \(M\begin{pmatrix}a\\c\end{pmatrix}=\begin{pmatrix}34\\61\end{pmatrix}\)✓ 1
M1 \(\begin{pmatrix}a\\c\end{pmatrix}=M^{-1}\begin{pmatrix}34\\61\end{pmatrix}\)✓ 1
A1 \(a=7\), \(c=6.50\) euros✓ 1

Common error: Multiply by \(M^{-1}\) on the left of the column vector: \(M^{-1}\mathbf{b}\), not \(\mathbf{b}M^{-1}\).

161IBMI3H-1-02-06Complex numbers in Cartesian, polar and exponential form; matrices, inverses and solving systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

Two streaming services, A and B, share 1000 subscribers. Each month 20% of A's subscribers switch to B and 30% of B's subscribers switch to A; nobody else joins or leaves. The state at the start of month \(n\) is \(\mathbf{s}_n=\begin{pmatrix}a_n\\b_n\end{pmatrix}\) and \(\mathbf{s}_{n+1}=T\mathbf{s}_n\), where \(T=\begin{pmatrix}0.8&0.3\\0.2&0.7\end{pmatrix}\). At the start, \(\mathbf{s}_0=\begin{pmatrix}400\\600\end{pmatrix}\).

(a)

Find the number of subscribers of each service after 1 month.

(2)
(b)

Find the number of subscribers of each service after 3 months.

(2)
(c)

By finding an eigenvector of \(T\) for the eigenvalue 1, find the number of subscribers of A in the long term.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(T\mathbf{s}_0\)✓ 1
A1 500 for A and 500 for B✓ 1
Part (b)
M1 \(T^{3}\mathbf{s}_0\) or repeated multiplication✓ 1
A1 575 for A and 425 for B✓ 1
Part (c)
M1 solving \((T-I)\mathbf{v}=\mathbf{0}\)✓ 1
A1 \(\mathbf{v}\propto\begin{pmatrix}3\\2\end{pmatrix}\)✓ 1
A1 \(\frac35\times1000=600\) subscribers✓ 1

Exam tip: The long-term state of a transition matrix is a multiple of the eigenvector for eigenvalue 1, scaled so that the total matches the population.

162IBMI3H-1-02-07Complex numbers in Cartesian, polar and exponential form; matrices, inverses and solving systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

The numbers of juvenile (\(j_n\)) and adult (\(a_n\)) lizards on an island in year \(n\) satisfy \(j_{n+1}=2a_n\) and \(a_{n+1}=0.3j_n+0.7a_n\). This is written \(\mathbf{x}_{n+1}=L\mathbf{x}_n\), where \(\mathbf{x}_n=\begin{pmatrix}j_n\\a_n\end{pmatrix}\) and \(L=\begin{pmatrix}0&2\\0.3&0.7\end{pmatrix}\).

(a)

Show that the eigenvalues of \(L\) satisfy \(\lambda^{2}-0.7\lambda-0.6=0\) and find both eigenvalues.

(3)
(b)

Find an eigenvector of \(L\) corresponding to the larger eigenvalue, in the form \(\begin{pmatrix}1\\k\end{pmatrix}\).

(2)
(c)

Interpret the larger eigenvalue and its eigenvector in the context of the lizard population.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(\det(L-\lambda I)=0\)✓ 1
A1 \(\lambda(\lambda-0.7)-0.6=0\) leading to \(\lambda^{2}-0.7\lambda-0.6=0\) (AG)✓ 1
A1 \(\lambda=1.2\) and \(\lambda=-0.5\)✓ 1
Part (b)
M1 \((L-1.2I)\mathbf{v}=\mathbf{0}\) giving \(-1.2+2k=0\)✓ 1
A1 \(k=0.6\)✓ 1
Part (c)
R1 in the long term the population grows by 20% each year✓ 1
R1 the ratio of juveniles to adults settles at \(1:0.6\) (that is 5:3)✓ 1

Exam tip: For interpretation questions, link the dominant eigenvalue to the growth factor per time step and the eigenvector to the long-term proportions.

163IBMI3H-1-02-08Complex numbers in Cartesian, polar and exponential form; matrices, inverses and solving systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

Six sensors are placed at the points in the complex plane (distances in metres) that satisfy \(z^{6}=64\).

(a)

Find all six solutions, giving each in the form \(re^{i\theta}\) with \(-\pi<\theta\le\pi\).

(3)
(b)

Show that \(1+i\sqrt3\) is one of the solutions.

(2)
(c)

Find the distance between the sensors at \(2\) and at \(1+i\sqrt3\).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(r=64^{1/6}=2\) and \(6\theta=2\pi k\)✓ 1
A1 \(\theta\) spaced \(\frac{\pi}{3}\) apart✓ 1
A1 \(2\), \(2e^{\pm i\pi/3}\), \(2e^{\pm 2i\pi/3}\), \(2e^{i\pi}\)✓ 1
Part (b)
M1 \(2e^{i\pi/3}=2\left(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}\right)\)✓ 1
A1 \(=1+i\sqrt3\) (AG)✓ 1
Part (c)
M1 \(|(1+i\sqrt3)-2|=|-1+i\sqrt3|\)✓ 1
A1 2 m✓ 1

Exam tip: The roots of \(z^n=c\) lie on a circle of radius \(|c|^{1/n}\) and are equally spaced by \(\frac{2\pi}{n}\) in argument.

164IBMI3H-1-02-09Complex numbers in Cartesian, polar and exponential form; matrices, inverses and solving systems; eigenvalues and eigenvectors in context Paper 1Medium6 marks

A student models the height \(y\) metres of a kicked ball at horizontal distance \(x\) metres from the kicker by \(y=ax^{2}+bx+c\). The ball is measured at height 2.0 m when \(x=0\), at 5.2 m when \(x=2\) and at 6.4 m when \(x=4\).

(a)

Write down three equations for \(a\), \(b\) and \(c\).

(2)
(b)

Solve the equations to find \(a\), \(b\) and \(c\).

(2)
(c)

Find the maximum height of the ball predicted by the model, to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(c=2\)✓ 1
A1 \(4a+2b+c=5.2\) and \(16a+4b+c=6.4\)✓ 1
Part (b)
M1 solving the system (technology)✓ 1
A1 \(a=-0.25\), \(b=2.1\), \(c=2\)✓ 1
Part (c)
M1 \(x=-\frac{b}{2a}=4.2\) or use of technology✓ 1
A1 6.41 m✓ 1

Common error: Substituting \(x=0\) first gives \(c\) immediately and reduces the system to two unknowns.

165IBMI3H-1-02-10Complex numbers in Cartesian, polar and exponential form; matrices, inverses and solving systems; eigenvalues and eigenvectors in context Paper 1Medium5 marks

A total of 12 000 euros is shared between two savings accounts. Account X pays 2.5% interest per year and account Y pays 4% interest per year. After one year the total interest is 375 euros. Let \(x\) and \(y\) be the amounts invested in X and Y.

(a)

Write down two equations for \(x\) and \(y\).

(2)
(b)

Find \(x\) and \(y\).

(2)
(c)

Find the overall interest rate on the 12 000 euros, as a percentage to 3 significant figures.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(x+y=12\,000\)✓ 1
A1 \(0.025x+0.04y=375\)✓ 1
Part (b)
M1 solving the simultaneous equations✓ 1
A1 \(x=7000\) and \(y=5000\)✓ 1
Part (c)
A1 3.13 %✓ 1

Exam tip: Convert percentages to decimals before forming the equation: 2.5% is 0.025.

166IBMI3H-1-02-11Complex numbers in Cartesian, polar and exponential form; matrices, inverses and solving systems; eigenvalues and eigenvectors in context Paper 1Medium6 marks

A feed mix for a research farm uses three grains. One portion of grain X, Y or Z provides the nutrient units shown.
Protein: 20, 10, 5
Carbohydrate: 5, 30, 10
Fibre: 4, 6, 15
The mix must provide 70 units of protein, 85 units of carbohydrate and 62 units of fibre. Let \(x\), \(y\) and \(z\) be the numbers of portions of X, Y and Z.

(a)

Write down three equations for \(x\), \(y\) and \(z\).

(3)
(b)

Solve the system.

(2)
(c)

The grains cost 0.90, 1.20 and 0.60 euros per portion respectively. Find the cost of the mix.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(20x+10y+5z=70\)✓ 1
A1 \(5x+30y+10z=85\)✓ 1
A1 \(4x+6y+15z=62\)✓ 1
Part (b)
M1 use of technology or elimination✓ 1
A1 \(x=2\), \(y=1.5\), \(z=3\)✓ 1
Part (c)
A1 5.40 euros✓ 1

Exam tip: Check a solution by substituting it back into all three original equations.

167IBMI3H-1-02b-05Complex numbers in Cartesian, polar and exponential form; matrices, inverses and solving systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

A farmer mixes x kg of maize, y kg of soya meal and z kg of grass meal to make a feed. Each kilogram contains the following.
Maize: 0.08 kg protein, 0.02 kg fibre, 14 MJ energy.
Soya meal: 0.44 kg protein, 0.06 kg fibre, 12 MJ energy.
Grass meal: 0.16 kg protein, 0.30 kg fibre, 9 MJ energy.
The mix must contain exactly 92 kg of protein, 57 kg of fibre and 6750 MJ of energy.

(a)

Write the conditions as a matrix equation Mv = b, where v = \(\begin{pmatrix} x \\ y \\ z \end{pmatrix}\).

(2)
(b)

Find det M and explain what its value tells you about the solutions.

(2)
(c)

Solve the equation to find the mass of each ingredient.

(2)
(d)

Maize costs €0.28 per kg, soya meal €0.52 per kg and grass meal €0.18 per kg. Find the cost of the mix.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 M = \(\begin{pmatrix} 0.08 & 0.44 & 0.16 \\ 0.02 & 0.06 & 0.30 \\ 14 & 12 & 9 \end{pmatrix}\)✓ 1
A1 b = \(\begin{pmatrix} 92 \\ 57 \\ 6750 \end{pmatrix}\)✓ 1
Part (b)
M1 det M = 1.428 (using a calculator)✓ 1
R1 the determinant is not zero, so M is invertible and there is a unique solution✓ 1
Part (c)
M1 v = M−1b✓ 1
A1 maize 300 kg, soya meal 100 kg, grass meal 150 kg✓ 1
Part (d)
A1 0.28 × 300 + 0.52 × 100 + 0.18 × 150 = €163.00✓ 1

Exam tip: Write the rows in the same order (protein, fibre, energy) in both M and b, otherwise the solution is meaningless.

168IBMI3H-1-02b-06Complex numbers in Cartesian, polar and exponential form; matrices, inverses and solving systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

The numbers of pollinating insects, in thousands, in two neighbouring orchards xn (orchard A) and yn (orchard B) in year n are modelled by \(\begin{pmatrix} x_{n+1} \\ y_{n+1} \end{pmatrix}\) = T\(\begin{pmatrix} x_n \\ y_n \end{pmatrix}\), where T = \(\begin{pmatrix} 1.0 & 0.2 \\ 0.2 & 0.7 \end{pmatrix}\).

(a)

Find the eigenvalues of T.

(3)
(b)

Find an eigenvector of T corresponding to the larger eigenvalue.

(2)
(c)

Explain what the larger eigenvalue shows about the long-term behaviour, and state the annual percentage growth of the total number of insects in the long term.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 det(T − λI) = 0✓ 1
A1 λ2 − 1.7λ + 0.66 = 0✓ 1
A1 λ = 1.1 and λ = 0.6✓ 1
Part (b)
M1 solves (T − 1.1I)v = 0, for example −0.1a + 0.2b = 0✓ 1
A1 \(\begin{pmatrix} 2 \\ 1 \end{pmatrix}\) (or any non-zero multiple)✓ 1
Part (c)
R1 the component along the eigenvector for 1.1 grows, while that for 0.6 decays, so the pollinator numbers settle into the ratio 2 : 1✓ 1
A1 10 % growth per year✓ 1

Common error: Eigenvectors are only defined up to a non-zero multiple, so (4, 2) is as correct as (2, 1); do not force a particular scale.

169IBMI3H-1-02b-07Complex numbers in Cartesian, polar and exponential form; matrices, inverses and solving systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

In an audio circuit the voltage is V = 36ei3π/20 volts and the impedance is Z = 9eiπ/20 ohms. The current is I = V ÷ Z.

(a)

Find I in the form reiθ.

(2)
(b)

Write I in the form a + bi, with a and b correct to 3 significant figures.

(2)
(c)

Find the smallest positive integer n for which In is a positive real number.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 divides the moduli (36 ÷ 9) and subtracts the arguments (3π/20 − π/20)✓ 1
A1 I = 4eiπ/10✓ 1
Part (b)
M1 4cos(π/10) and 4sin(π/10)✓ 1
A1 3.80 + 1.24i✓ 1
Part (c)
M1 In = 4neinπ/10✓ 1
R1 requires nπ/10 = 2πk for an integer k✓ 1
A1 n = 20✓ 1

Exam tip: For quotients, divide the moduli and subtract the arguments; for powers, raise the modulus to the power and multiply the argument.

170IBMI3H-1-02b-08Complex numbers in Cartesian, polar and exponential form; matrices, inverses and solving systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

The flow rates a, b and c (litres per second) in three irrigation channels satisfy
2a + b = P1
a + kb + c = P2
b + 2c = P3
where k depends on a valve setting and P1, P2, P3 are pump pressures. The equations are written as M\(\begin{pmatrix} a \\ b \\ c \end{pmatrix}\) = \(\begin{pmatrix} P_1 \\ P_2 \\ P_3 \end{pmatrix}\).

(a)

Write down M and show that det M = 4k − 4.

(3)
(b)

Find the value of k for which M has no inverse, and state what this means for the flow rates.

(2)
(c)

For k = 3, the pressures are P1 = 10, P2 = 20 and P3 = 14. Find the flow rates.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 M = \(\begin{pmatrix} 2 & 1 & 0 \\ 1 & k & 1 \\ 0 & 1 & 2 \end{pmatrix}\)✓ 1
M1 expands: 2(2k − 1) − 1(2 − 0) + 0✓ 1
A1 4k − 2 − 2 = 4k − 4✓ 1
Part (b)
A1 k = 1✓ 1
R1 there is no unique solution for the flow rates (either none or infinitely many)✓ 1
Part (c)
M1 M−1 applied to the pressure column (det M = 8)✓ 1
A1 a = 3, b = 4, c = 5 litres per second✓ 1

Exam tip: A zero determinant means the matrix has no inverse, so you cannot solve by multiplying by M⁻¹; say whether the system is inconsistent or has infinitely many solutions only after checking.

171IBMI3H-1-02b-12Complex numbers in Cartesian, polar and exponential form; matrices, inverses and solving systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

The fuel burn rate F kg per minute of a light aircraft at airspeed v m s−1 is modelled by F = avb, where a and b are constants. At 50 m s−1 the burn rate is 1.8 kg per minute and at 80 m s−1 it is 3.9 kg per minute.

(a)

Show that ln F = ln a + b ln v.

(2)
(b)

Find the values of a and b, giving b to 3 significant figures and a to 3 significant figures.

(3)
(c)

Use the model to estimate the burn rate at 65 m s−1. Use unrounded values of a and b.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 takes natural logarithms of both sides: ln F = ln(avb)✓ 1
A1 uses ln(pq) = ln p + ln q and ln(vb) = b ln v✓ 1
Part (b)
M1 b = (ln 3.9 − ln 1.8) ÷ (ln 80 − ln 50)✓ 1
A1 b = 1.65✓ 1
A1 a = 0.00289 (using the unrounded value of b)✓ 1
Part (c)
M1 F = a × 65b✓ 1
A1 2.77 kg per minute✓ 1

Exam tip: Rounding a and b before predicting changes the answer in the second or third significant figure: store the full values on your calculator.

172IBMI3H-1-02b-14Complex numbers in Cartesian, polar and exponential form; matrices, inverses and solving systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

The interval between two notes with frequencies f1 and f2 (with f2 > f1) is measured in cents: c = 1200 log2(f2 ÷ f1).

(a)

Two notes have frequencies 440 Hz and 466.16 Hz. Show that the interval between them is 100 cents, to the nearest whole cent.

(2)
(b)

A just-tuned major third has frequency ratio 5 : 4. Find its size in cents, and find by how many cents it differs from an equal-tempered major third of 400 cents.

(3)
(c)

A note is 850 cents above a note of frequency 261.6 Hz. Find its frequency, to the nearest hertz.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 1200 log2(466.16 ÷ 440), using log2 x = ln x ÷ ln 2✓ 1
A1 99.99, which is 100 to the nearest whole cent✓ 1
Part (b)
M1 1200 log2(5 ÷ 4)✓ 1
A1 386.3 cents✓ 1
A1 400 − 386.3 = 13.7 cents✓ 1
Part (c)
M1 850 = 1200 log2(f ÷ 261.6), so f = 261.6 × 2850/1200✓ 1
A1 427 Hz✓ 1

Exam tip: To undo a base-2 logarithm, raise 2 to the power: if log₂ x = y then x = 2y.

173IBMI3H-1-03r7-05Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

Aphids in two neighbouring fields are monitored weekly. The populations (in thousands) in field 1 and field 2 in week \(n\) are \(x_n\) and \(y_n\), and they are modelled by
\(\begin{pmatrix}x_{n+1}\\y_{n+1}\end{pmatrix}=\mathbf{A}\begin{pmatrix}x_n\\y_n\end{pmatrix}\), where \(\mathbf{A}=\begin{pmatrix}1.2&0.4\\0.1&0.9\end{pmatrix}\).

(a)

By solving \(\det(\mathbf{A}-\lambda\mathbf{I})=0\), find the eigenvalues of \(\mathbf{A}\).

(3)
(b)

Find an eigenvector of \(\mathbf{A}\) corresponding to the larger eigenvalue.

(2)
(c)

In the long term, the populations settle into a fixed ratio \(x_n:y_n\) and both increase by the same percentage each week. State this ratio, and find the percentage weekly increase.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \((1.2-\lambda)(0.9-\lambda)-0.04=0\)✓ 1
A1 \(\lambda^2-2.1\lambda+1.04=0\)✓ 1
A1 \(\lambda=1.3\) and \(\lambda=0.8\)✓ 1
Part (b)
M1 attempt to solve \((\mathbf{A}-1.3\mathbf{I})\mathbf{v}=\mathbf{0}\)✓ 1
A1 for example \(\begin{pmatrix}4\\1\end{pmatrix}\) (any non-zero multiple)✓ 1
Part (c)
A1 \(x_n:y_n=4:1\) (ECF from (b))✓ 1
A1 30% per week (ECF from (a))✓ 1

Exam tip: An eigenvector is only defined up to a non-zero multiple, so any scalar multiple of your vector is also a correct answer.

174IBMI3H-1-03r7-08Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium6 marks

In a recording studio, the voltage across a speaker circuit is modelled by the complex number \(V=48e^{0.4i}\) volts and the current by \(I=6e^{-0.3i}\) amperes. The impedance \(Z\) in ohms satisfies \(V=IZ\).

(a)

Find \(Z\) in the form \(re^{i\theta}\).

(2)
(b)

Write \(Z\) in the form \(R+iX\), where \(R\) is the resistance and \(X\) the reactance. Give \(R\) and \(X\) correct to 3 significant figures.

(2)
(c)

The voltage is changed to \(V=60e^{0.9i}\) volts with the same impedance. Find the amplitude of the new current, in amperes.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(Z=\frac{V}{I}\) with moduli divided and arguments subtracted✓ 1
A1 \(Z=8e^{0.7i}\)✓ 1
Part (b)
M1 \(R=8\cos0.7\) and \(X=8\sin0.7\) (ECF from (a))✓ 1
A1 \(R=6.12\ \Omega\), \(X=5.15\ \Omega\)✓ 1
Part (c)
M1 \(|I|=\frac{|V|}{|Z|}=\frac{60}{8}\) (ECF from (a))✓ 1
A1 7.5 A✓ 1

Exam tip: In exponential form, divide by dividing the moduli and subtracting the arguments; this is far quicker than converting to Cartesian form first.

175IBMI3H-1-04r10-02Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

A cyclist's tyre valve moves on a wheel of radius 0.35 m. Taking the axle as the origin of the complex plane (distances in metres), the position of the valve at time t seconds is z = 0.35ei(π/2 − 8t).

(a)

State the angular speed of the wheel and whether it turns clockwise or anticlockwise.

(2)
(b)

Find the position of the valve when t = 0.5, in the form a + bi with a and b to 3 significant figures. Use radians.

(3)
(c)

Find the time for the wheel to make one complete turn. Give your answer to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
angular speed 8 rad s−1✓ 1
clockwise (the argument decreases with time)✓ 1
Part (b)
argument = π/2 − 4 = −2.429...✓ 1
a = 0.35 cos(−2.429...) = −0.265✓ 1
b = 0.35 sin(−2.429...) = −0.229✓ 1
Part (c)
8T = 2π✓ 1
T = 0.785 s✓ 1

Common error: Check your calculator is in radian mode whenever the argument contains π or is given without a degree sign.

176IBMI3H-1-04r10-03Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

A data centre has three server models, X, Y and Z. Each X uses x kW, each Y uses y kW and each Z uses z kW. Rack 1 holds 2 X, 1 Y and 1 Z and draws 5.7 kW. Rack 2 holds 1 X, 3 Y and 2 Z and draws 8.6 kW. Rack 3 holds 1 X and 4 Z only, and draws 11.2 kW.

(a)

Write the information as a matrix equation of the form Av = b, where v is the column vector (x, y, z).

(2)
(b)

Show that A has an inverse and use it to find x, y and z.

(3)
(c)

A fourth rack holds 3 X, 2 Y and 4 Z. Find its power draw in kW.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A = rows (2, 1, 1), (1, 3, 2), (1, 0, 4)✓ 1
b = column (5.7, 8.6, 11.2)✓ 1
Part (b)
det A = 19 ≠ 0, so the inverse exists✓ 1
v = A−1b used✓ 1
x = 1.2, y = 0.8, z = 2.5 (kW)✓ 1
Part (c)
3(1.2) + 2(0.8) + 4(2.5) (ECF from previous part)✓ 1
15.2 kW✓ 1

Exam tip: A system has a unique solution only when the determinant of its coefficient matrix is non-zero.

177IBMI3H-1-04r10-05Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

A city bike scheme has two docking stations, A and B. Each morning, 80% of bikes at A stay at A and 20% move to B, while 30% of bikes at B move to A and 70% stay at B. The transition matrix is M = rows (0.8, 0.3), (0.2, 0.7), acting on column vectors (bikes at A, bikes at B).

(a)

Show that the eigenvalues of M are 1 and 0.5.

(3)
(b)

Find an eigenvector corresponding to the eigenvalue 1.

(2)
(c)

Hence find the long-term percentage of bikes at station A.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
characteristic equation (0.8 − λ)(0.7 − λ) − 0.06 = 0✓ 1
λ2 − 1.5λ + 0.5 = 0✓ 1
(λ − 1)(λ − 0.5) = 0, so λ = 1 or 0.5 (AG)✓ 1
Part (b)
(M − I)v = 0 gives −0.2x + 0.3y = 0✓ 1
eigenvector (3, 2) (or any multiple)✓ 1
Part (c)
long-term distribution is proportional to (3, 2)✓ 1
3/(3 + 2) = 60%✓ 1

Exam tip: The long-term distribution of a transition matrix is the eigenvector for eigenvalue 1, scaled so its entries sum to the total.

178IBMI3H-1-04r10-07Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium8 marks

In a data centre power supply, two components have complex impedances (in ohms) Z1 = 6 + 8i and Z2 = 3 − 2i.

(a)

The components are connected in series, so the total impedance is Z1 + Z2. Write the total in the form reiθ, giving r and θ to 3 significant figures.

(3)
(b)

The components are now connected in parallel, so the total impedance is Zp = Z1Z2 / (Z1 + Z2). Find Zp in the form a + bi, with a and b to 3 significant figures.

(3)
(c)

Find |Zp| to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Z = 9 + 6i✓ 1
r = √117 = 10.8✓ 1
θ = arctan(6/9) = 0.588✓ 1
Part (b)
Z1Z2 = 34 + 12i✓ 1
multiplies by the conjugate of 9 + 6i (or uses GDC) to divide✓ 1
Zp = 3.23 − 0.821i✓ 1
Part (c)
|Zp| = |Z1||Z2| / |Z1 + Z2| = √1300 / √117✓ 1
3.33 Ω✓ 1

Exam tip: The modulus of a quotient is the quotient of the moduli, so |Zp| can be found without first writing Zp in Cartesian form.

179IBMI3H-1-04r10-08Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium8 marks

Five postholes of a prehistoric round-house are found at the roots of the equation z5 = 3125, where z is a complex number whose real and imaginary parts are east and north coordinates in metres from the central hearth.

(a)

Write down the modulus of each root and the argument of each root, for 0 ≤ arg < 2π.

(3)
(b)

Find the root with the smallest positive argument in the form a + bi, with a and b to 3 significant figures.

(2)
(c)

Find the distance between two adjacent postholes. Give your answer to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
modulus = 31251/5 = 5✓ 1
arguments 0, 2π/5, 4π/5, 6π/5, 8π/5✓ 1
uses 5θ = 2πk (or De Moivre)✓ 1
Part (b)
5cos(2π/5), 5sin(2π/5)✓ 1
1.55 + 4.76i✓ 1
Part (c)
the angle between adjacent roots is 2π/5 at the centre✓ 1
distance = 2 × 5 × sin(π/5) (cosine rule or equivalent)✓ 1
distance = 5.88 m✓ 1

Exam tip: The n-th roots of a complex number are equally spaced around a circle, separated by an angle of 2π/n.

180IBMI3H-1-04r10-10Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium8 marks

A cycling club buys energy bars at €x each and gels at €y each. Shop A charges €11 for 2 bars and k gels. Shop B charges €14.70 for 3 bars and 6 gels. The prices are modelled by 2x + ky = 11 and 3x + 6y = 14.7, where k is a positive integer.

(a)

Find an expression for the determinant of the coefficient matrix in terms of k.

(1)
(b)

Find the value of k for which the system does not have a unique solution.

(2)
(c)

For this value of k, show that the equations have no solution.

(2)
(d)

Given instead that k = 5, find the price of one energy bar.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
det = 2 × 6 − 3k = 12 − 3k✓ 1
Part (b)
12 − 3k = 0✓ 1
k = 4✓ 1
Part (c)
2x + 4y = 11 gives 3x + 6y = 16.5✓ 1
this contradicts 3x + 6y = 14.7, so no solution✓ 1
Part (d)
inverse or elimination used (determinant = −3)✓ 1
x = (11 × 6 − 5 × 14.7)/(12 − 15) = 2.5✓ 1
€2.50 (and y = 1.20)✓ 1

Exam tip: A zero determinant means the matrix has no inverse; the equations then have either no solution or infinitely many, and the constants decide which.

181IBMI3H-1-04r10-13Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

Archaeologists date a wooden beam using carbon-14, which has a half-life of 5730 years. The fraction of carbon-14 remaining after t years is f = e−kt.

(a)

Show that k = 1.21 × 10−4 to 3 significant figures.

(2)
(b)

Find the percentage of carbon-14 remaining after 9000 years, to 3 significant figures.

(2)
(c)

The beam contains 62% of its original carbon-14. Find its age, to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
e−5730k = 0.5✓ 1
k = ln 2 / 5730 = 1.21 × 10−4 (AG)✓ 1
Part (b)
f = e−9000k = 0.3366✓ 1
33.7%✓ 1
Part (c)
0.62 = e−kt✓ 1
t = ln(1/0.62) × 5730 / ln 2✓ 1
3950 years✓ 1

Exam tip: Use the unrounded decay constant ln 2 / 5730 in later calculations to avoid a small error in the final age.

182IBMI3H-1-03r11-03Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium5 marks

A water company draws \(x\), \(y\) and \(z\) thousand litres per hour from wells A, B and C respectively. The wells have different properties:
• the total volume drawn is 90 thousand litres per hour;
• pumping uses 2 kWh per thousand litres at A, 1 kWh at B and nothing at C (gravity fed), giving 85 kWh in total each hour;
• the water contains 0, 1 and 3 mg of a contaminant per thousand litres at A, B and C respectively, giving 130 mg in total each hour.

(a)

Write the information as a matrix equation of the form \(\mathbf{M}\mathbf{v} = \mathbf{c}\), where \(\mathbf{v} = \begin{pmatrix} x\\ y\\ z\end{pmatrix}\).

(2)
(b)

Use the inverse matrix to find the amount drawn from each well.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 coefficient matrix \(\begin{pmatrix}1&1&1\\2&1&0\\0&1&3\end{pmatrix}\)✓ 1
A1 right-hand vector \(\begin{pmatrix}90\\85\\130\end{pmatrix}\)✓ 1
Part (b)
M1 \(\mathbf{v} = \mathbf{M}^{-1}\mathbf{c}\) (or equivalent method using a calculator)✓ 1
A1 \(x = 30\) and \(y = 25\)✓ 1
A1 \(z = 35\) thousand litres per hour✓ 1

Exam tip: Check a solution by substituting it back into all three original equations: here \(2(30)+25 = 85\) and \(25+3(35)=130\).

183IBMI3H-1-03r11-04Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium8 marks

A game has two servers, A and B. Each week, 20% of the players on A move to B and 30% of the players on B move to A; nobody else moves or joins. Let \(\mathbf{x}_n = \begin{pmatrix} a_n\\ b_n\end{pmatrix}\) be the numbers of players on A and B after \(n\) weeks, so that \(\mathbf{x}_{n+1} = \mathbf{T}\mathbf{x}_n\) with \(\mathbf{T} = \begin{pmatrix}0.8&0.3\\0.2&0.7\end{pmatrix}\). Initially there are 5100 players on A and 2900 on B.

(a)

Find the number of players on each server after 4 weeks, to the nearest whole number.

(3)
(b)

Find the long-term number of players on each server by finding the eigenvector of \(\mathbf{T}\) for eigenvalue 1.

(3)
(c)

Find the other eigenvalue of \(\mathbf{T}\) and explain how it shows that the numbers approach the values in part (b).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to find \(\mathbf{T}^4\mathbf{x}_0\) (or iterate four times)✓ 1
A1 \(a_4 = 4818.75 \approx 4819\)✓ 1
A1 \(b_4 = 3181.25 \approx 3181\)✓ 1
Part (b)
M1 solve \(\mathbf{T}\mathbf{x} = \mathbf{x}\)✓ 1
A1 eigenvector in ratio \(3 : 2\)✓ 1
A1 4800 players on A and 3200 on B (total stays 8000)✓ 1
Part (c)
A1 other eigenvalue is \(0.5\) (trace \(1.5\) minus \(1\))✓ 1
R1 \(|0.5| < 1\), so the part of the population that differs from the steady state is halved each week and the populations tend to 4800 and 3200✓ 1

Exam tip: The total number of players is unchanged when each column of the transition matrix sums to 1, which is a useful check on every vector you calculate.

184IBMI3H-1-03r11-05Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

A drug is given as a single dose. Let \(b\) and \(t\) be the amounts (mg) in the blood and in body tissue. Each hour the amounts change according to \(b_{n+1} = 0.6b_n + 0.2t_n\) and \(t_{n+1} = 0.3b_n + 0.7t_n\). The matrix of this system is \(\mathbf{M} = \begin{pmatrix}0.6&0.2\\0.3&0.7\end{pmatrix}\).

(a)

Find the eigenvalues of \(\mathbf{M}\).

(3)
(b)

Find an eigenvector corresponding to the larger eigenvalue.

(2)
(c)

Hence find the percentage of the drug remaining in the body that is in the blood after a long time, and the percentage of the remaining drug eliminated each hour in the long term.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 form \(\det(\mathbf{M} - \lambda\mathbf{I}) = 0\)✓ 1
A1 \(\lambda^2 - 1.3\lambda + 0.36 = 0\)✓ 1
A1 \(\lambda = 0.9\) and \(\lambda = 0.4\)✓ 1
Part (b)
M1 solve \((\mathbf{M} - 0.9\mathbf{I})\mathbf{v} = \mathbf{0}\), e.g. \(-0.3b + 0.2t = 0\)✓ 1
A1 eigenvector \(\begin{pmatrix}2\\3\end{pmatrix}\) (or any multiple)✓ 1
Part (c)
A1 proportion in blood \(= \frac{2}{2+3} = 40\%\)✓ 1
R1 in the long term the total amount is multiplied by 0.9 each hour, so 10% is eliminated each hour✓ 1

Common error: Using the smaller eigenvalue for the long-term behaviour. Components with eigenvalue 0.4 die away faster, so the 0.9 eigenvector dominates.

185IBMI3H-1-03r11-09Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium8 marks

A patient takes a 200 mg dose of a drug every 8 hours, the first dose at time 0. After each 8-hour interval, 30% of the drug present at the start of the interval is still in the body. Let \(A_n\) be the amount (mg) in the body immediately after the \(n\)th dose.

(a)

Find \(A_3\).

(2)
(b)

Find the greatest amount that can be in the body immediately after a dose, to 3 significant figures, if the doses continue indefinitely.

(2)
(c)

Find the smallest value of \(n\) for which \(A_n > 284\).

(3)
(d)

Find the time, in hours after the first dose, at which this \(n\)th dose is taken.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(A_3 = 200 + 200(0.3) + 200(0.3)^2\)✓ 1
A1 \(A_3 = 278\) mg✓ 1
Part (b)
M1 \(\frac{200}{1 - 0.3}\)✓ 1
A1 \(286\) mg✓ 1
Part (c)
M1 \(A_n = \frac{200(1 - 0.3^n)}{0.7}\) and set \(> 284\)✓ 1
M1 solve, e.g. \(0.3^n < 0.006\) so \(n > 4.25\) (check: \(A_4 = 283.4\), \(A_5 = 285.0\))✓ 1
A1 \(n = 5\)✓ 1
Part (d)
A1 \(8(5 - 1) = 32\) hours (ECF from (c))✓ 1

Exam tip: Check your inequality answer by evaluating the sequence at \(n\) and \(n-1\): one value must be below the target and the other above it.

186IBMI3H-1-03r11-10Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium6 marks

Chlorine in a water tank decays according to \(C = 2.0e^{-kt}\), where \(C\) is the concentration in mg/L, \(t\) is the time in hours and \(k\) is a positive constant. After 6 hours the concentration is 1.2 mg/L.

(a)

Find the value of \(k\), to 3 significant figures.

(3)
(b)

The tank must be refilled when the concentration reaches 0.5 mg/L. Use the unrounded value of \(k\) to find the time at which this happens, to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(1.2 = 2.0e^{-6k}\)✓ 1
M1 \(-6k = \ln 0.6\) (or equivalent)✓ 1
A1 \(k = 0.0851\)✓ 1
Part (b)
M1 \(0.5 = 2.0e^{-kt}\)✓ 1
M1 \(t = \frac{\ln 4}{k}\)✓ 1
A1 \(t = 16.3\) hours✓ 1

Common error: Rounding \(k\) before the second part. Use the unrounded value (or store it in your calculator) so that your final answer is accurate to 3 significant figures.

187IBMI3H-1-03r11-11Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

A cylindrical water tank has radius 2.5 m and height 4.0 m, each measured to the nearest 0.1 m. The volume is \(V = \pi r^2 h\).

(a)

Calculate the volume using the measured values, to 4 significant figures.

(1)
(b)

Find the lower and upper bounds for the true volume, to 4 significant figures.

(3)
(c)

The percentage error in using the volume from part (a) is \(\frac{|V_{\text{est}} - V_{\text{true}}|}{V_{\text{true}}} \times 100\). Use the unrounded bounds to find the greatest possible percentage error, to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(V = \pi(2.5)^2(4.0) = 78.54\) m3✓ 1
Part (b)
A1 \(2.45 \le r < 2.55\) and \(3.95 \le h < 4.05\)✓ 1
A1 lower bound \(\pi(2.45)^2(3.95) = 74.49\) m3✓ 1
A1 upper bound \(\pi(2.55)^2(4.05) = 82.73\) m3✓ 1
Part (c)
M1 percentage error at the lower bound \(= \frac{78.54-74.49}{74.49} \times 100 = 5.44\%\)✓ 1
M1 percentage error at the upper bound \(= 5.07\%\)✓ 1
A1 greatest possible percentage error: 5.44%✓ 1

Exam tip: Percentage error is measured against the true value, so the lower bound (smaller denominator) can give a larger error than the upper bound even though both are about the same size.

188IBMI3H-1-03r11-12Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium6 marks

A pump motor with impedance \(Z_1 = (30 + 40i)\ \Omega\) is connected in series with a correction unit of impedance \(Z_2 = (20 - 15i)\ \Omega\). The supply has amplitude 230 V.

(a)

Find the total impedance \(Z = Z_1 + Z_2\).

(1)
(b)

Find \(|Z|\) in ohms and \(\arg Z\) in degrees, each to 3 significant figures.

(3)
(c)

The current amplitude is \(I = \frac{V}{|Z|}\). Find \(I\) in amperes, to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(Z = (50 + 25i)\ \Omega\)✓ 1
Part (b)
M1 \(|Z| = \sqrt{50^2 + 25^2}\)✓ 1
A1 \(|Z| = 55.9\ \Omega\)✓ 1
A1 \(\arg Z = \tan^{-1}\frac{25}{50} = 26.6^\circ\)✓ 1
Part (c)
M1 \(\frac{230}{55.9}\) (or unrounded \(|Z|\))✓ 1
A1 \(I = 4.11\) A✓ 1

Exam tip: The modulus of an impedance is used for current amplitude, and the argument gives the phase shift between voltage and current.

189IBMI3H-1-03r13-03Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium5 marks

In a coffee-roasting drum, a sensor circuit has voltage V = 3 + 4i volts and current I = 1 + i amperes (both complex phasors). The impedance is Z = V ÷ I, in ohms.

(a)

Find Z in the form a + bi.

(3)
(b)

Find |Z|, giving your answer to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 multiply numerator and denominator by 1 − i (the conjugate)✓ 1
A1 numerator (3 + 4i)(1 − i) = 7 + i and denominator 2✓ 1
A1 Z = 3.5 + 0.5i✓ 1
Part (b)
M1 attempt to use |Z| = √(3.52 + 0.52) or 5/√2✓ 1
A1 |Z| = 3.54 Ω (3.5355…)✓ 1

Exam tip: Dividing complex numbers: multiply top and bottom by the conjugate of the denominator so the denominator becomes real.

190IBMI3H-1-03r13-04Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium4 marks

Each time a bridge-monitoring unit samples, its phasor is multiplied by z = 2eπi/12, so after n samples the phasor is zn, where n is a positive integer.

(a)

Find the smallest value of n for which zn is a negative real number.

(3)
(b)

Write down the value of zn for this value of n.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 use of zn = 2neinπ/12✓ 1
M1 set nπ/12 = π (or π + 2kπ) for a negative real value✓ 1
A1 n = 12 (smallest, since nπ/12 = π gives the first negative real value)✓ 1
Part (b)
A1 z12 = −4096✓ 1

Exam tip: A complex number is a negative real number when its argument is π (plus any multiple of 2π); use De Moivre's theorem to link the power to the argument.

191IBMI3H-1-03r13-06Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium5 marks

A coffee roaster sells two blends. Blend 1 contains 4 kg of Arabica and 3 kg of Robusta and costs €48.50. Blend 2 contains 2 kg of Arabica and 2 kg of Robusta and costs €27.00. Let the price per kilogram of Arabica be €a and of Robusta be €r.

(a)

Write the information as a matrix equation of the form Ax = b.

(1)
(b)

Find the inverse of the matrix A.

(2)
(c)

Hence find a and r.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 [[4, 3],[2, 2]] × [a, r] = [48.5, 27] (matrix equation with correct matrices)✓ 1
Part (b)
M1 determinant 4×2 − 3×2 = 2 and swap/negate entries✓ 1
A1 A−1 = ½[[2, −3],[−2, 4]]✓ 1
Part (c)
M1 attempt to compute A−1b (any valid method)✓ 1
A1 a = 8.00 and r = 5.50 (€ per kg)✓ 1

Exam tip: Check a matrix solution by substituting back into both original equations: 4(8) + 3(5.5) = 48.5 and 2(8) + 2(5.5) = 27.

192IBMI3H-1-03r13-07Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium5 marks

A probe carries three instruments with power draws x, y and z watts. Running the instruments in three different combinations gives these total power demands:
2x + y + z = 10.5
x + 3y + 2z = 17.5
x + y + 4z = 12.5

(a)

Write the system as a matrix equation AX = B.

(2)
(b)

Solve the system to find x, y and z.

(2)
(c)

Find the total power draw when all three instruments run together.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 A = [[2, 1, 1],[1, 3, 2],[1, 1, 4]]✓ 1
A1 X = [x, y, z] and B = [10.5, 17.5, 12.5]✓ 1
Part (b)
M1 attempt to solve (inverse matrix or technology)✓ 1
A1 x = 2.5, y = 4, z = 1.5✓ 1
Part (c)
A1 total = 8.0 W (ECF from (b))✓ 1

Exam tip: Keep the unknowns in the same order in every equation so the coefficient matrix is read off correctly.

193IBMI3H-1-03r13-08Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

A probe operates in two modes, scanning (S) and transmitting (T), and its mode is reviewed once an hour. From S, it stays in S with probability 0.6 and switches to T with probability 0.4. From T, it switches to S with probability 0.2 and stays in T with probability 0.8. The transition matrix, acting on the column vector (S, T), is M = [[0.6, 0.2],[0.4, 0.8]].

(a)

Show that the eigenvalues of M are 1 and 0.4.

(2)
(b)

Find an eigenvector of M corresponding to λ = 1.

(2)
(c)

Hence find the long-term proportion of hours the probe spends transmitting, giving your answer to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 set det(M − λI) = 0, i.e. (0.6 − λ)(0.8 − λ) − 0.08 = 0✓ 1
A1 λ2 − 1.4λ + 0.4 = 0, so λ = 1 or λ = 0.4✓ 1
Part (b)
M1 solve (M − I)v = 0, i.e. −0.4s + 0.2t = 0✓ 1
A1 eigenvector (1, 2) (or any multiple)✓ 1
Part (c)
R1 the component associated with λ = 0.4 shrinks to zero because 0.4n → 0✓ 1
M1 scale (1, 2) so the entries sum to 1✓ 1
A1 proportion = 2/3 = 0.667✓ 1

Common error: The steady-state vector must be scaled so its entries sum to 1 (the proportions), not left as the raw eigenvector.

194IBMI3H-1-04r14-05Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium6 marks

A sprinkler arm at the centre of a circular field is modelled in the complex plane with the pivot at the origin (units: metres). The nozzle at the end of the arm starts at \(z_0=30e^{i\pi/3}\). At each step the arm turns \(15^\circ\) anticlockwise, so that \(z_{n+1}=e^{i\pi/12}z_n\).

(a)

Find the position \(z_7\) after 7 steps, in the form \(a+bi\) with \(a\) and \(b\) to 3 significant figures.

(3)
(b)

Find the smallest number of steps \(n\) after which the nozzle points due south, i.e. \(\arg z_n=\frac{3\pi}{2}\) (equivalently \(-\frac{\pi}{2}\)).

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(z_7=30e^{i\pi/3}\times\left(e^{i\pi/12}\right)^7\)✓ 1
A1 \(z_7=30e^{i\,11\pi/12}\)✓ 1
A1 \(z_7=-29.0+7.76i\)✓ 1
Part (b)
M1 \(\frac{\pi}{3}+\frac{n\pi}{12}=\frac{3\pi}{2}\) (accept adding multiples of \(2\pi\))✓ 1
M1 \(\frac{n\pi}{12}=\frac{7\pi}{6}\)✓ 1
A1 \(n=14\)✓ 1

Exam tip: Multiplying by \(e^{i\theta}\) rotates a point by \(\theta\) without changing its distance from the origin; add the arguments and keep the modulus.

195IBMI3H-1-04r14-07Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium6 marks

Four radio beacons are placed at the positions in the complex plane (units: km, airport tower at the origin) given by the solutions of \(z^4=-64\).

(a)

Write \(-64\) in the form \(re^{i\theta}\) with \(r>0\) and \(0\le\theta<2\pi\).

(1)
(b)

Find the four solutions in the form \(a+bi\).

(3)
(c)

The beacons are the vertices of a square. Find its area in km2.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(-64=64e^{i\pi}\)✓ 1
Part (b)
M1 modulus \(|z|=64^{1/4}=2\sqrt{2}\)✓ 1
M1 arguments \(\frac{\pi+2k\pi}{4}\) for \(k=0,1,2,3\)✓ 1
A1 \(z=2+2i,\ -2+2i,\ -2-2i,\ 2-2i\)✓ 1
Part (c)
M1 side length is the distance between adjacent beacons, e.g. 4 km✓ 1
A1 16 km2✓ 1

Exam tip: The \(n\) roots of a complex number are equally spaced around a circle, separated by \(\frac{2\pi}{n}\) in argument.

196IBMI3H-1-04r14-09Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

The pitch deviation \(\theta\) degrees of an aircraft, \(t\) seconds after a gust, satisfies a model with characteristic equation \(\lambda^2+6\lambda+25=0\). The solution has the form \(\theta=e^{pt}(A\cos qt+B\sin qt)\), where the roots of the characteristic equation are \(p\pm qi\).

(a)

Solve \(\lambda^2+6\lambda+25=0\).

(2)
(b)

Find the period of the oscillation, to 3 significant figures.

(2)
(c)

The size of the oscillations is bounded by the envelope \(e^{-3t}\) times the initial size. Find the time for the envelope to fall to 5% of its initial value, to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 uses the quadratic formula or completes the square: \(\frac{-6\pm\sqrt{36-100}}{2}\)✓ 1
A1 \(\lambda=-3\pm4i\)✓ 1
Part (b)
M1 \(q=4\), period \(=\frac{2\pi}{4}\)✓ 1
A1 1.57 s✓ 1
Part (c)
M1 \(e^{-3t}=0.05\)✓ 1
M1 takes natural logarithms: \(-3t=\ln0.05\)✓ 1
A1 \(t=0.999\) s✓ 1

Exam tip: The real part of a complex root controls decay or growth; the imaginary part controls the frequency of oscillation.

197IBMI3H-1-04r14-11Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

In an audio equaliser, the ratio of output voltage to input voltage (as phasors) is \(H=\dfrac{6+2i}{2-i}\).

(a)

Write \(H\) in the form \(a+bi\).

(3)
(b)

Find the gain \(|H|\) in exact form, and the phase shift \(\arg H\) in radians.

(2)
(c)

An input signal has amplitude 1.5 V. Find the output amplitude, to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 multiplies numerator and denominator by \(2+i\)✓ 1
A1 numerator \(10+10i\) and denominator 5✓ 1
A1 \(H=2+2i\)✓ 1
Part (b)
A1 \(|H|=2\sqrt{2}\)✓ 1
A1 \(\arg H=\frac{\pi}{4}\)✓ 1
Part (c)
M1 \(1.5\times2\sqrt{2}\)✓ 1
A1 4.24 V✓ 1

Common error: Multiply by the conjugate of the denominator (\(2+i\)), not the denominator itself, so the denominator becomes real.

198IBMI3H-1-04r14-12Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium6 marks

A farmer mixes corn (\(x\) kg), soya (\(y\) kg) and barley (\(z\) kg) to make 1000 kg of feed. Protein contents are 9%, 44% and 12% by mass respectively, and the mix must contain 169 kg of protein. The costs are €0.30, €0.55 and €0.25 per kg respectively, and the total ingredient cost is €335.

(a)

Write down three equations for \(x\), \(y\) and \(z\).

(3)
(b)

Solve the equations to find \(x\), \(y\) and \(z\).

(2)
(c)

Find the percentage of protein in the mix.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(x+y+z=1000\)✓ 1
A1 \(0.09x+0.44y+0.12z=169\)✓ 1
A1 \(0.30x+0.55y+0.25z=335\)✓ 1
Part (b)
M1 uses a solver or matrix inverse✓ 1
A1 \(x=500\), \(y=200\), \(z=300\)✓ 1
Part (c)
A1 \(\frac{169}{1000}\times100=16.9\%\)✓ 1

Exam tip: Define each variable with units, then use a calculator's simultaneous-equation solver and check the solution in the original equations.

199IBMI3H-1-04r15-03Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium5 marks

The tip of a robot arm is at the point \(z_0 = 2.5e^{0.7i}\) in the complex plane (metres). Each command rotates the arm about the origin by multiplying the current position by \(e^{1.1i}\) (the angle is in radians).

(a)

Write down the argument of the arm tip after one command.

(1)
(b)

Find the smallest number of commands after which the arm tip has turned through a total angle of at least \(2\pi\) radians from its starting position.

(2)
(c)

Find the position of the arm tip after one command in the form \(a + bi\), with a and b to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 1.8 (rad)✓ 1
Part (b)
M1 \(1.1n \geq 2\pi\), so \(n \geq 5.71\ldots\)✓ 1
A1 \(n = 6\)✓ 1
Part (c)
M1 \(2.5(\cos1.8 + i\sin1.8)\)✓ 1
A1 \(-0.568 + 2.43i\)✓ 1

Exam tip: Make sure your calculator is in radian mode when the angle is given without a degree sign.

200IBMI3H-1-04r15-04Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium6 marks

A fashion designer places the vertices of a motif at the complex numbers that solve \(z^2 - 6z + 25 = 0\), drawn on an Argand diagram with 1 unit = 1 cm.

(a)

Solve the equation, giving your answers in the form \(a + bi\).

(2)
(b)

Find the modulus and the argument of the root with positive imaginary part. Give the argument in radians to 3 significant figures.

(2)
(c)

Write this root in the form \(re^{i\theta}\).

(1)
(d)

Find the distance between the two vertices.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 quadratic formula or completing the square, \(z = \frac{6 \pm \sqrt{36-100}}{2}\)✓ 1
A1 \(z = 3 \pm 4i\)✓ 1
Part (b)
A1 modulus 5✓ 1
A1 argument \(\tan^{-1}\!\left(\tfrac{4}{3}\right) = 0.927\)✓ 1
Part (c)
A1 \(5e^{0.927i}\) (ECF from (b))✓ 1
Part (d)
A1 8 cm✓ 1

Common error: Complex roots of a real quadratic come in conjugate pairs, so the two vertices are mirror images in the real axis.

201IBMI3H-1-04r15-07Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium6 marks

Three fishing trawlers each use three types of net, X, Y and Z, for different numbers of hours. The nets catch x, y and z kilograms of fish per hour respectively.
Trawler 1 uses X for 2 h, Y for 3 h and Z for 1 h and catches 530 kg.
Trawler 2 uses X for 1 h, Y for 2 h and Z for 4 h and catches 480 kg.
Trawler 3 uses X for 3 h, Y for 1 h and Z for 2 h and catches 540 kg.

(a)

Write the information as a matrix equation \(\mathbf{A}\mathbf{v} = \mathbf{b}\).

(2)
(b)

Find \(\det\mathbf{A}\), and explain what this tells you about the solution.

(2)
(c)

Hence find the catch rate for each type of net.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(\mathbf{A} = \begin{pmatrix}2&3&1\\1&2&4\\3&1&2\end{pmatrix}\)✓ 1
A1 \(\mathbf{v} = \begin{pmatrix}x\\y\\z\end{pmatrix}\) and \(\mathbf{b} = \begin{pmatrix}530\\480\\540\end{pmatrix}\)✓ 1
Part (b)
A1 \(\det\mathbf{A} = 25\)✓ 1
R1 non-zero determinant, so A is invertible and there is a unique solution✓ 1
Part (c)
M1 \(\mathbf{v} = \mathbf{A}^{-1}\mathbf{b}\) (using technology)✓ 1
A1 X: 120 kg h−1, Y: 80 kg h−1, Z: 50 kg h−1✓ 1

Exam tip: Pre-multiply by the inverse: \(\mathbf{v} = \mathbf{A}^{-1}\mathbf{b}\), not \(\mathbf{b}\mathbf{A}^{-1}\).

202IBMI3H-1-04r15-08Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium5 marks

The control software for a robot gripper solves the system
\(kx + 3y = 6\)
\(4x + (k+1)y = 9\)
where k is a calibration constant.

(a)

Find the two values of k for which the system does not have a unique solution.

(3)
(b)

Given that \(k = 2\), find the solution.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(\det = k(k+1) - 12 = 0\)✓ 1
A1 \(k^2 + k - 12 = 0\)✓ 1
A1 \(k = 3\) or \(k = -4\)✓ 1
Part (b)
M1 matrix inverse or elimination: \(2x + 3y = 6\), \(4x + 3y = 9\)✓ 1
A1 \(x = 1.5\), \(y = 1\)✓ 1

Common error: A zero determinant means no unique solution; the system could have either no solutions or infinitely many.

203IBMI3H-1-04r15-10Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium6 marks

A robot's movement controller applies the linear map represented by \(\mathbf{M} = \begin{pmatrix}3&1\\2&2\end{pmatrix}\) to position vectors in the plane (units are metres).

(a)

Show that the eigenvalues of M are 1 and 4.

(2)
(b)

Find an eigenvector for each eigenvalue.

(3)
(c)

A point on the line \(y = x\) is mapped by M. Describe the image of the point.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(\det(\mathbf{M} - \lambda\mathbf{I}) = (3-\lambda)(2-\lambda) - 2 = 0\)✓ 1
A1 \(\lambda^2 - 5\lambda + 4 = 0\), giving \(\lambda = 1\) or \(4\) (AG)✓ 1
Part (b)
M1 solve \((\mathbf{M} - \lambda\mathbf{I})\mathbf{v} = \mathbf{0}\) for one eigenvalue✓ 1
A1 \(\lambda = 4\): \(\begin{pmatrix}1\\1\end{pmatrix}\) (or any multiple)✓ 1
A1 \(\lambda = 1\): \(\begin{pmatrix}1\\-2\end{pmatrix}\) (or any multiple)✓ 1
Part (c)
R1 the image lies on the same line \(y = x\), at four times the distance from the origin✓ 1

Common error: Eigenvectors are only defined up to a scalar multiple, so any non-zero multiple of the vector is correct.

204IBMI3H-1-04r15-11Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium6 marks

Customers of an online fashion store choose between brand A and brand B each week. The numbers choosing each brand are modelled by \(\begin{pmatrix}a_{n+1}\\b_{n+1}\end{pmatrix} = \mathbf{T}\begin{pmatrix}a_n\\b_n\end{pmatrix}\) with \(\mathbf{T} = \begin{pmatrix}0.8&0.3\\0.2&0.7\end{pmatrix}\), where \(a_n\) and \(b_n\) are the numbers of customers choosing A and B in week n. In week 0 there are 400 customers for A and 600 for B.

(a)

Find the number of customers choosing each brand in week 2.

(2)
(b)

Show that 1 is an eigenvalue of T and find a corresponding eigenvector.

(2)
(c)

Hence find the long-term number of customers choosing brand A, assuming the total remains 1000.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(\mathbf{T}^2\begin{pmatrix}400\\600\end{pmatrix}\) or two successive multiplications✓ 1
A1 550 for A and 450 for B✓ 1
Part (b)
M1 \(\det(\mathbf{T} - \mathbf{I}) = (-0.2)(-0.3) - 0.3(0.2) = 0\)✓ 1
A1 e.g. \(\begin{pmatrix}3\\2\end{pmatrix}\) (or any multiple)✓ 1
Part (c)
M1 scale the eigenvector so the total is 1000, i.e. \(\tfrac{3}{5}\times1000\)✓ 1
A1 600✓ 1

Exam tip: The long-term distribution is the eigenvector for eigenvalue 1, scaled so its entries add up to the known total.

205IBMI3H-1-04r15-14Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium6 marks

A fashion designer invests £2500 in an account that pays a nominal annual interest rate of 3.6%, compounded monthly.

(a)

Calculate the value of the investment after 5 years. Give your answer to the nearest penny.

(3)
(b)

Find the smallest whole number of months after which the value first exceeds £4000.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 monthly rate \(3.6/12 = 0.3\%\)✓ 1
M1 \(2500\left(1+\dfrac{0.036}{12}\right)^{60}\)✓ 1
A1 £2992.24✓ 1
Part (b)
M1 \(2500 \times 1.003^n > 4000\)✓ 1
M1 \(n > \dfrac{\ln 1.6}{\ln 1.003} = 156.9\ldots\) (or table on GDC)✓ 1
A1 157 months✓ 1

Common error: With monthly compounding use \(n = 60\) periods and a monthly rate of \(3.6/12\) per cent, not 5 periods at 3.6%.

206IBMI3H-1-04r17-04Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

The storage capacity of a data centre doubles every 18 months. Its capacity \(t\) years after installation is \(C=C_0\times2^{t/1.5}\), where the initial capacity is \(C_0=2.5\) petabytes (PB).

(a)

Find the capacity after 6 years.

(2)
(b)

Show that the model can be written \(C=C_0e^{kt}\) and find \(k\) to 3 significant figures.

(2)
(c)

Find the time taken for the capacity to reach 40 times its initial value. Give your answer in years to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(2.5\times2^{6/1.5}=2.5\times2^4\)✓ 1
A1 \(40\) PB✓ 1
Part (b)
M1 \(2^{t/1.5}=e^{(t/1.5)\ln 2}\)✓ 1
A1 \(k=\frac{\ln2}{1.5}=0.462\)✓ 1
Part (c)
M1 sets \(2^{t/1.5}=40\)✓ 1
M1 \(t=1.5\log_2 40\) (or \(\frac{\ln40}{k}\))✓ 1
A1 \(t=7.98\) years✓ 1

Exam tip: Doubling models convert easily between base 2 and base \(e\) using \(2^x=e^{x\ln2}\); keep the exact \(k=\frac{\ln 2}{1.5}\) in later work.

207IBMI3H-1-04r17-07Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium8 marks

A cyclist rides two legs of a route, each leg described by a complex number whose modulus is the distance in km and whose argument is the angle anticlockwise from due east. Leg 1 is \(z_1=15e^{i\pi/6}\) and leg 2 is \(z_2=9e^{2i\pi/3}\).

(a)

Write \(z_1\) and \(z_2\) in the form \(a+bi\), with \(a\) and \(b\) to 3 significant figures.

(2)
(b)

Find the modulus and argument of the total displacement \(z_1+z_2\). Give the argument in radians, to 3 significant figures.

(3)
(c)

The cyclist wants to return directly to the start. Find the bearing, measured clockwise from north, of the direct outward line from the start to the finish. Give your answer to the nearest tenth of a degree.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(z_1=13.0+7.50i\)✓ 1
A1 \(z_2=-4.50+7.79i\)✓ 1
Part (b)
M1 \(z_1+z_2=8.49+15.3i\) (unrounded: \(8.4903\ldots+15.2942\ldots i\))✓ 1
A1 modulus \(17.5\) km✓ 1
A1 argument \(\tan^{-1}\frac{15.294}{8.490}=1.06\) radians✓ 1
Part (c)
M1 converts the argument to degrees, 61.0°, measured anticlockwise from east✓ 1
M1 bearing \(=90^\circ-61.0\ldots^\circ\)✓ 1
A1 \(029.0^\circ\) (29.0°)✓ 1

Common error: A complex argument is measured anticlockwise from the positive real axis, whereas a bearing is measured clockwise from north, so convert with \(90^\circ-\theta\).

208IBMI3H-1-04r17-11Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium8 marks

Each morning, the employees of a data centre travel by bike (B), bus (U) or on foot (W). The probabilities of an employee's choice on one day, given the choice on the previous day, are in the transition matrix $\(\mathbf{T}=\begin{pmatrix}0.7&0.2&0.1\\0.2&0.6&0.3\\0.1&0.2&0.6\end{pmatrix}\)$ with rows and columns in the order B, U, W (each column sums to 1, and the entry in row $i$, column $j$ is the probability of moving from state $j$ to state $i$). On Monday 50% cycle, 30% take the bus and 20% walk.

(a)

Find the proportions using each mode on Tuesday.

(2)
(b)

Find the proportions using each mode on the Saturday of the same week, giving your answers to 3 significant figures.

(2)
(c)

Find the long-term proportion of employees who travel by bike, giving your answer to 3 significant figures.

(4)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(\mathbf{T}\begin{pmatrix}0.5\\0.3\\0.2\end{pmatrix}\)✓ 1
A1 bike 0.43, bus 0.34, walk 0.23✓ 1
Part (b)
M1 \(\mathbf{T}^5\begin{pmatrix}0.5\\0.3\\0.2\end{pmatrix}\)✓ 1
A1 bike 0.353, bus 0.376, walk 0.271✓ 1
Part (c)
M1 states that the long-term vector \(\mathbf{v}\) satisfies \(\mathbf{T}\mathbf{v}=\mathbf{v}\) (eigenvalue 1)✓ 1
M1 uses \(B+U+W=1\)✓ 1
A1 solves the equations (for example \(B=\frac{10}{29}\), \(U=\frac{11}{29}\), \(W=\frac{8}{29}\))✓ 1
A1 \(0.345\)✓ 1

Exam tip: The long-term distribution is the eigenvector of the transition matrix for eigenvalue 1, scaled so its entries sum to 1.

209IBMI3H-1-04r18-02Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

Two components in a hospital monitor are connected in series. Their complex impedances are \(Z_1=30+40i\ \Omega\) and \(Z_2=20-10i\ \Omega\). The monitor is supplied with a voltage \(V=230\) V, taken as a real number (phase 0). The current is \(I=\dfrac{V}{Z}\), where \(Z\) is the total impedance.

(a)

Find the total impedance \(Z=Z_1+Z_2\).

(1)
(b)

Write \(Z\) in the form \(re^{i\theta}\), with \(r\) and \(\theta\) to 3 significant figures.

(2)
(c)

Find the magnitude of the current in amperes, and write the current in the form \(Ae^{i\phi}\), with \(A\) and \(\phi\) to 3 significant figures.

(3)
(d)

Interpret the sign of \(\phi\) for the timing of the current relative to the voltage.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(Z=50+30i\ \Omega\)✓ 1
Part (b)
A1 \(r=\sqrt{50^2+30^2}=58.3\)✓ 1
A1 \(\theta=\arctan\frac{30}{50}=0.540\) rad, so \(Z=58.3e^{0.540i}\)✓ 1
Part (c)
M1 \(I=\dfrac{230}{58.3e^{0.540i}}\) (divides moduli, subtracts arguments)✓ 1
A1 \(A=3.94\) A✓ 1
A1 \(\phi=-0.540\) rad, so \(I=3.94e^{-0.540i}\)✓ 1
Part (d)
R1 the argument is negative, so the current lags the voltage by 0.540 radians✓ 1

Exam tip: To divide in exponential form, divide the moduli and subtract the arguments. Keep unrounded values from earlier parts in your calculator to protect the third significant figure.

210IBMI3H-1-04r18-03Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

A beekeeper places five hives at the points in the complex plane that are solutions of \(z^5=243\), with the centre of the apiary at the origin and distances in metres.

(a)

Write down the five solutions in the form \(re^{i\theta}\) with \(-\pi<\theta\le\pi\), giving \(\theta\) in terms of \(\pi\).

(3)
(b)

Show that the distance between two neighbouring hives is \(6\sin\frac{\pi}{5}\) metres, and find this distance to 3 significant figures.

(3)
(c)

The beekeeper fences the pentagon formed by joining the five hives. Find the total length of fencing, to 3 significant figures.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 uses \(z=243^{1/5}e^{2\pi ik/5}\) for \(k=0,1,2,3,4\) (from \(243=243e^{2\pi ik}\))✓ 1
A1 \(r=3\)✓ 1
A1 \(\theta=0,\ \pm\frac{2\pi}{5},\ \pm\frac{4\pi}{5}\)✓ 1
Part (b)
M1 neighbouring hives are 3 m from the centre with an angle \(\frac{2\pi}{5}\) between them (isosceles triangle or \(|z_1-z_2|\))✓ 1
A1 distance \(=2\times3\times\sin\frac{\pi}{5}\) (AG) or equivalent cosine rule working✓ 1
A1 \(3.53\) m✓ 1
Part (c)
A1 \(5\times6\sin\frac{\pi}{5}=17.6\) m (ECF from (b), using the unrounded value)✓ 1

Exam tip: The \(n\) roots of \(z^n=a\) are equally spaced around a circle of radius \(a^{1/n}\), so adjacent roots are separated by an angle \(\frac{2\pi}{n}\).

211IBMI3H-1-04r18-06Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium6 marks

A beekeeper mixes two supplementary feeds. Each kilogram of feed X contains 0.6 kg of sugar and 0.1 kg of pollen substitute. Each kilogram of feed Y contains 0.2 kg of sugar and 0.5 kg of pollen substitute. The hives need exactly 8.4 kg of sugar and 7.0 kg of pollen substitute. Let \(x\) and \(y\) be the masses in kg of feed X and feed Y used.

(a)

Write the requirements as a matrix equation of the form \(M\begin{pmatrix}x\\y\end{pmatrix}=\mathbf{c}\).

(1)
(b)

Find \(\det M\) and \(M^{-1}\).

(2)
(c)

Hence find \(x\) and \(y\).

(2)
(d)

The beekeeper has only 20 kg of feed available. Determine whether there is enough.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(\begin{pmatrix}0.6&0.2\\0.1&0.5\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}8.4\\7.0\end{pmatrix}\)✓ 1
Part (b)
A1 \(\det M=0.6(0.5)-0.2(0.1)=0.28\)✓ 1
A1 \(M^{-1}=\frac{1}{0.28}\begin{pmatrix}0.5&-0.2\\-0.1&0.6\end{pmatrix}\)✓ 1
Part (c)
M1 \(\begin{pmatrix}x\\y\end{pmatrix}=M^{-1}\begin{pmatrix}8.4\\7.0\end{pmatrix}\)✓ 1
A1 \(x=10\) kg, \(y=12\) kg✓ 1
Part (d)
R1 total required is \(10+12=22\) kg, which exceeds 20 kg, so there is not enough✓ 1

Common error: Multiplying in the wrong order. The solution is \(M^{-1}\mathbf{c}\), not \(\mathbf{c}M^{-1}\), and the matrix product is not commutative.

212IBMI3H-1-04r18-10Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium6 marks

Two sinusoidal signals on a hospital monitor are represented by the complex numbers \(z_1=4e^{i\pi/3}\) and \(z_2=3e^{i\pi/2}\). The resultant signal is represented by their sum.

(a)

Write \(z_1\) in the form \(a+bi\), giving exact values.

(1)
(b)

Find \(z_1+z_2\) in Cartesian form.

(1)
(c)

Find the amplitude and phase of the resultant signal, to 3 significant figures.

(2)
(d)

The second signal is delayed so that it is represented by \(z_2e^{-i\pi/6}\). Find the new amplitude of the resultant signal.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(z_1=2+2\sqrt3\,i\)✓ 1
Part (b)
A1 \(z_1+z_2=2+(2\sqrt3+3)i\)✓ 1
Part (c)
A1 amplitude \(=|z_1+z_2|=6.77\)✓ 1
A1 phase \(=\arctan\frac{2\sqrt3+3}{2}=1.27\) rad✓ 1
Part (d)
M1 \(z_2e^{-i\pi/6}=3e^{i\pi/3}\), so both signals have the same phase✓ 1
A1 \(4e^{i\pi/3}+3e^{i\pi/3}=7e^{i\pi/3}\), amplitude 7✓ 1

Exam tip: Convert to Cartesian form to add complex numbers, and to polar or exponential form to multiply or divide. Then convert back when an amplitude or phase is asked for.

213IBMI3H-1-04r18-11Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

A scout bee's search path is modelled by points in the complex plane (units: metres) with \(z_0=1\) and \(z_{n+1}=(1+i)z_n\) for \(n=0,1,2,\dots\), where \(z_n\) is the bee's position after \(n\) turns.

(a)

Write \(1+i\) in the form \(re^{i\theta}\).

(2)
(b)

Hence write down an expression for \(z_n\) in terms of \(n\).

(1)
(c)

Find the smallest value of \(n\) for which the bee is more than 50 m from the origin.

(2)
(d)

Find \(z_{12}\) and state how far the bee is from the origin after 12 turns.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(r=\sqrt2\)✓ 1
A1 \(\theta=\frac{\pi}{4}\), so \(1+i=\sqrt2\,e^{i\pi/4}\)✓ 1
Part (b)
A1 \(z_n=(\sqrt2)^n e^{in\pi/4}\)✓ 1
Part (c)
M1 \((\sqrt2)^n>50\), i.e. \(2^{n/2}>50\) (so \(n>11.3\))✓ 1
A1 \(n=12\)✓ 1
Part (d)
M1 \(e^{12i\pi/4}=e^{3\pi i}=-1\)✓ 1
A1 \(z_{12}=-64\), a distance of 64 m✓ 1

Exam tip: Multiplying by \(re^{i\theta}\) scales a point by \(r\) and rotates it by \(\theta\). Repeated multiplication therefore produces powers of \(r\) and multiples of \(\theta\).

214IBMI3H-1-04r18-13Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium6 marks

A hospital pharmacy buys three types of intravenous bag: A, B and C. The prices per bag, in £, are unknown. The costs of three ward orders are:
Ward 1: 5 bags of A, 8 of B, 2 of C, total £190.10
Ward 2: 3 bags of A, 10 of B, 4 of C, total £223.50
Ward 3: 6 bags of A, 2 of B, 7 of C, total £259.40
Let the prices of one bag of A, B and C be \(a\), \(b\) and \(c\) pounds.

(a)

Write down three equations for \(a\), \(b\) and \(c\).

(2)
(b)

Solve the system to find the price of each type of bag.

(2)
(c)

A new order contains 4 bags of A, 5 of B and 3 of C, and the pharmacy receives a 10% discount on the whole order. Find the price the pharmacy pays.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 forms at least one correct equation, e.g. \(5a+8b+2c=190.10\)✓ 1
A1 all three correct: \(3a+10b+4c=223.50\) and \(6a+2b+7c=259.40\)✓ 1
Part (b)
M1 solves with a calculator or by elimination✓ 1
A1 \(a=14.50\), \(b=9.20\), \(c=22.00\)✓ 1
Part (c)
M1 \(4(14.50)+5(9.20)+3(22.00)=170.00\) then \(\times0.9\)✓ 1
A1 £153.00✓ 1

Exam tip: Solve simultaneous equations on your calculator, but write the equations you entered, since marks are given for forming the system correctly.

215IBMI3H-1-05r19-05Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium8 marks

The end of a robotic arm is at position \(z = 2.4\,e^{0.7i}\) metres from its base, where the angle is in radians. The arm control program moves the end to a new position \(w\) by rotating it anticlockwise through 0.5 radians about the base and extending its distance from the base by a factor of 1.25.

(a)

Write \(z\) in the form \(a+bi\), giving \(a\) and \(b\) to 3 significant figures.

(3)
(b)

Find \(w\) in the form \(r\,e^{i\theta}\).

(2)
(c)

Find the distance between the two positions, in metres, to 3 significant figures. Use unrounded values throughout.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to use \(a=2.4\cos0.7\) or \(b=2.4\sin0.7\)✓ 1
A1 \(a=1.84\) (1.8356…)✓ 1
A1 \(b=1.55\) (1.5461…)✓ 1
Part (b)
M1 multiply moduli and add arguments, \(r=1.25\times2.4\), \(\theta=0.7+0.5\)✓ 1
A1 \(w=3\,e^{1.2i}\)✓ 1
Part (c)
M1 convert \(w\) to Cartesian form: \(1.0870\ldots+2.7961\ldots i\)✓ 1
M1 \(|w-z|=\sqrt{(1.0870-1.8356)^2+(2.7961-1.5461)^2}\)✓ 1
A1 1.46 m (1.4570…)✓ 1

Exam tip: Keep the calculator in radian mode when an angle is given without a degree sign; \(e^{i\theta}\) in exponential form always uses radians.

216IBMI3H-1-05r19-06Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium6 marks

The stability of a wheel-speed controller for a small robot is analysed using the quadratic equation \(z^2+pz+q=0\), where \(p\) and \(q\) are real numbers. One root of the equation is \(2-3i\).

(a)

Write down the other root.

(1)
(b)

Find the value of \(p\) and the value of \(q\).

(3)
(c)

Write \(2-3i\) in the form \(r\,e^{i\theta}\) with \(-\pi<\theta\le\pi\), giving \(r\) and \(\theta\) to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(2+3i\) (complex conjugate)✓ 1
Part (b)
M1 use of sum of roots \(=-p\) or product of roots \(=q\) (or expanding \((z-(2-3i))(z-(2+3i))\))✓ 1
A1 \(p=-4\)✓ 1
A1 \(q=13\)✓ 1
Part (c)
A1 \(r=\sqrt{13}=3.61\)✓ 1
A1 \(\theta=-\tan^{-1}\frac{3}{2}=-0.983\) (−0.9828…)✓ 1

Common error: A negative imaginary part means the argument is negative (fourth quadrant) when it is written in the range \(-\pi<\theta\le\pi\); a positive angle from the calculator would describe the conjugate.

217IBMI3H-1-05r19-10Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

A fishery authority sets an annual quota for a species of tuna. The quota in year 1 is 840 tonnes and it is reduced by 6% each year, so the quotas form a geometric sequence.

(a)

Find the quota in year 4, to the nearest tonne.

(2)
(b)

Find the first year in which the quota is below 500 tonnes.

(3)
(c)

Find the total quota allowed over the first 10 years, to the nearest tonne.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(840\times0.94^3\)✓ 1
A1 698 tonnes (697.69…)✓ 1
Part (b)
M1 \(840\times0.94^{n-1}<500\)✓ 1
M1 solving, \(n-1>8.38\) (or comparing year 9: 512.04 and year 10: 481.32)✓ 1
A1 year 10✓ 1
Part (c)
M1 \(S_{10}=\frac{840(1-0.94^{10})}{1-0.94}\)✓ 1
A1 6459 tonnes (6459.39…)✓ 1

Common error: A reduction of 6% means the common ratio is 0.94, not 0.06; the exponent in the n-th term is \(n-1\), so year 4 uses \(0.94^3\).

218IBMI3H-1-05r19-11Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

An engineering firm buys an inspection robot for £48000 using a loan. The loan is repaid in equal monthly payments, made at the end of each month, over 6 years. Interest is charged at a nominal annual rate of 4.8%, compounded monthly.

(a)

Find the monthly repayment, to the nearest penny.

(3)
(b)

Find the total interest paid over the 6 years. Use the monthly repayment £768.59.

(2)
(c)

Find the balance still owed after 3 years (36 payments). Use the monthly repayment £768.59 and give your answer to the nearest penny.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 use of finance solver / TVM with \(N=72\), PV \(=48000\)✓ 1
M1 \(I\%=4.8\), FV \(=0\), \(P/Y=C/Y=12\)✓ 1
A1 £768.59 (768.5916…)✓ 1
Part (b)
M1 \(72\times768.59-48000\)✓ 1
A1 £7338.48✓ 1
Part (c)
M1 TVM with \(N=36\), PV \(=48000\), PMT \(=-768.59\) solving for FV, or \(48000(1.004)^{36}-768.59\times\frac{1.004^{36}-1}{0.004}\)✓ 1
A1 £25721.65✓ 1

Exam tip: On a finance solver, enter money you receive as positive and money you pay as negative, and set payments per year and compounding periods per year to 12 for monthly payments.

219IBMI3H-1-05r19-12Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

An inspection robot is bought for €25000. Its value \(V\) euros after \(t\) years is modelled by \(V=25000\times0.82^{t}\).

(a)

Find the value of the robot after 3 years, to the nearest cent.

(2)
(b)

Find how long it takes for the value to fall to €8000, in years to 3 significant figures.

(3)
(c)

The model can also be written as \(V=25000\,e^{kt}\). Find the value of \(k\), to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(25000\times0.82^3\)✓ 1
A1 €13784.20✓ 1
Part (b)
M1 \(25000\times0.82^t=8000\)✓ 1
M1 \(t=\frac{\ln0.32}{\ln0.82}\) or use of a GDC solver✓ 1
A1 \(t=5.74\) (5.7416…)✓ 1
Part (c)
M1 \(e^k=0.82\) or \(k=\ln0.82\)✓ 1
A1 \(k=-0.198\) (−0.19845…)✓ 1

Exam tip: To solve for an exponent, isolate the power first, then take logarithms of both sides; the answer must be given in the units of \(t\), here years.

220IBMI3H-1-05r20-05Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

Calculator

In an arcade game a spinning drone starts at \(z_0=2\) on the complex plane (one unit is one tenth of the screen width). At every frame its position is multiplied by \(w=1.5e^{i\pi/6}\), so after \(n\) frames its position is \(z_n=2w^n\).

(a)

Find the modulus and argument of \(z_5\). Give the modulus to 4 significant figures and the argument in radians to 4 significant figures.

(3)
(b)

The drone leaves the screen when \(|z_n|>100\). Find the number of frames it takes for the drone to leave the screen.

(2)
(c)

Write \(z_5\) in Cartesian form, to 3 significant figures. Use the exact modulus \(2(1.5)^5\) and argument \(\frac{5\pi}{6}\), not rounded values.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(z_5=2(1.5)^5e^{i5\pi/6}\)✓ 1
A1 \(|z_5|=15.19\)✓ 1
A1 \(\arg z_5=\frac{5\pi}{6}=2.618\)✓ 1
Part (b)
M1 \(2(1.5)^n>100\), i.e. \(n>\frac{\ln 50}{\ln 1.5}=9.65\), or the values \(|z_9|=76.9\) and \(|z_{10}|=115\)✓ 1
A1 \(n=10\)✓ 1
Part (c)
M1 \(2(1.5)^5\cos\frac{5\pi}{6}\) and \(2(1.5)^5\sin\frac{5\pi}{6}\)✓ 1
A1 \(z_5=-13.2+7.59i\)✓ 1

Common error: Rounding up is essential in part (b): \(n\) must be a whole number of frames, and at \(n=9\) the inequality is not yet satisfied.

221IBMI3H-1-05r20-06Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium6 marks

Calculator

A water treatment plant blends water from three boreholes X, Y and Z. The flow rates are \(x\), \(y\) and \(z\) litres per second. The plant needs a total flow of 60 L s−1. The boreholes contain nitrate at 0.4, 0.9 and 1.5 mg per litre respectively and the blend must carry 50.5 mg of nitrate per second. Pumping costs are €0.80, €0.50 and €0.30 per litre per second for X, Y and Z respectively, and the total pumping cost is €34.50 per second.

(a)

Write down three equations connecting \(x\), \(y\) and \(z\).

(3)
(b)

Solve the system to find the flow rate from each borehole.

(2)
(c)

Borehole Z can supply at most 18 L s−1. State, with a reason, whether the plant can meet this blend.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(x+y+z=60\)✓ 1
A1 \(0.4x+0.9y+1.5z=50.5\)✓ 1
A1 \(0.8x+0.5y+0.3z=34.5\)✓ 1
Part (b)
M1 use of technology (or matrix inverse) to solve the system✓ 1
A1 \(x=25\), \(y=20\), \(z=15\) (all three)✓ 1
Part (c)
R1 yes, since \(z=15\le 18\)✓ 1

Exam tip: Enter the coefficients carefully into your calculator's simultaneous-equation solver and substitute the solution back into all three equations to check it.

222IBMI3H-1-05r20-07Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

Calculator

A drug is modelled in two compartments, the blood and the tissue. The amounts (in mg) at the start of hour \(n\) are written as \(\mathbf{x}_n=\begin{pmatrix}b_n\\t_n\end{pmatrix}\) and
\(\mathbf{x}_{n+1}=\mathbf{T}\mathbf{x}_n,\qquad \mathbf{T}=\begin{pmatrix}0.8&0.3\\0.1&0.6\end{pmatrix}.\)

(a)

Show that the eigenvalues of \(\mathbf{T}\) are 0.9 and 0.5.

(3)
(b)

Find an eigenvector of \(\mathbf{T}\) corresponding to the eigenvalue 0.9.

(2)
(c)

Explain what the eigenvector and eigenvalue from part (b) tell you about the drug in the long term.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(\det(\mathbf{T}-\lambda\mathbf{I})=0\)✓ 1
A1 \((0.8-\lambda)(0.6-\lambda)-0.03=0\), i.e. \(\lambda^2-1.4\lambda+0.45=0\)✓ 1
A1 \((\lambda-0.9)(\lambda-0.5)=0\) so \(\lambda=0.9\) or \(0.5\)✓ 1
Part (b)
M1 \((0.8-0.9)b+0.3t=0\) or equivalent✓ 1
A1 e.g. \(\begin{pmatrix}3\\1\end{pmatrix}\), accept any non-zero multiple✓ 1
Part (c)
R1 the ratio blood : tissue tends to 3 : 1 (the proportions settle to the eigenvector)✓ 1
R1 the amount in each compartment, and the total, is multiplied by 0.9 each hour, a fall of 10% per hour✓ 1

Exam tip: An eigenvalue is a per-step multiplier and its eigenvector is the pattern that is preserved. Always translate both back into the context in a final comment.

223IBMI3H-1-05r20-08Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

Calculator

Each week a player of an online game is either Active or Lapsed. A fraction \(a\) of Active players become Lapsed the following week, and 30% of Lapsed players return to Active. With the state vector \(\begin{pmatrix}A_n\\L_n\end{pmatrix}\) giving the proportions in week \(n\):
\(\begin{pmatrix}A_{n+1}\\L_{n+1}\end{pmatrix}=\mathbf{T}\begin{pmatrix}A_n\\L_n\end{pmatrix},\qquad\mathbf{T}=\begin{pmatrix}1-a&0.3\\a&0.7\end{pmatrix}.\)
In the long term, 60% of players are Active and 40% are Lapsed.

(a)

Find the value of \(a\).

(3)
(b)

Using this value of \(a\), find the eigenvalues of \(\mathbf{T}\).

(3)
(c)

Explain the significance of the eigenvalue \(\lambda=1\) for the long-term proportions.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 steady state satisfies \(\mathbf{T}\mathbf{v}=\mathbf{v}\) with \(\mathbf{v}=\begin{pmatrix}0.6\\0.4\end{pmatrix}\)✓ 1
M1 \((1-a)(0.6)+0.3(0.4)=0.6\) or equivalent✓ 1
A1 \(a=0.2\)✓ 1
Part (b)
M1 \(\det(\mathbf{T}-\lambda\mathbf{I})=0\)✓ 1
A1 \(\lambda^2-1.5\lambda+0.5=0\)✓ 1
A1 \(\lambda=1\) and \(\lambda=0.5\)✓ 1
Part (c)
R1 its eigenvector \(\begin{pmatrix}0.6\\0.4\end{pmatrix}\) is unchanged by \(\mathbf{T}\), so it gives the steady-state proportions✓ 1

Exam tip: For a transition matrix whose columns sum to 1, the steady state is the eigenvector for eigenvalue 1, scaled so its entries sum to 1. Use this rather than repeated multiplication.

224IBMI3H-1-05r20-09Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

Calculator

A game engine updates the position of a sprite on screen by multiplying its coordinate vector by the matrix \(\mathbf{R}=\begin{pmatrix}3&-2\\2&3\end{pmatrix}\) at every frame.

(a)

Find the eigenvalues of \(\mathbf{R}\), giving them in the form \(p+qi\).

(3)
(b)

For the eigenvalue \(3+2i\), find the modulus and the argument in radians, giving each to 3 significant figures.

(2)
(c)

The matrix \(\mathbf{R}\) has the effect of an anticlockwise rotation combined with an enlargement. Use your answers to part (b) to state the scale factor and the angle of rotation in degrees, to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(\det(\mathbf{R}-\lambda\mathbf{I})=0\)✓ 1
A1 \((3-\lambda)^2+4=0\), i.e. \(\lambda^2-6\lambda+13=0\)✓ 1
A1 \(\lambda=3\pm2i\)✓ 1
Part (b)
A1 modulus \(=\sqrt{13}=3.61\)✓ 1
A1 argument \(=\arctan\frac{2}{3}=0.588\)✓ 1
Part (c)
A1 scale factor \(3.61\)✓ 1
A1 angle \(=33.7^\circ\)✓ 1

Exam tip: When a real matrix has complex eigenvalues, the modulus gives the enlargement and the argument gives the rotation. Convert radians to degrees only at the final step.

225IBMI3H-1-05r20-10Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

Calculator

A patient takes a 250 mg dose of a drug every 6 hours, the first dose being at time \(t=0\). Just before each new dose, 40% of the drug that was in the body immediately after the previous dose is still present. Let \(S_n\) be the amount of drug in the body, in mg, immediately after the \(n\)th dose.

(a)

Show that \(S_3=390\) mg.

(2)
(b)

A doctor wants the amount immediately after a dose to exceed 410 mg. Find the smallest number of doses after which this is true.

(3)
(c)

Find the value that \(S_n\) approaches in the long term, to 3 significant figures, and hence state whether 410 mg can be exceeded.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(S_3=250+250(0.4)+250(0.4)^2\)✓ 1
A1 \(=250+100+40=390\) (AG)✓ 1
Part (b)
M1 \(S_n=\frac{250(1-0.4^n)}{1-0.4}\)✓ 1
M1 solving \(S_n>410\), e.g. \(0.4^n<0.016\) or comparing \(S_4=406\) and \(S_5=412.4\)✓ 1
A1 \(n=5\)✓ 1
Part (c)
A1 \(S_\infty=\frac{250}{1-0.4}=417\) mg✓ 1
R1 yes, since \(417>410\)✓ 1

Exam tip: The sum to infinity formula \(\frac{u_1}{1-r}\) applies only when \(|r|<1\). Here \(r=0.4\), so the amount settles to a finite limit.

226IBMI3H-1-05r20-11Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium5 marks

Calculator

The concentration of an antibiotic in a patient's blood is modelled by \(C=C_0e^{-kt}\), where \(C\) is in mg L−1, \(t\) is in hours after the peak and \(k>0\) is a constant. The half-life of the antibiotic is 5.5 hours.

(a)

Find the value of \(k\) to 3 significant figures.

(2)
(b)

At the peak the concentration is 12 mg L−1. The antibiotic stops being effective when the concentration falls to 2.5 mg L−1. Find the time, to 3 significant figures, for which it is effective. Use the unrounded value of \(k\).

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(e^{-5.5k}=0.5\), i.e. \(k=\frac{\ln 2}{5.5}\)✓ 1
A1 \(k=0.126\)✓ 1
Part (b)
M1 \(2.5=12e^{-kt}\)✓ 1
M1 \(t=\frac{\ln(12/2.5)}{k}\)✓ 1
A1 \(t=12.4\) hours✓ 1

Exam tip: Take the natural logarithm of both sides to bring the exponent down, and keep the unrounded value of \(k\) in your calculator memory to avoid rounding errors.

227IBMI3H-1-04r21-03Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

Three beacons in a game world stand at the points representing the solutions of z3 = 8i in the complex plane. One unit represents 100 m.

(a)

Find the three solutions in the form reiθ, with −π < θ ≤ π.

(3)
(b)

Write the three solutions in exact Cartesian form.

(2)
(c)

Find the distance between two neighbouring beacons, in units, to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 write 8i = 8eiπ/2✓ 1
M1 cube root of modulus gives 2, and arguments (π/2 + 2πk)/3✓ 1
A1 2eiπ/6, 2ei5π/6, 2e−iπ/2✓ 1
Part (b)
A1 √3 + i and −√3 + i✓ 1
A1 −2i✓ 1
Part (c)
M1 distance between two solutions, for example |(√3 + i) − (−√3 + i)|✓ 1
A1 2√3 = 3.46 units✓ 1

Exam tip: The solutions of zⁿ = w are equally spaced on a circle, separated by an angle of 2π/n. Use this to check your arguments.

228IBMI3H-1-04r21-05Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium6 marks

A pharmacist prepares 100 mL of a mixture from three stock solutions, using x mL, y mL and z mL respectively. Drug P is present at 2, 1 and 0 mg per mL in the three solutions, and drug Q at 0, 1 and 3 mg per mL. The mixture must contain 70 mg of drug P and 180 mg of drug Q.

(a)

Write the information as a matrix equation of the form AX = B.

(2)
(b)

Show that the matrix A has an inverse.

(1)
(c)

Hence find the volume of each stock solution to be used.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 A = \(\begin{pmatrix}1&1&1\\2&1&0\\0&1&3\end{pmatrix}\) and X = \(\begin{pmatrix}x\\y\\z\end{pmatrix}\)✓ 1
A1 B = \(\begin{pmatrix}100\\70\\180\end{pmatrix}\)✓ 1
Part (b)
A1 det A = −1 ≠ 0✓ 1
Part (c)
M1 X = A−1B✓ 1
A1 x = 20 mL and y = 30 mL✓ 1
A1 z = 50 mL✓ 1

Exam tip: Check your solution by substituting it into all three original equations. A single slip in a matrix entry is easy to spot this way.

229IBMI3H-1-04r21-07Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

In a 2D game, every point of a map is transformed by the matrix \(\mathbf{M} = \begin{pmatrix}3&1\\2&4\end{pmatrix}\), so that a point with position vector p moves to Mp. Positions are in map units.

(a)

Find det M.

(1)
(b)

Find M−1.

(2)
(c)

A treasure chest is now at the point (11, 14) after the transformation. Find its original position.

(2)
(d)

A triangular region had area 4.5 square map units before the transformation. Find its area after the transformation.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 10✓ 1
Part (b)
M1 swap diagonal entries and change the signs of the others✓ 1
A1 \(\frac{1}{10}\begin{pmatrix}4&-1\\-2&3\end{pmatrix}\)✓ 1
Part (c)
M1 M−1\(\begin{pmatrix}11\\14\end{pmatrix}\)✓ 1
A1 (3, 2)✓ 1
Part (d)
M1 multiply by |det M|✓ 1
A1 45 square units✓ 1

Common error: Multiplying the vector on the wrong side of the matrix. To undo Mp you must pre-multiply by M⁻¹.

230IBMI3H-1-04r21-08Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium8 marks

Two storage tanks, A and B, share a water supply. Each day, 30% of the water in tank A is pumped to tank B and 20% of the water in tank B is pumped to tank A. No water is lost or added. The volumes in the tanks after n days are an and bn in m3.

(a)

Write down the matrix T such that \(\begin{pmatrix}a_{n+1}\\b_{n+1}\end{pmatrix} = \mathbf{T}\begin{pmatrix}a_n\\b_n\end{pmatrix}\).

(2)
(b)

Find the eigenvalues of T.

(2)
(c)

Find an eigenvector of T corresponding to the eigenvalue 1.

(2)
(d)

The tanks hold a total of 5000 m3. Find the volume that tank B will hold in the long run.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 first row 0.7, 0.2✓ 1
A1 second row 0.3, 0.8✓ 1
Part (b)
M1 use of det(T − λI) = 0, giving λ2 − 1.5λ + 0.5 = 0✓ 1
A1 λ = 1 and λ = 0.5✓ 1
Part (c)
M1 solve (T − I)v = 0, for example −0.3a + 0.2b = 0✓ 1
A1 any non-zero multiple of \(\begin{pmatrix}2\\3\end{pmatrix}\)✓ 1
Part (d)
M1 share 5000 in the ratio 2 : 3✓ 1
A1 3000 m3✓ 1

Exam tip: A steady state of a process like this is an eigenvector for the eigenvalue 1. Scale it so the quantities add up to the known total.

231IBMI3H-1-04r22-03Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium8 marks

Four navigation beacons are placed around a circular track. On a map with the centre of the track at the origin and distances measured in km, the beacons are at the points represented by the complex numbers that satisfy \(z^4 = -16\).

(a)

Write \(-16\) in the form \(re^{i\theta}\) with \(r > 0\) and \(0 \le \theta < 2\pi\).

(2)
(b)

Find the four solutions of \(z^4 = -16\) in the form \(re^{i\theta}\) with \(-\pi < \theta \le \pi\).

(3)
(c)

Write the solution in the first quadrant in the form a + bi, giving exact values.

(1)
(d)

Find the distance between two adjacent beacons. Give your answer in km to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(r = 16\)✓ 1
A1 \(\theta = \pi\), so \(-16 = 16e^{i\pi}\)✓ 1
Part (b)
M1 \(z = 16^{1/4}e^{i(\pi + 2k\pi)/4}\), giving modulus 2✓ 1
A1 arguments \(\dfrac{\pi}{4}\), \(\dfrac{3\pi}{4}\) (for \(k = 0, 1\))✓ 1
A1 \(2e^{i\pi/4}\), \(2e^{3i\pi/4}\), \(2e^{-3i\pi/4}\), \(2e^{-i\pi/4}\)✓ 1
Part (c)
A1 \(2\cos\dfrac{\pi}{4} + 2i\sin\dfrac{\pi}{4} = \sqrt{2} + \sqrt{2}\,i\)✓ 1
Part (d)
M1 distance between \(\sqrt{2} + \sqrt{2}i\) and \(-\sqrt{2} + \sqrt{2}i\) (or \(|2\sqrt{2}|\); or chord \(2 \times 2\sin 45^{\circ}\))✓ 1
A1 2.83 km✓ 1

Exam tip: The roots of \(z^n = w\) are equally spaced around a circle, separated by an angle of \(2\pi/n\); use this to check you have found all n of them.

232IBMI3H-1-04r22-05Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

A stadium sells adult tickets for €x, student tickets for €y and child tickets for €z. Three groups pay the following amounts. Group 1 buys 2 adult, 1 student and 3 child tickets for €71. Group 2 buys 1 adult, 4 student and 2 child tickets for €83. Group 3 buys 3 adult, 2 student and 1 child ticket for €86.50.

(a)

Write the information as a matrix equation \(A\mathbf{v} = \mathbf{b}\), where \(\mathbf{v} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}\).

(2)
(b)

Use your calculator to find \(\det A\) and the inverse matrix \(A^{-1}\). Give each entry of \(A^{-1}\) as an exact fraction or as a decimal.

(2)
(c)

Hence find the price of each type of ticket.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(A = \begin{pmatrix} 2 & 1 & 3 \\ 1 & 4 & 2 \\ 3 & 2 & 1 \end{pmatrix}\)✓ 1
A1 \(\mathbf{b} = \begin{pmatrix} 71 \\ 83 \\ 86.5 \end{pmatrix}\)✓ 1
Part (b)
A1 \(\det A = -25\)✓ 1
A1 \(A^{-1} = \dfrac{1}{25}\begin{pmatrix} 0 & -5 & 10 \\ -5 & 7 & 1 \\ 10 & 1 & -7 \end{pmatrix}\) (rows (0, -0.2, 0.4), (-0.2, 0.28, 0.04), (0.4, 0.04, -0.28) in decimals)✓ 1
Part (c)
M1 \(\mathbf{v} = A^{-1}\mathbf{b}\)✓ 1
A1 adult €18✓ 1
A1 student €12.50 and child €7.50✓ 1

Exam tip: Use \(\mathbf{v} = A^{-1}\mathbf{b}\) with the inverse on the left; multiplying on the right is not defined for these shapes.

233IBMI3H-1-05r22-04Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

In a telecoms relay chain, each relay multiplies the signal phasor by \(0.9\,e^{i\pi/6}\) (the amplitude falls by 10 % and the phase advances by \(\frac{\pi}{6}\)). The signal entering the first relay is \(z_0=5e^{i\pi/12}\) volts. Let zn be the signal after n relays.

(a)

Write z3 in the form \(re^{i\theta}\), with \(-\pi<\theta\le\pi\).

(2)
(b)

Find the smallest number of relays after which the amplitude is below 1 volt.

(3)
(c)

Find the real part of z3 in volts, to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(5(0.9)^3e^{i(\pi/12+3\pi/6)}\)✓ 1
A1 \(3.645\,e^{i7\pi/12}\)✓ 1
Part (b)
M1 amplitude \(5(0.9)^n\)✓ 1
M1 solving \(5(0.9)^n<1\) using logarithms or a table✓ 1
A1 n = 16✓ 1
Part (c)
M1 \(3.645\cos\left(\frac{7\pi}{12}\right)\)✓ 1
A1 −0.943 V✓ 1

Exam tip: Multiplying complex numbers in polar form multiplies the moduli and adds the arguments.

234IBMI3H-1-05r22-06Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

A mountain-hut cooperative deposits €18 000 in an account that pays a nominal annual interest rate of 3.2 %, compounded monthly.

(a)

Find the value of the account after 5 years, to the nearest euro.

(2)
(b)

Find how many years it takes for the account to reach €25 000. Give your answer to 3 significant figures.

(3)
(c)

A second account pays 3.2 % per year, compounded annually. Find how many years it takes €18 000 to reach €25 000 in this account, to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(18000\left(1+\frac{0.032}{12}\right)^{60}\) or use of TVM solver✓ 1
A1 €21 119✓ 1
Part (b)
M1 \(25000=18000\left(1+\frac{0.032}{12}\right)^{12t}\)✓ 1
M1 correct method to solve for t (logarithms or solver)✓ 1
A1 10.3 years✓ 1
Part (c)
M1 \(25000=18000(1.032)^t\)✓ 1
A1 10.4 years✓ 1

Exam tip: When solving for the number of periods, convert back to years at the end: here n = 12t for monthly compounding.

235IBMI3H-1-05r22-08Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

A food scientist monitors 100 fermentation vats. Each day a vat is either active (A) or stalled (S). An active vat stays active the next day with probability 0.9. A stalled vat becomes active the next day with probability 0.4. Let \(\mathbf{x}_n=\begin{pmatrix}A_n\\S_n\end{pmatrix}\) be the numbers of active and stalled vats on day n.

(a)

Write down the transition matrix T such that \(\mathbf{x}_{n+1}=T\mathbf{x}_n\).

(2)
(b)

On day 0 all 100 vats are active. Find the expected number of active vats on day 2.

(2)
(c)

In the long run the numbers settle to a steady state. By finding an eigenvector of T for the eigenvalue 1, find the long-term percentage of vats that are active.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 first column (0.9, 0.1)✓ 1
A1 second column (0.4, 0.6), giving \(T=\begin{pmatrix}0.9&0.4\\0.1&0.6\end{pmatrix}\)✓ 1
Part (b)
M1 \(T^2\begin{pmatrix}100\\0\end{pmatrix}\) or two applications of T✓ 1
A1 85✓ 1
Part (c)
M1 \((T-I)\mathbf{v}=\mathbf{0}\), e.g. \(-0.1a+0.4s=0\)✓ 1
A1 a : s = 4 : 1✓ 1
A1 80 %✓ 1

Common error: The steady-state eigenvector must be scaled so that its entries represent a total of 100 % (or 100 vats), not left as (4, 1).

236IBMI3H-1-05r22-13Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium6 marks

Signal power in an optical cable is measured in decibels. The loss in decibels over a cable is \(10\log_{10}\left(\frac{P_{\text{in}}}{P_{\text{out}}}\right)\). A cable loses 0.35 dB per kilometre. A signal of 40 mW enters a 12 km cable.

(a)

Find the total loss over 12 km.

(1)
(b)

Find the output power, in mW, to 3 significant figures.

(3)
(c)

The output power must be at least 5 mW. Find the greatest cable length, in km, to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 4.2 dB✓ 1
Part (b)
M1 \(10\log_{10}\left(\frac{40}{P_{\text{out}}}\right)=4.2\)✓ 1
M1 \(P_{\text{out}}=40\times10^{-0.42}\)✓ 1
A1 15.2 mW✓ 1
Part (c)
M1 \(10\log_{10}8=0.35L\)✓ 1
A1 25.8 km✓ 1

Exam tip: Convert a decibel loss back to a power ratio with \(10^{\text{loss}/10}\); do not subtract decibels from milliwatts.

237IBMI3H-1-04r24-03Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

Two wave components in a bay are modelled by the complex numbers \(z_1=5e^{i\pi/3}\) and \(z_2=8e^{-i\pi/6}\), where each modulus is an amplitude in centimetres.

(a)

Write \(z_1\) and \(z_2\) in the form \(a+bi\), giving the values to 4 significant figures.

(2)
(b)

Find \(z_1+z_2\) in the form \(a+bi\) to 4 significant figures.

(1)
(c)

Find the argument of \(z_1+z_2\) in radians to 3 significant figures. Use the unrounded values of \(z_1\) and \(z_2\).

(2)
(d)

Find the amplitude of the combined wave \(|z_1+z_2|\) to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(z_1=2.500+4.330i\)✓ 1
A1 \(z_2=6.928-4.000i\)✓ 1
Part (b)
A1 \(9.428+0.3301i\) (ECF from (a))✓ 1
Part (c)
M1 \(\arg=\tan^{-1}\!\left(\dfrac{0.3301}{9.428}\right)\)✓ 1
A1 \(0.0350\) rad✓ 1
Part (d)
M1 \(\sqrt{9.428^2+0.3301^2}\) or \(\sqrt{25+64}\) because the components are perpendicular✓ 1
A1 9.43 cm✓ 1

Exam tip: Use the unrounded real and imaginary parts when finding a modulus or argument, to protect the third significant figure.

238IBMI3H-1-04r24-04Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium6 marks

Three support columns of a pavilion are to be placed at the points in an Argand diagram (lengths in metres) that represent the solutions of \(z^3=8i\).

(a)

Find the three solutions in the form \(re^{i\theta}\), with \(-\pi<\theta\le\pi\).

(4)
(b)

The columns form an equilateral triangle. Find the distance between two adjacent columns, to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 write \(8i=8e^{i\pi/2}\) and take the cube root of the modulus✓ 1
A1 \(2e^{i\pi/6}\)✓ 1
A1 \(2e^{5i\pi/6}\)✓ 1
A1 \(2e^{-i\pi/2}\)✓ 1
Part (b)
M1 e.g. distance between \(\sqrt{3}+i\) and \(-\sqrt{3}+i\), or use side \(=2\sqrt{3}\) for circumradius 2✓ 1
A1 3.46 m✓ 1

Exam tip: The \(n\) roots of a complex number are equally spaced around a circle, separated by an angle of \(\dfrac{2\pi}{n}\).

239IBMI3H-1-04r24-07Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

A port charges handling fees of £\(x\), £\(y\) and £\(z\) per container for container types X, Y and Z. On three days the bills were:
Day 1: 2 of X, 1 of Y and 3 of Z, total £470.
Day 2: 1 of X, 4 of Y and 1 of Z, total £490.
Day 3: 3 of X, 2 of Y and 5 of Z, total £770.

(a)

Write down a system of three equations for \(x\), \(y\) and \(z\) and express it as a matrix equation.

(2)
(b)

Solve the system to find the fee for each container type.

(3)
(c)

Find the total bill for a day with 5 containers of each type.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(2x+y+3z=470\), \(x+4y+z=490\), \(3x+2y+5z=770\)✓ 1
A1 \(\begin{pmatrix}2&1&3\\1&4&1\\3&2&5\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}470\\490\\770\end{pmatrix}\)✓ 1
Part (b)
M1 use the inverse matrix or the GDC to solve✓ 1
A1 \(x=120\) and \(y=80\)✓ 1
A1 \(z=50\)✓ 1
Part (c)
M1 \(5(x+y+z)\) (ECF from (b))✓ 1
A1 £1250✓ 1

Exam tip: State the matrix equation clearly before using the GDC; the setup carries marks even if the solution has an error.

240IBMI3H-1-04r24-08Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

The numbers of juvenile and adult fish in a lake in year \(n\) are \(J_n\) and \(A_n\). The populations satisfy \(\binom{J_{n+1}}{A_{n+1}}=\mathbf{L}\binom{J_n}{A_n}\), where \(\mathbf{L}=\begin{pmatrix}0&3\\0.4&0.6\end{pmatrix}\).

(a)

Show that the eigenvalues of \(\mathbf{L}\) satisfy \(\lambda^2-0.6\lambda-1.2=0\).

(2)
(b)

Find the positive eigenvalue, to 4 significant figures, and interpret it in context.

(2)
(c)

Find an eigenvector for this eigenvalue, and hence the long-term ratio of juveniles to adults, giving the ratio \(J:A\) as \(k:1\) with \(k\) to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(\det(\mathbf{L}-\lambda\mathbf{I})=(-\lambda)(0.6-\lambda)-3(0.4)\)✓ 1
A1 \(\lambda^2-0.6\lambda-1.2=0\) (AG)✓ 1
Part (b)
A1 \(\lambda=1.436\)✓ 1
R1 in the long term the total population grows by about 43.6% each year✓ 1
Part (c)
M1 \(-\lambda J+3A=0\) from the first row✓ 1
A1 eigenvector \(\propto\binom{3}{\lambda}\)✓ 1
A1 \(k=\dfrac{3}{1.436\ldots}=2.09\)✓ 1

Common error: In a growth matrix the eigenvalue is the factor by which the population multiplies each step: 1.436 means growth of 43.6%, not 143.6%.

241IBMI3H-1-04r24-11Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

In the plan of a spiral tower, the \(n\)th marker is placed at \(P_n=w^n\) for \(n=0,1,2,\ldots\), where \(w=1.1e^{i\pi/12}\) and lengths are in metres from the centre.

(a)

Write down the modulus and argument of \(P_n\) in terms of \(n\).

(2)
(b)

Find \(P_{12}\) in the form \(a+bi\), giving \(a\) and \(b\) to 3 significant figures.

(2)
(c)

The tower’s outermost marker is the first with \(|P_n|>5\). Find \(n\) for this marker and its distance from the centre, to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(|P_n|=1.1^n\)✓ 1
A1 \(\arg P_n=\dfrac{n\pi}{12}\)✓ 1
Part (b)
M1 \(P_{12}=1.1^{12}e^{i\pi}\)✓ 1
A1 \(-3.14+0i\)✓ 1
Part (c)
M1 solve \(1.1^n>5\), giving \(n>\dfrac{\ln5}{\ln1.1}=16.9\ldots\)✓ 1
A1 \(n=17\)✓ 1
A1 \(|P_{17}|=1.1^{17}=5.05\) m✓ 1

Exam tip: Moduli multiply and arguments add when you take powers, so \(|w^n|=|w|^n\) and \(\arg w^n=n\arg w\).

242IBMI3H-1-04r24-12Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium8 marks

A hatchery mixes \(x\) kg of feed X and \(y\) kg of feed Y. The protein content and energy content requirements give the system
\(2x+ky=b_1\)
\(3x+6y=b_2\)
where \(k\) is a constant and \(b_1\), \(b_2\) are the required totals.

(a)

Find the determinant of the coefficient matrix in terms of \(k\).

(1)
(b)

Find the value of \(k\) for which the system does not have a unique solution.

(2)
(c)

With \(k=4\), \(b_1=16\) and \(b_2=21\), show that the system has no solution.

(2)
(d)

With \(k=5\), \(b_1=16\) and \(b_2=21\), find \(x\) and \(y\).

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(12-3k\)✓ 1
Part (b)
M1 set \(12-3k=0\)✓ 1
A1 \(k=4\)✓ 1
Part (c)
M1 divide to obtain \(x+2y=8\) and \(x+2y=7\)✓ 1
R1 these contradict each other, so the equations are inconsistent✓ 1
Part (d)
M1 use the inverse matrix, Cramer’s rule or the GDC✓ 1
A1 \(x=3\)✓ 1
A1 \(y=2\)✓ 1

Common error: A zero determinant can mean either no solutions or infinitely many; test the right-hand side to decide which.

243IBMI3H-1-04r25-04Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

Five standing stones of an ancient circle lie at the solutions of z5 = 1024 on an Argand diagram, where one unit represents one metre and the origin is the centre of the circle.

(a)

Show that the stones lie on a circle of radius 4 m.

(2)
(b)

Write down all five solutions in the form reiθ with −π < θ ≤ π.

(3)
(c)

Find the distance between two adjacent stones. Give your answer to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 |z|5 = 1024✓ 1
A1 |z| = 10241/5 = 4✓ 1
Part (b)
M1 1024 = 1024e2πki and taking fifth roots✓ 1
A1 arguments 2πk/5 for k = 0, ±1, ±2✓ 1
A1 4e0, 4e±2πi/5, 4e±4πi/5✓ 1
Part (c)
M1 adjacent stones separated by an angle of 2π/5 at the centre (e.g. chord length 2 × 4 × sin(π/5), or |4 − 4e2πi/5|)✓ 1
A1 4.70 m✓ 1

Exam tip: The n solutions of z^n = c are equally spaced on a circle, separated by an angle of 2π/n; use this to find geometric quantities without listing every root.

244IBMI3H-1-04r25-05Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

A cycle shop sells helmets, lights and locks at fixed prices of £h, £l and £k respectively. Three customers pay as follows:
Customer 1: 2 helmets, 3 lights and 1 lock, total £176.
Customer 2: 1 helmet, 2 lights and 3 locks, total £177.
Customer 3: 3 helmets, 1 light and 2 locks, total £217.

(a)

Write the information as a matrix equation of the form Ax = b.

(2)
(b)

Show that this system has a unique solution.

(1)
(c)

Hence find the price of each item, and the cost of buying one of each item.

(4)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 A = [[2, 3, 1], [1, 2, 3], [3, 1, 2]] with x = (h, l, k)T✓ 1
A1 b = (176, 177, 217)T✓ 1
Part (b)
R1 det A = 18 ≠ 0 (so A is invertible)✓ 1
Part (c)
M1 x = A−1b (or solve using the GDC)✓ 1
A1 h = 45✓ 1
A1 l = 18 and k = 32✓ 1
A1 total £95✓ 1

Exam tip: Check a matrix solution by substituting back into one of the original equations; it takes ten seconds and catches arithmetic slips.

245IBMI3H-1-04r25-06Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

A data centre scrambles each pair of whole numbers (x, y) by applying the matrix M = [[2, 1], [3, 2]] to the column vector (x, y)T.

(a)

Find M−1.

(2)
(b)

A scrambled pair is (47, 78). Find the original pair (x, y).

(3)
(c)

A second data centre uses N = [[2, 1], [3, c]]. Find the value of c for which N has no inverse.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 det M = 4 − 3 = 1✓ 1
A1 M−1 = [[2, −1], [−3, 2]]✓ 1
Part (b)
M1 multiply (47, 78)T by M−1✓ 1
A1 x = 2 × 47 − 78 = 16✓ 1
A1 y = −3 × 47 + 2 × 78 = 15✓ 1
Part (c)
M1 det N = 2c − 3 set equal to 0✓ 1
A1 c = 1.5✓ 1

Common error: When multiplying a matrix by a column vector, the matrix goes on the left: M⁻¹ times the vector, not the vector times M⁻¹.

246IBMI3H-1-04r25-07Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

Two cycling clubs, X and Y, compete for members. Each year, the numbers of members (xn, yn) change according to
xn+1 = 1.2xn + 0.4yn
yn+1 = 0.3xn + 1.1yn
This can be written as vn+1 = Mvn with M = [[1.2, 0.4], [0.3, 1.1]].

(a)

Find the eigenvalues of M.

(3)
(b)

Find an eigenvector corresponding to the larger eigenvalue.

(2)
(c)

In the long run, the ratio of members is determined by this eigenvector. Find the long-term proportion of all members who belong to club X. Give your answer to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 det(M − λI) = 0, i.e. (1.2 − λ)(1.1 − λ) − 0.12 = 0✓ 1
A1 λ2 − 2.3λ + 1.2 = 0✓ 1
A1 λ = 1.5 and λ = 0.8✓ 1
Part (b)
M1 solve (M − 1.5I)v = 0, e.g. −0.3x + 0.4y = 0✓ 1
A1 eigenvector (4, 3) (or any multiple)✓ 1
Part (c)
R1 the dominant eigenvalue 1.5 means total membership grows by 50% a year and the ratio approaches 4 : 3 (ECF from (b))✓ 1
A1 4/7 = 0.571✓ 1

Exam tip: The long-term behaviour of a repeated matrix multiplication is governed by the eigenvector of the largest-magnitude eigenvalue; the eigenvalue itself gives the yearly growth factor.

247IBMI3H-1-04r25-09Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

In a data centre, a clock signal is represented by a phasor that is multiplied at each cycle by the complex number w = 1.2eiπ/12. The initial phasor is 1, so after n cycles it is wn.

(a)

Write w in the form a + bi, giving a and b to 3 significant figures.

(2)
(b)

Find the modulus and argument of w6, and hence state the first value of n for which the phasor is purely imaginary.

(2)
(c)

Find the smallest value of n for which wn is a real number, and give that number to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 1.2(cos(π/12) + i sin(π/12))✓ 1
A1 1.16 + 0.311i✓ 1
Part (b)
A1 |w6| = 1.26 = 2.99 and arg = π/2✓ 1
R1 n = 6 is the first value, since the argument nπ/12 first reaches π/2 at n = 6✓ 1
Part (c)
M1 nπ/12 = π (de Moivre), so n = 12✓ 1
A1 w12 = 1.212eiπ✓ 1
A1 −8.92✓ 1

Common error: With de Moivre's theorem, the modulus is raised to the power n and the argument is multiplied by n; forgetting to raise the modulus to the power is a frequent slip.

248IBMI3H-1-04r25-11Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium6 marks

A cycling club buys a fleet of e-bikes for €18 000 using a loan with a nominal annual interest rate of 5.2%, compounded monthly. The loan is repaid in equal payments at the end of each month for 4 years.

(a)

Find the monthly payment, to the nearest cent.

(2)
(b)

Find the total interest paid over the 4 years, using the payment from part (a) rounded to the nearest cent.

(2)
(c)

Find the amount still owed immediately after the 24th payment, using the unrounded monthly payment. Give your answer to the nearest euro.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 use of the finance solver with N = 48, I% = 5.2, PV = 18000, FV = 0, P/Y = C/Y = 12✓ 1
A1 monthly rate 0.052/12, payment €416.16✓ 1
Part (b)
M1 48 × payment − 18 000✓ 1
A1 €1975.68✓ 1
Part (c)
M1 balance after 24 payments from the finance solver, or 18 000 × (1 + 0.052/12)24 − payment × ((1 + 0.052/12)24 − 1)/(0.052/12)✓ 1
A1 €9467✓ 1

Exam tip: With the finance solver, enter the loan amount as positive and the payment as negative (or vice versa) and leave FV as 0 when the loan is fully repaid.

249IBMI3H-1-04r25-13Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

The mass of carbon-14 in organic material decreases according to N = N0 × 0.5t/5730, where t is the time in years since the organism died and N0 is the initial mass.

(a)

An archaeologist finds that a bone contains 62% of its original carbon-14. Find the age of the bone, to 3 significant figures.

(3)
(b)

Show that the model can be written as N = N0e−kt and find k to 3 significant figures.

(2)
(c)

A second sample died 12 000 years ago. Find the percentage of its carbon-14 that remains, to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 0.5t/5730 = 0.62✓ 1
M1 t = 5730 × ln 0.62 / ln 0.5✓ 1
A1 3950 years (3 s.f.; unrounded 3951.75)✓ 1
Part (b)
M1 0.5 = e−k × 5730, so k = ln 2 / 5730✓ 1
A1 k = 1.21 × 10−4✓ 1
Part (c)
M1 0.512000/5730✓ 1
A1 23.4%✓ 1

Exam tip: To solve for an exponent, take logarithms of both sides first; keep the unrounded intermediate values on your GDC.

250IBMI3H-1-08r27-02Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium8 marks

A designer places five beads on a dress so that, on an Argand diagram with 1 unit representing 1 cm, their positions are the solutions of z5 = 1024.

(a)

Solve z5 = 1024, giving the solutions in the form reiθ with −π < θ ≤ π.

(3)
(b)

Write the solution with argument 2π/5 in the form a + bi, with a and b to 3 significant figures.

(2)
(c)

Find the distance between the bead at z = 4 and the bead with argument 2π/5, in centimetres, to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 write 1024 = 1024e2kπi and take fifth roots✓ 1
A1 r = 4✓ 1
A1 z = 4, 4e±2πi/5, 4e±4πi/5✓ 1
Part (b)
A1 a = 4cos(2π/5) = 1.24✓ 1
A1 b = 4sin(2π/5) = 3.80✓ 1
Part (c)
M1 distance = |4 − 4e2πi/5| (or use of the cosine rule with angle 72°)✓ 1
A1 √((4 − 1.236)2 + 3.8042) (or 8 sin 36°)✓ 1
A1 4.70 cm✓ 1

Exam tip: All nth roots of a complex number have the same modulus and are spaced 2π ÷ n apart in argument, so five roots form a regular pentagon.

251IBMI3H-1-08r27-04Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium6 marks

A fish farm mixes three feeds, 1, 2 and 3, to meet a daily requirement. Each kilogram of feed 1 supplies 0.40 kg of protein, 0.10 kg of fat and 0.05 kg of vitamin mix. Each kilogram of feed 2 supplies 0.30 kg of protein, 0.15 kg of fat and 0.02 kg of vitamin mix. Each kilogram of feed 3 supplies 0.20 kg of protein, 0.25 kg of fat and 0.10 kg of vitamin mix. The farm needs exactly 33 kg of protein, 14.5 kg of fat and 5.1 kg of vitamin mix per day. Let x, y and z be the masses, in kg, of feeds 1, 2 and 3 used per day.

(a)

Write the requirements as a matrix equation of the form Av = b.

(2)
(b)

Solve the matrix equation to find x, y and z.

(2)
(c)

Feed 1 costs €1.20 per kg, feed 2 costs €0.95 per kg and feed 3 costs €2.10 per kg. Find the daily cost of the mixture.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 A = [[0.40, 0.30, 0.20], [0.10, 0.15, 0.25], [0.05, 0.02, 0.10]] (rows listed in turn)✓ 1
A1 v = (x, y, z)T and b = (33, 14.5, 5.1)T✓ 1
Part (b)
M1 v = A−1b (or GDC simultaneous equations solver)✓ 1
A1 x = 50, y = 30, z = 20✓ 1
Part (c)
M1 1.20 × 50 + 0.95 × 30 + 2.10 × 20✓ 1
A1 €130.50✓ 1

Exam tip: On a GDC, enter the coefficient matrix and the constants column separately, then compute A⁻¹b; check by substituting back into one of the original equations.

252IBMI3H-1-08r27-05Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium8 marks

The population of a fish stock is modelled in two age classes, juveniles J and adults A (in thousands). Each year
Jn+1 = 2.4An and An+1 = 0.5Jn + 0.6An.
This can be written as xn+1 = Lxn with xn = (Jn, An)T and L = [[0, 2.4], [0.5, 0.6]] (rows listed in turn).

(a)

Find the eigenvalues of L, to 3 significant figures.

(3)
(b)

Find an eigenvector corresponding to the larger eigenvalue, and hence the long-term ratio of juveniles to adults, J : A, in the form k : 1 with k to 3 significant figures.

(3)
(c)

In the long term, by what percentage does the total population increase each year? Give your answer to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 det(L − λI) = 0✓ 1
A1 λ2 − 0.6λ − 1.2 = 0✓ 1
A1 λ = 1.44 and λ = -0.836✓ 1
Part (b)
M1 (L − λI)v = 0 with the larger eigenvalue✓ 1
A1 −λJ + 2.4A = 0, so J ÷ A = 2.4 ÷ λ✓ 1
A1 k = 1.67✓ 1
Part (c)
M1 growth factor per year is the dominant eigenvalue, so increase = λ − 1✓ 1
A1 43.6%✓ 1

Exam tip: In the long run the population vector behaves like the dominant eigenvector, and each year it is multiplied by the dominant eigenvalue; negative eigenvalues cannot be dominant here.

253IBMI3H-1-08r27-07Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium8 marks

In the control circuit of a robot motor, two components in series have complex impedances Z1 = 30 + 40i ohms and Z2 = 20 − 15i ohms. The total impedance is Z = Z1 + Z2. A voltage V = 240 volts (with argument 0) is applied, and the current is I = V ÷ Z.

(a)

Write down Z in the form a + bi.

(1)
(b)

Find the modulus of Z and the argument of Z in radians. Give both to 3 significant figures.

(3)
(c)

Find the amplitude of the current, in amperes, to 3 significant figures.

(2)
(d)

The argument of I is the argument of V minus the argument of Z. Find the angle, in degrees to 3 significant figures, by which the current lags behind the voltage.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 Z = 50 + 25i✓ 1
Part (b)
M1 modulus = √(502 + 252) or argument = tan−1(25 ÷ 50)✓ 1
A1 |Z| = 55.9 ohms✓ 1
A1 arg Z = 0.464 rad✓ 1
Part (c)
M1 |I| = |V| ÷ |Z| = 240 ÷ 55.9…✓ 1
A1 4.29 A✓ 1
Part (d)
M1 arg I = 0 − arg Z (use of arg(V ÷ Z) = arg V − arg Z)✓ 1
A1 lag = 26.6°✓ 1

Exam tip: When dividing complex numbers in polar form, divide the moduli and subtract the arguments; this is quicker than multiplying by the conjugate.

254IBMI3H-1-04r30-05Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium6 marks

The vertical motion of a railway carriage on its suspension is described by an equation whose characteristic equation is \(z^2+5z+9=0\), where \(z\) is a complex number.

(a)

Solve \(z^2+5z+9=0\). Give the solutions in the form \(a\pm bi\) with \(a\) and \(b\) correct to 3 significant figures.

(3)
(b)

Let \(z_1\) be the solution with positive imaginary part. Find \(|z_1|\).

(1)
(c)

Find \(\arg z_1\) in radians, to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 use of quadratic formula or solver✓ 1
A1 \(-2.5\) (real part)✓ 1
A1 \(\pm1.66i\) (imaginary part \(\sqrt{2.75}\))✓ 1
Part (b)
A1 \(|z_1|=3\)✓ 1
Part (c)
M1 \(\pi-\arctan\left(\frac{\sqrt{2.75}}{2.5}\right)\) (second quadrant)✓ 1
A1 2.56✓ 1

Common error: For a number such as \(-2.5+1.66i\) the arctan alone gives a negative angle; add \(\pi\) to place it in the second quadrant.

255IBMI3H-1-04r30-06Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

Three relay satellites orbit a planet. In a model, their positions in the orbital plane, in thousands of kilometres from the planet's centre, are the three solutions of \(z^3=-27\).

(a)

Write \(-27\) in the form \(re^{i\theta}\) with \(\theta=\pi\), and hence find the three solutions of \(z^3=-27\) in the form \(re^{i\theta}\), with \(-\pi<\theta\le\pi\).

(3)
(b)

Write the solution in the fourth quadrant in the form \(a+bi\), with \(a\) and \(b\) to 3 significant figures.

(2)
(c)

Find the distance between two adjacent satellites, in thousands of kilometres, correct to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 modulus \(27^{1/3}=3\)✓ 1
M1 arguments \(\frac{\pi+2k\pi}{3}\)✓ 1
A1 \(3e^{i\pi/3}\), \(3e^{i\pi}\), \(3e^{-i\pi/3}\)✓ 1
Part (b)
M1 \(3\cos\left(-\frac{\pi}{3}\right)\) and \(3\sin\left(-\frac{\pi}{3}\right)\)✓ 1
A1 \(1.50-2.60i\)✓ 1
Part (c)
M1 equilateral triangle with side \(2\times3\sin\frac{\pi}{3}\) (or modulus of difference of two solutions)✓ 1
A1 5.20 (5.196...)✓ 1

Exam tip: The roots of \(z^n=w\) are equally spaced around a circle, separated by an angle of \(2\pi/n\); a quick sketch helps you check the arguments.

256IBMI3H-1-04r30-07Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium6 marks

Two vibration sources act on the floor of a rail carriage. Their vertical displacements, in mm, are \(x_1=5\cos(\omega t)\) and \(x_2=8\cos\left(\omega t+\frac{\pi}{3}\right)\). Each is the real part of a complex number: \(x_1=\mathrm{Re}(z_1e^{i\omega t})\) and \(x_2=\mathrm{Re}(z_2e^{i\omega t})\), where \(z_1=5\) and \(z_2=8e^{i\pi/3}\). The combined displacement is \(x_1+x_2=\mathrm{Re}\left((z_1+z_2)e^{i\omega t}\right)\).

(a)

Find \(z_1+z_2\) in the form \(a+bi\), with \(a\) and \(b\) to 3 significant figures.

(2)
(b)

Find the amplitude \(R=|z_1+z_2|\) of the combined vibration, in mm, to 3 significant figures.

(2)
(c)

Find the phase \(\phi=\arg(z_1+z_2)\) in radians, to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(z_2=8\cos\frac{\pi}{3}+8i\sin\frac{\pi}{3}=4+6.93i\)✓ 1
A1 \(9+6.93i\)✓ 1
Part (b)
M1 \(\sqrt{9^2+(4\sqrt3)^2}\)✓ 1
A1 11.4 mm✓ 1
Part (c)
M1 \(\arctan\left(\frac{4\sqrt3}{9}\right)\)✓ 1
A1 0.656✓ 1

Exam tip: Adding oscillations of the same frequency is much easier in complex form: add the complex amplitudes first, then read off the modulus and argument.

257IBMI3H-1-04r30-08Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

A sports scientist designs a recovery meal from three foods, A, B and C. The protein, carbohydrate and fat content, in grams per serving, is shown in the matrix \(M\), where the rows are protein, carbohydrate and fat and the columns are foods A, B and C: \(M=\begin{pmatrix}25&10&4\\5&40&12\\3&2&20\end{pmatrix}\). A meal of \(x\) servings of A, \(y\) servings of B and \(z\) servings of C must contain exactly 67 g of protein, 76 g of carbohydrate and 19 g of fat.

(a)

Write the requirements as a matrix equation.

(1)
(b)

Find \(\det M\) and state what this tells you about the system.

(1)
(c)

Use \(M^{-1}\) to find the number of servings of each food.

(3)
(d)

Protein and carbohydrate each provide 4 kcal per gram and fat provides 9 kcal per gram. Calculate the energy content of the meal.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(M\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}67\\76\\19\end{pmatrix}\)✓ 1
Part (b)
A1 \(\det M=18320\), which is non-zero so \(M^{-1}\) exists and the solution is unique✓ 1
Part (c)
M1 \(\begin{pmatrix}x\\y\\z\end{pmatrix}=M^{-1}\begin{pmatrix}67\\76\\19\end{pmatrix}\)✓ 1
A1 any two of \(x=2\), \(y=1.5\), \(z=0.5\) correct✓ 1
A1 all three correct✓ 1
Part (d)
M1 \(4\times67+4\times76+9\times19\)✓ 1
A1 743 kcal✓ 1

Common error: \(M^{-1}b\) is the solution; if you enter the matrices in the wrong order the calculator will report a dimension error.

258IBMI3H-1-04r30-09Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

Commuters in a city choose each month between two rail lines, A and B. Each month 20% of the commuters who used line A switch to line B, and 30% of those who used line B switch to line A. Nobody else changes line. The monthly changes are modelled by \(\mathbf{x}_{n+1}=T\mathbf{x}_n\) with \(T=\begin{pmatrix}0.8&0.3\\0.2&0.7\end{pmatrix}\), where \(\mathbf{x}_n\) gives the numbers using lines A and B after n months. Initially 400 commuters use line A and 600 use line B.

(a)

Find the number of commuters on each line after 2 months.

(2)
(b)

Find the eigenvalues of \(T\).

(3)
(c)

Find an eigenvector of \(T\) corresponding to the eigenvalue 1 and hence find the number of commuters on line B in the long term.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(T^2\begin{pmatrix}400\\600\end{pmatrix}\) (or two successive multiplications)✓ 1
A1 550 on line A and 450 on line B✓ 1
Part (b)
M1 set up \(\det(T-\lambda I)=0\): \(\lambda^2-1.5\lambda+0.5=0\)✓ 1
A1 \(\lambda=1\)✓ 1
A1 \(\lambda=0.5\)✓ 1
Part (c)
M1 eigenvector \(\begin{pmatrix}3\\2\end{pmatrix}\) (or multiple) and total of 1000 shared in ratio 3 : 2✓ 1
A1 400 on line B (600 on line A)✓ 1

Exam tip: In a transition-matrix model, the long-term distribution is the eigenvector for eigenvalue 1, scaled so that its entries add to the total number of individuals.

259IBMI3H-1-04r30-10Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

A sports scientist models two indices of an athlete's weekly condition: a performance index \(p\) and a recovery index \(q\). From one week to the next, \(\begin{pmatrix}p_{n+1}\\q_{n+1}\end{pmatrix}=L\begin{pmatrix}p_n\\q_n\end{pmatrix}\) where \(L=\begin{pmatrix}0.5&1.2\\0.3&0\end{pmatrix}\).

(a)

Show that the eigenvalues of \(L\) satisfy \(\lambda^2-0.5\lambda-0.36=0\), and find the two eigenvalues.

(3)
(b)

Find an eigenvector of \(L\) for the eigenvalue 0.9.

(2)
(c)

At week 0, \(p_0=30\) and \(q_0=10\). Find the performance index \(p_{10}\) at week 10, to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(\det(L-\lambda I)=(0.5-\lambda)(-\lambda)-1.2\times0.3\)✓ 1
A1 \(\lambda^2-0.5\lambda-0.36=0\) (AG)✓ 1
A1 \(\lambda=0.9\) and \(\lambda=-0.4\)✓ 1
Part (b)
M1 solve \((L-0.9I)\mathbf{v}=\mathbf{0}\), e.g. \(-0.4a+1.2b=0\)✓ 1
A1 \(\begin{pmatrix}3\\1\end{pmatrix}\) (or any non-zero multiple)✓ 1
Part (c)
M1 \(\begin{pmatrix}30\\10\end{pmatrix}=10\begin{pmatrix}3\\1\end{pmatrix}\) is an eigenvector, so \(p_{10}=30\times0.9^{10}\) (or \(L^{10}\mathbf{x}_0\))✓ 1
A1 10.5✓ 1

Exam tip: If the starting vector is an eigenvector, each step just multiplies it by the eigenvalue, so \(L^n\mathbf{x}_0=\lambda^n\mathbf{x}_0\).

260IBMI3H-1-04r30-11Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

A rail operator sells three types of ticket: adult (€\(a\)), student (€\(s\)) and senior (€\(n\)). On three days the number of passengers of each type and the total revenue were: day 1: 40 adults, 25 students, 10 seniors, €740; day 2: 55 adults, 30 students, 20 seniors, €1020; day 3: 30 adults, 45 students, 15 seniors, €810.

(a)

Write down three equations for \(a\), \(s\) and \(n\).

(2)
(b)

Solve the system to find the three fares.

(3)
(c)

On day 4 the operator carries 50 adults, 20 students and 30 seniors. Find the revenue.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(40a+25s+10n=740\)✓ 1
A1 \(55a+30s+20n=1020\) and \(30a+45s+15n=810\)✓ 1
Part (b)
M1 use of solver or matrix method✓ 1
A1 any two fares correct✓ 1
A1 adult €12, student €8, senior €6✓ 1
Part (c)
M1 \(50\times12+20\times8+30\times6\) (ECF from (b))✓ 1
A1 €940✓ 1

Common error: Keep the order of the unknowns the same in every equation; swapping the order of \(s\) and \(n\) in one line gives a wrong solution.

261IBMI3H-1-06r30-04Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium6 marks

Two signals from a space probe are modelled by the complex numbers z = 6eiπ/3 and w = 2.5e−iπ/4.

(a)

Find zw in the form reiθ, giving θ as an exact fraction of π.

(2)
(b)

Find z3 and write it as a real number.

(2)
(c)

Find the real part of zw, correct to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 multiplies the moduli and adds the arguments✓ 1
A1 15eiπ/12✓ 1
Part (b)
M1 cubes the modulus and triples the argument: 216eiπ✓ 1
A1 −216✓ 1
Part (c)
M1 real part = 15 cos(π/12)✓ 1
A1 14.5✓ 1

Exam tip: In exponential form, multiplication multiplies moduli and adds arguments, so no expansion is needed.

262IBMI3H-1-06r30-05Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

A coffee roaster makes batches of light, medium and dark roast, in numbers x, y and z respectively. Each batch uses a fixed amount of three kinds of green bean. Using all of the 43 kg of Ethiopian beans, 49 kg of Brazilian beans and 58 kg of Sumatran beans gives the equations
2x + y + 3z = 43
x + 3y + z = 49
3x + 2y + 2z = 58

(a)

Write the system in the form Mv = c, where v is the column vector of x, y and z.

(2)
(b)

Solve the system to find the number of each type of batch.

(3)
(c)

Light, medium and dark batches sell for €24, €30 and €38 respectively. Find the total income from selling all the batches.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 M = \(\begin{pmatrix}2&1&3\\1&3&1\\3&2&2\end{pmatrix}\)✓ 1
A1 c = \(\begin{pmatrix}43\\49\\58\end{pmatrix}\)✓ 1
Part (b)
M1 uses v = M−1c or a solver✓ 1
A1 x = 8 and y = 12✓ 1
A1 z = 5✓ 1
Part (c)
M1 8 × 24 + 12 × 30 + 5 × 38✓ 1
A1 €742✓ 1

Exam tip: State the matrix equation clearly before using a calculator, so that the method mark is secure even if the numbers are mis-entered.

263IBMI3H-1-06r30-06Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

An instrument on a space probe is either Active or on Standby at the start of each orbit. If it is Active, then the probability that it is Active on the next orbit is 0.8. If it is on Standby, then the probability that it is Active on the next orbit is 0.3. The transition matrix, acting on the column vector (Active, Standby), is T = \(\begin{pmatrix}0.8&0.3\\0.2&0.7\end{pmatrix}\).

(a)

Show that the eigenvalues of T satisfy λ2 − 1.5λ + 0.5 = 0, and find both eigenvalues.

(3)
(b)

Find an eigenvector of T corresponding to the eigenvalue 1.

(2)
(c)

Hence find the long-term proportion of orbits for which the instrument is Active.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 sets det(T − λI) = 0✓ 1
A1 (0.8 − λ)(0.7 − λ) − 0.06 = λ2 − 1.5λ + 0.5 = 0✓ 1
A1 λ = 1 and λ = 0.5✓ 1
Part (b)
M1 solves (T − I)v = 0, for example −0.2a + 0.3b = 0✓ 1
A1 \(\begin{pmatrix}3\\2\end{pmatrix}\) or any non-zero multiple✓ 1
Part (c)
M1 scales the eigenvector so that its components add to 1: 3/(3 + 2)✓ 1
A1 0.6✓ 1

Exam tip: The long-term state of a transition matrix is the eigenvector for eigenvalue 1, scaled so the probabilities sum to 1.

264IBMI3H-1-06r30-07Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium8 marks

A coffee plantation manager models plants as young or mature. Each year, every mature plant produces 3 new young plants and does not produce any other new plants, 40% of young plants become mature, and 80% of mature plants survive as mature. Young plants do not produce new plants. The numbers (young, mature) in successive years are related by L = \(\begin{pmatrix}0&3\\0.4&0.8\end{pmatrix}\).

(a)

Show that the eigenvalues of L satisfy λ2 − 0.8λ − 1.2 = 0.

(2)
(b)

Find the long-term annual percentage growth in the number of plants, to 3 significant figures.

(2)
(c)

Find an eigenvector for this eigenvalue, in the form (1, k), with k given to 3 significant figures.

(2)
(d)

Hence find the long-term proportion of plants that are young, as a percentage to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 det(L − λI) = (−λ)(0.8 − λ) − 3(0.4)✓ 1
A1 λ2 − 0.8λ − 1.2 = 0✓ 1
Part (b)
M1 takes the positive root λ = (0.8 + √5.44)/2 = 1.5662✓ 1
A1 56.6% per year✓ 1
Part (c)
M1 uses −λ + 3k = 0 so k = λ/3✓ 1
A1 k = 0.522✓ 1
Part (d)
M1 1/(1 + k) using unrounded k = 0.52206✓ 1
A1 65.7%✓ 1

Common error: Always use the unrounded eigenvalue when finding the eigenvector and proportions, to avoid accumulated rounding error.

265IBMI3H-1-06r30-08Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

A bridge-monitoring model requires a complex number z to satisfy z3 = 8i.

(a)

Write 8i in the form reiθ.

(1)
(b)

Find all three solutions, in the form reiθ with −π < θ ≤ π.

(3)
(c)

The three solutions are plotted on an Argand diagram and joined to form a triangle. Find the area of this triangle, to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 8eiπ/2✓ 1
Part (b)
M1 r3 = 8 so r = 2, and 3θ = π/2 + 2kπ✓ 1
A1 2eiπ/6 and 2ei5π/6✓ 1
A1 2e−iπ/2✓ 1
Part (c)
M1 recognises an equilateral triangle with vertices on a circle of radius 2, or finds side length 2√3 = 3.46✓ 1
M1 area = 3 × ½ × 22 × sin 120°, or ½ × (2√3)2 × sin 60°✓ 1
A1 5.20 square units✓ 1

Exam tip: The n roots of a complex number lie equally spaced around a circle centred at the origin.

266IBMI3H-1-06r30-09Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium8 marks

The forces x, y and z, in kN, in three cables of a bridge model satisfy
x + 2y + z = 4
2x + ky + 3z = 10
x + y + 2z = 5
where k is a constant.

(a)

Show that the determinant of the coefficient matrix is k − 3.

(3)
(b)

Show that when k = 3 the system has no solution.

(2)
(c)

Find x, y and z when k = 5.

(2)
(d)

Hence find the total force in the three cables when k = 5.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 expands the determinant of \(\begin{pmatrix}1&2&1\\2&k&3\\1&1&2\end{pmatrix}\) along a row or column✓ 1
A1 1(2k − 3) − 2(4 − 3) + 1(2 − k)✓ 1
A1 = k − 3✓ 1
Part (b)
M1 subtracts the first equation from the second to get x + y + 2z = 6✓ 1
A1 the third equation says x + y + 2z = 5, which is a contradiction✓ 1
Part (c)
M1 uses the inverse matrix or a solver on the system with k = 5✓ 1
A1 x = 1.5, y = 0.5, z = 1.5✓ 1
Part (d)
A1 1.5 + 0.5 + 1.5 = 3.5 kN✓ 1

Common error: A zero determinant means there is no inverse, but the system may have no solution or infinitely many; check the equations to decide which.

267IBMI3H-1-06r30-14Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

A probe fires its thrusters in two manoeuvres. The thruster settings are in the columns of a matrix X and the resulting impulses along two axes, in N s, are in the columns of B = AX, where A = \(\begin{pmatrix}3&-1\\2&4\end{pmatrix}\) and B = \(\begin{pmatrix}6.5&5\\9&-6\end{pmatrix}\).

(a)

Find det A.

(1)
(b)

Find A−1.

(2)
(c)

Hence find X.

(3)
(d)

Find the sum of the four entries of X.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 3×4 − (−1)×2 = 14✓ 1
Part (b)
M1 swaps the leading-diagonal entries and changes signs of the others✓ 1
A1 (1/14)\(\begin{pmatrix}4&1\\-2&3\end{pmatrix}\)✓ 1
Part (c)
M1 X = A−1B in this order✓ 1
A1 at least two correct entries✓ 1
A1 X = \(\begin{pmatrix}2.5&1\\1&-2\end{pmatrix}\)✓ 1
Part (d)
A1 2.5 + 1 + 1 − 2 = 2.5✓ 1

Common error: Matrix multiplication is not commutative: to solve AX = B, multiply by A⁻¹ on the left, giving X = A⁻¹B.

268IBMI3H-1-05r32-02Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium6 marks

Two signals picked up from avalanche rescue beacons are modelled by the complex numbers \(z = 8e^{i\pi/3}\) and \(w = 2e^{i\pi/6}\).

(a)

Find \(\dfrac{z}{w}\) in the form \(re^{i\theta}\).

(2)
(b)

Write \(\dfrac{z}{w}\) in the form \(a+bi\), giving exact values of \(a\) and \(b\).

(2)
(c)

Show that \(z^3\) is a real number and state its value.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 divides the moduli and subtracts the arguments✓ 1
A1 \(4e^{i\pi/6}\)✓ 1
Part (b)
M1 \(4\cos\frac{\pi}{6}+4i\sin\frac{\pi}{6}\) (ECF from (a))✓ 1
A1 \(2\sqrt{3}+2i\)✓ 1
Part (c)
M1 \(z^3 = 8^3e^{i\pi}\) (cubes the modulus, multiplies the argument by 3)✓ 1
A1 \(z^3=-512\) (real since \(e^{i\pi}=-1\))✓ 1

Exam tip: In exponential form, multiplication and division act separately on modulus and argument; powers raise the modulus to the power and multiply the argument.

269IBMI3H-1-05r32-03Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium6 marks

A climber packs two types of energy bar. Each bar of type P contains 12 g of protein and 40 g of carbohydrate. Each bar of type Q contains 8 g of protein and 30 g of carbohydrate. The climber needs a total of exactly 200 g of protein and 690 g of carbohydrate. Let \(p\) and \(q\) be the numbers of bars of type P and type Q.

(a)

Write the conditions \(12p+8q=200\) and \(40p+30q=690\) as a matrix equation.

(2)
(b)

Find the inverse of the matrix of coefficients.

(2)
(c)

Hence find the number of bars of each type.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 matrix of coefficients multiplied by the column vector \((p, q)\)✓ 1
A1 \(\begin{pmatrix}12&8\\40&30\end{pmatrix}\begin{pmatrix}p\\q\end{pmatrix}=\begin{pmatrix}200\\690\end{pmatrix}\)✓ 1
Part (b)
M1 determinant \(12\times30-8\times40=40\)✓ 1
A1 \(\frac{1}{40}\begin{pmatrix}30&-8\\-40&12\end{pmatrix}\)✓ 1
Part (c)
M1 multiplies the inverse by \(\begin{pmatrix}200\\690\end{pmatrix}\) (ECF from (b))✓ 1
A1 \(p=12\), \(q=7\)✓ 1

Exam tip: Pre-multiply by the inverse: \(X = A^{-1}B\). Multiplying in the other order is not defined for these sizes.

270IBMI3H-1-05r32-04Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

On a trek, climbers spend each night in one of two huts, A or B. Each morning 30% of the climbers who slept in A move to B and 20% of those who slept in B move to A; all others stay where they are. The numbers are modelled by \(\begin{pmatrix}A_{n+1}\\B_{n+1}\end{pmatrix}=M\begin{pmatrix}A_n\\B_n\end{pmatrix}\) where \(M=\begin{pmatrix}0.7&0.2\\0.3&0.8\end{pmatrix}\).

(a)

Find the eigenvalues of \(M\).

(3)
(b)

Find an eigenvector of \(M\) corresponding to the eigenvalue \(1\).

(2)
(c)

In the long run, what proportion of the climbers sleep in hut A?

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 sets \(\det(M-\lambda I)=0\)✓ 1
A1 \(\lambda^2-1.5\lambda+0.5=0\)✓ 1
A1 \(\lambda=1\) and \(\lambda=0.5\)✓ 1
Part (b)
M1 solves \((M-I)\mathbf{v}=\mathbf{0}\), e.g. \(-0.3x+0.2y=0\)✓ 1
A1 \(\begin{pmatrix}2\\3\end{pmatrix}\) or any non-zero multiple✓ 1
Part (c)
M1 \(\frac{2}{2+3}\) (ECF from (b))✓ 1
A1 \(0.4\)✓ 1

Exam tip: The eigenvector for eigenvalue 1 of a transition matrix gives the long-run split once it is scaled so the entries sum to 1.

271IBMI3H-1-05r32-06Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

A food scientist blends three ingredients, A, B and C, to make exactly 100 g of a mixture, using \(x\) g of A, \(y\) g of B and \(z\) g of C. Ingredient A is 20% protein and 5% fat, ingredient B is 5% protein and 30% fat, and ingredient C is 10% protein and 50% fat (by mass). The mixture must contain exactly 12.25 g of protein and 25 g of fat.

(a)

Write down a matrix equation \(AX=B\) that represents this situation.

(2)
(b)

Use the inverse of \(A\) to find \(x\), \(y\) and \(z\).

(3)
(c)

The scientist now makes 2.5 kg of the same mixture. Find the mass of ingredient B needed.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 correct matrix form with \(X=(x, y, z)^T\)✓ 1
A1 \(\begin{pmatrix}1&1&1\\0.2&0.05&0.1\\0.05&0.3&0.5\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}100\\12.25\\25\end{pmatrix}\)✓ 1
Part (b)
M1 \(X=A^{-1}B\)✓ 1
A1 \(x=40\) and \(y=35\)✓ 1
A1 \(z=25\)✓ 1
Part (c)
M1 scales by \(\frac{2500}{100}=25\) (ECF from (b))✓ 1
A1 \(875\) g✓ 1

Exam tip: Check a solved system by substituting back into all three original equations, not just one.

272IBMI3H-1-05r32-08Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium5 marks

A telecoms company invests £250 000 in upgrading a mast network. The investment earns interest at a nominal annual rate of 4.8%, compounded quarterly.

(a)

Find the value of the investment after 5 years, to the nearest £.

(2)
(b)

The company instead wants an investment of £250 000 to grow to £350 000 after 7 years with interest compounded quarterly. Find the nominal annual interest rate required, as a percentage to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(250000\left(1+\frac{4.8}{400}\right)^{20}\)✓ 1
A1 £317 359✓ 1
Part (b)
M1 \(250000\left(1+\frac{r}{400}\right)^{28}=350000\)✓ 1
M1 \(r=400\left(1.4^{1/28}-1\right)\) (or solver)✓ 1
A1 \(4.84\%\)✓ 1

Exam tip: With quarterly compounding the number of periods is \(4\times\) years and the rate per period is the nominal rate divided by 4.

273IBMI3H-1-05r32-09Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium5 marks

In a long communications cable, each repeater passes on 88% of the signal power it receives. A signal of 500 mW enters the first repeater.

(a)

Find the power of the signal after it has passed through 6 repeaters, to 3 significant figures.

(2)
(b)

A different cable has a constant fraction \(r\) of power retained by each repeater. Starting from 500 mW, the power is exactly 10 mW after 30 repeaters. Find the percentage of power retained by each repeater, to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(500\times0.88^6\)✓ 1
A1 \(232\) mW✓ 1
Part (b)
M1 \(500r^{30}=10\)✓ 1
M1 \(r=\left(\frac{10}{500}\right)^{1/30}\)✓ 1
A1 \(87.8\%\)✓ 1

Exam tip: Each repeater multiplies by a constant factor, so the situation is a geometric sequence with common ratio equal to the fraction retained.

274IBMI3H-1-05r32-13Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

A telecoms company scrambles pairs of data values before transmission. A pair \((a, b)\), written as a column vector, is multiplied by \(E=\begin{pmatrix}3&2\\1&1\end{pmatrix}\) to give the transmitted pair.

(a)

Find \(\det E\).

(1)
(b)

Find \(E^{-1}\).

(2)
(c)

The transmitted pair is \(\begin{pmatrix}18\\7\end{pmatrix}\). Find the original pair.

(2)
(d)

A faulty encoder uses \(F=\begin{pmatrix}3&k\\1&1\end{pmatrix}\). Find the value of \(k\) for which the original pair cannot be recovered from the transmitted pair.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(3\times1-2\times1=1\)✓ 1
Part (b)
M1 swaps the leading diagonal entries and changes the signs of the others, dividing by the determinant✓ 1
A1 \(\begin{pmatrix}1&-2\\-1&3\end{pmatrix}\)✓ 1
Part (c)
M1 \(E^{-1}\begin{pmatrix}18\\7\end{pmatrix}\) (ECF from (b))✓ 1
A1 \((4, 3)\)✓ 1
Part (d)
M1 \(\det F=3-k=0\)✓ 1
A1 \(k=3\)✓ 1

Common error: A matrix has no inverse exactly when its determinant is zero; the original data can then not be recovered uniquely.

275IBMI3H-1-08r33-02Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium6 marks

In a theatre, seven spotlights are fixed above a circular stage. Taking the centre of the stage as the origin of an Argand diagram, with 1 unit = 1 metre, the positions of the lights are the solutions of z7 = 78125.

(a)

Find all seven solutions, giving them in the form reiθ with 0 ≤ θ < 2π.

(3)
(b)

Find the distance between two neighbouring spotlights, to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 |z| = 781251/7 = 5✓ 1
M1 argument of 78125 is 0 (or 2πk), so θ = 2πk/7✓ 1
A1 z = 5e2πki/7 for k = 0, 1, …, 6✓ 1
Part (b)
M1 distance = |5 − 5e2πi/7| or chord of a circle of radius 5 subtending 2π/7 at the centre✓ 1
M1 10 sin(π/7) (or √(5² + 5² − 2 × 5 × 5 × cos(2π/7)))✓ 1
A1 4.34 m✓ 1

Exam tip: The n-th roots of a positive real number are equally spaced around a circle, separated by an angle of 2π/n; check that you list n roots.

276IBMI3H-1-08r33-03Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

A cargo ship leaves port at the origin of an Argand diagram, with the real axis pointing east and the imaginary axis pointing north (1 unit = 1 km). It sails 40 km on a bearing of 060°, then 30 km on a bearing of 150°.

(a)

Write the displacement for each leg as a complex number in Cartesian form, to 3 significant figures.

(2)
(b)

Find the final distance of the ship from the port.

(2)
(c)

Find the bearing of the ship from the port, to the nearest 0.1°.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 first leg 40eiπ/6 = 34.6 + 20.0i (bearing 060° is 30° above the east direction)✓ 1
A1 second leg 30e−iπ/3 = 15.0 − 26.0i✓ 1
Part (b)
M1 sum 49.64 − 5.981i and its modulus, or Pythagoras since the legs are perpendicular✓ 1
A1 50.0 km✓ 1
Part (c)
M1 argument of the final position is −arctan(5.981/49.64) = −6.87° (north-of-east angle)✓ 1
M1 bearing = 90° − (−6.87°)✓ 1
A1 096.9° (accept 96.9°)✓ 1

Common error: A bearing is measured clockwise from north, but the argument is measured anticlockwise from the real axis; convert with bearing = 90° − argument.

277IBMI3H-1-08r33-05Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium5 marks

On a revolving stage a prop sits at position z = 6eiπ/6 (metres from the centre of the stage in an Argand diagram). The stage rotates anticlockwise at 3 revolutions per minute, so the position of the prop after t seconds is z(t) = 6ei(π/6 + πt/10).

(a)

Find the position of the prop after 4 seconds in Cartesian form, to 3 significant figures.

(2)
(b)

Find the first time after t = 0 at which the real part of z(t) is −3.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 argument = π/6 + 4π/10 = 17π/30 (102°) and conversion to Cartesian form✓ 1
A1 −1.25 + 5.87i✓ 1
Part (b)
M1 6 cos(π/6 + πt/10) = −3, so cos θ = −0.5✓ 1
M1 smallest θ > π/6 is 2π/3, then π/6 + πt/10 = 2π/3✓ 1
A1 t = 5 seconds✓ 1

Exam tip: Choose the smallest solution of the trigonometric equation that is larger than the starting argument, since the angle only increases with time.

278IBMI3H-1-08r33-09Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium8 marks

The daily emission rates of gas and ash (in tonnes per day, relative to a baseline) from a volcanic vent are modelled over a short period by xn+1 = Mxn, where xn = (gas, ash) as a column vector and M has rows (1.1, k) and (0.2, 0.8). One eigenvalue of M is 1.2.

(a)

Find the value of k.

(3)
(b)

Find the other eigenvalue of M.

(2)
(c)

Find an eigenvector for the eigenvalue 1.2. Hence, in the long term, if the gas emission is 480 tonnes per day, find the ash emission.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 det(M − 1.2I) = 0✓ 1
M1 (1.1 − 1.2)(0.8 − 1.2) − 0.2k = 0, i.e. 0.04 − 0.2k = 0✓ 1
A1 k = 0.2✓ 1
Part (b)
M1 sum of eigenvalues equals the trace 1.9, or solving the characteristic equation✓ 1
A1 0.7✓ 1
Part (c)
M1 solve (M − 1.2I)v = 0, e.g. −0.1a + 0.2b = 0✓ 1
A1 eigenvector (2, 1) (any multiple)✓ 1
A1 long-term ash emission = 240 tonnes per day (the dominant eigenvalue fixes the gas to ash ratio at 2 : 1)✓ 1

Exam tip: The long-term direction is set by the eigenvector of the eigenvalue with the largest absolute value, because its component grows fastest.

279IBMI3H-1-08r33-11Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium5 marks

A stage designer maps points on a backdrop using the matrix M with rows (1.2, 0.4) and (−0.3, 0.9), where a point (x, y) in metres is sent to M(x, y)T. A triangular panel of area 6 m2 is transformed by M.

(a)

Find the area of the image of the panel.

(2)
(b)

The point P is mapped to P′ = (2.8, 0.3). Find the coordinates of P.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 area scale factor |det M| = 1.08 + 0.12 = 1.2✓ 1
A1 7.2 m2✓ 1
Part (b)
M1 M−1 = (1/1.2) × rows (0.9, −0.4) and (0.3, 1.2)✓ 1
M1 P = M−1(2.8, 0.3)T✓ 1
A1 P = (2, 1)✓ 1

Exam tip: The area scale factor of a transformation matrix is the absolute value of its determinant; a negative determinant only reverses orientation.

280IBMI3H-1-08r33-12Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium5 marks

Two seismic waves arrive at a station. In the complex-number model, the first has amplitude 4 and phase π/3, and is represented by z1 = 4eiπ/3. The second has amplitude R and phase π/2, and is represented by z2 = Reiπ/2. The combined wave is represented by z1 + z2.

(a)

Write z1 in Cartesian form.

(2)
(b)

The combined wave has amplitude 7. Find the value of R to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 4cos(π/3) + 4i sin(π/3)✓ 1
A1 2 + 2√3i (2 + 3.46i)✓ 1
Part (b)
M1 z1 + z2 = 2 + (2√3 + R)i and modulus equation 22 + (2√3 + R)2 = 49✓ 1
M1 2√3 + R = √45 (positive root since R > 0)✓ 1
A1 R = 3.24✓ 1

Exam tip: To add waves with different phases, convert each to Cartesian form, add the real and imaginary parts separately, then take the modulus.

281IBMI3H-1-08r33-13Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium6 marks

A shipping company buys a container crane for €420000 using a loan with an interest rate of 4.8% per year, compounded monthly. The loan is repaid by equal payments at the end of each month for 12 years.

(a)

Find the monthly repayment, giving your answer to the nearest cent.

(2)
(b)

Find the total interest paid over the 12 years. Use the unrounded monthly repayment.

(2)
(c)

Find the amount still owed after 5 years, to the nearest cent. Use the unrounded monthly repayment.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 use of a finance solver with N = 144, I% = 4.8, PV = 420000, FV = 0, P/Y = C/Y = 12 (or the annuity formula with r = 0.004)✓ 1
A1 €3842.54✓ 1
Part (b)
M1 144 × 3842.5355 − 420000✓ 1
A1 €133325.11✓ 1
Part (c)
M1 finance solver with N = 60 and the unrounded payment, or balance = 420000 × 1.00460 − payment × (1.00460 − 1) ÷ 0.004✓ 1
A1 €273682.46✓ 1

Common error: Use the unrounded repayment from your calculator in later parts; rounding it to the nearest cent first can change the final cents.

282IBMI3H-1-08r34-03Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

Three spotlights are positioned on a circular rig above a stage. Their positions, in metres, are represented on an Argand diagram by the three solutions of \(z^3 = 8i\).

(a)

Write \(8i\) in the form \(re^{i\theta}\) with \(r>0\) and \(0\le\theta<2\pi\).

(2)
(b)

Find the three solutions of \(z^3=8i\), giving them in the form \(re^{i\theta}\) with \(-\pi<\theta\le\pi\).

(3)
(c)

Hence find the distance between the two spotlights at \(2e^{i\pi/6}\) and \(2e^{i5\pi/6}\), to three significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 modulus 8 identified✓ 1
A1 \(8e^{i\pi/2}\)✓ 1
Part (b)
M1 cube root of the modulus, \(r=2\), and argument \(\frac{\pi/2 + 2k\pi}{3}\)✓ 1
A1 \(2e^{i\pi/6}\) and \(2e^{i5\pi/6}\)✓ 1
A1 \(2e^{-i\pi/2}\)✓ 1
Part (c)
M1 Cartesian forms \(\sqrt3 + i\) and \(-\sqrt3 + i\) (or the chord length \(2\times 2\sin60°\))✓ 1
A1 \(2\sqrt3 = 3.46\) m✓ 1

Common error: Forgetting to add 2kπ before dividing the argument by 3 loses two of the three roots.

283IBMI3H-1-08r34-04Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium6 marks

On a volcanic island, each day is classified as quiet or active. After a quiet day, the next day is quiet with probability 0.8. After an active day, the next day is active with probability 0.7. The transition matrix is \(T = \begin{pmatrix}0.8 & 0.3\\0.2 & 0.7\end{pmatrix}\), acting on the column vector (proportion quiet, proportion active).

(a)

Show that the eigenvalues of T are 1 and 0.5.

(3)
(b)

Find an eigenvector of T for \(\lambda=1\), and hence find the long-term proportion of days that are quiet.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(\det(T-\lambda I)=0\)✓ 1
A1 \(\lambda^2 - 1.5\lambda + 0.5 = 0\)✓ 1
A1 \(\lambda = 1\) and \(\lambda = 0.5\) (AG)✓ 1
Part (b)
M1 solve \((T - I)\mathbf{v} = \mathbf{0}\), e.g. \(-0.2q + 0.3a = 0\)✓ 1
A1 eigenvector proportional to \(\begin{pmatrix}3\\2\end{pmatrix}\)✓ 1
A1 scale so the entries sum to 1: quiet proportion \(\frac{3}{3+2} = 0.6\)✓ 1

Exam tip: A long-term (steady-state) vector is an eigenvector for λ = 1 scaled so that its entries sum to 1.

284IBMI3H-1-08r34-05Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium8 marks

A colony of seabirds on a volcanic island has J juveniles and A adults. Each year, every adult produces on average 2 juveniles, 50% of juveniles survive to become adults, and 80% of adults survive. So \(\begin{pmatrix}J_{n+1}\\A_{n+1}\end{pmatrix} = L\begin{pmatrix}J_n\\A_n\end{pmatrix}\) where \(L = \begin{pmatrix}0 & 2\\0.5 & 0.8\end{pmatrix}\).

(a)

Show that the eigenvalues of L satisfy \(\lambda^2 - 0.8\lambda - 1 = 0\).

(3)
(b)

Find the dominant (largest) eigenvalue of L, to three significant figures.

(2)
(c)

Interpret the dominant eigenvalue in the context of the colony.

(1)
(d)

Find the long-term ratio of juveniles to adults, \(J:A\), in the form \(k:1\). Give k to three significant figures, using the unrounded eigenvalue.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(\det(L - \lambda I) = 0\)✓ 1
A1 \((-\lambda)(0.8-\lambda) - 2\times0.5 = 0\)✓ 1
A1 \(\lambda^2 - 0.8\lambda - 1 = 0\) (AG)✓ 1
Part (b)
M1 solve the quadratic (formula or GDC)✓ 1
A1 1.48 (1.4770…)✓ 1
Part (c)
R1 in the long term, the total population grows by a factor of about 1.48 each year (about 48% growth per year)✓ 1
Part (d)
M1 first row of \(L\mathbf{v}=\lambda\mathbf{v}\) gives \(2A = \lambda J\)✓ 1
A1 \(k = \frac{2}{\lambda} = 1.35\) (1.3541…)✓ 1

Exam tip: Use the unrounded eigenvalue when finding the eigenvector; rounding first can change the third significant figure.

285IBMI3H-1-08r34-07Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

A theatre sells stalls, circle and balcony tickets at fixed prices of €s, €c and €b respectively. Over three performances the sales were:
Friday: 100 stalls, 60 circle, 40 balcony tickets, total takings €7140.
Saturday: 80 stalls, 90 circle, 70 balcony tickets, total takings €7740.
Sunday: 120 stalls, 50 circle, 100 balcony tickets, total takings €8800.

(a)

Write down a matrix equation that represents this information.

(2)
(b)

Use the inverse of the coefficient matrix to find the three ticket prices.

(3)
(c)

The next performance sells 90 stalls, 70 circle and 60 balcony tickets. Find the total takings.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 coefficient matrix with rows \((100,60,40)\), \((80,90,70)\), \((120,50,100)\)✓ 1
A1 \(\begin{pmatrix}100&60&40\\80&90&70\\120&50&100\end{pmatrix}\begin{pmatrix}s\\c\\b\end{pmatrix}=\begin{pmatrix}7140\\7740\\8800\end{pmatrix}\)✓ 1
Part (b)
M1 \(\begin{pmatrix}s\\c\\b\end{pmatrix}=A^{-1}\begin{pmatrix}7140\\7740\\8800\end{pmatrix}\) (GDC)✓ 1
A1 stalls €45✓ 1
A1 circle €32 and balcony €18✓ 1
Part (c)
M1 \(90\times45 + 70\times32 + 60\times18\) (ECF from (b))✓ 1
A1 €7370✓ 1

Exam tip: Write the matrix in the same row order as the equations and check one row against your answers.

286IBMI3H-1-08r34-10Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

In a theatre, a sound pulse of unit amplitude is followed by an infinite series of echoes. Each echo has 0.8 times the amplitude of the previous one and is delayed in phase by \(\frac{\pi}{4}\). The combined signal is represented by the complex series \(1 + r + r^2 + r^3 + \cdots\) where \(r = 0.8e^{i\pi/4}\).

(a)

Show that the series converges.

(2)
(b)

Find the sum to infinity in the form \(a + bi\), giving a and b to three significant figures.

(3)
(c)

Find the amplitude of the combined signal, to three significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(|r| = 0.8\)✓ 1
R1 \(|r| < 1\) so the geometric series converges✓ 1
Part (b)
M1 \(S = \frac{1}{1 - r}\) with \(r = 0.5657 + 0.5657i\)✓ 1
M1 multiply numerator and denominator by the conjugate of \(1 - r\) (or use GDC)✓ 1
A1 \(S = 0.854 + 1.11i\) (imaginary part 1.1122…)✓ 1
Part (c)
M1 \(|S| = \sqrt{a^2 + b^2}\) using unrounded values, or \(\frac{1}{|1-r|}\)✓ 1
A1 1.40✓ 1

Common error: The modulus of a sum is not the sum of the moduli: find the complex sum first, then take its modulus.

287IBMI3H-1-08r34-12Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium6 marks

A shipping company borrows €250 000 to refit a vessel. The loan has a nominal annual interest rate of 4.8%, compounded monthly, and is repaid with equal payments at the end of each month for 8 years.

(a)

Use your graphic display calculator to find the monthly payment, to the nearest cent.

(2)
(b)

Using your rounded answer to (a), find the total amount repaid.

(2)
(c)

Find the total interest paid over the 8 years.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 TVM entries: \(N=96\), \(I\%=4.8\), \(PV=250000\), \(FV=0\), \(P/Y=12\), \(C/Y=12\)✓ 1
A1 €3141.23 (3141.2298… before rounding)✓ 1
Part (b)
M1 \(96 \times\) monthly payment✓ 1
A1 €301558.08✓ 1
Part (c)
M1 total repaid \(- 250000\)✓ 1
A1 €51558.08✓ 1

Common error: Entering 4.8 as the interest rate but leaving P/Y and C/Y at their default values gives the wrong monthly payment.

288IBMI3H-1-08r34-13Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

Successive eruptions of a volcano deposit layers of ash. The thickness of the first layer is 24 mm and each later layer is 65% as thick as the layer before it.

(a)

Find the thickness of the third layer, in mm.

(2)
(b)

Find the least number of eruptions needed for the total thickness of ash to exceed 60 mm.

(3)
(c)

Find the greatest possible total thickness if the eruptions continued indefinitely, to three significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(24\times0.65^2\)✓ 1
A1 10.14 mm✓ 1
Part (b)
M1 \(S_n = \frac{24(1 - 0.65^n)}{1 - 0.65}\)✓ 1
M1 solve \(S_n > 60\) (GDC or logarithms), or compare values of \(S_n\)✓ 1
A1 5 eruptions (\(S_4 = 56.33\), \(S_5 = 60.62\))✓ 1
Part (c)
M1 \(S_\infty = \frac{24}{1 - 0.65}\)✓ 1
A1 68.6 mm✓ 1

Exam tip: For 'least n' questions, state the values either side of the threshold to justify the integer answer.

289IBMI3H-1-06r35-02Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

Two seismometers record ground motion from the same tremor. The recordings are modelled by the complex numbers z1 = 4ei π/6 and z2 = 3ei 2π/3, where the modulus is the amplitude in mm and the argument is the phase in radians.

(a)

Write z1 and z2 in the form a + bi, with a and b to 3 significant figures.

(2)
(b)

Find z1 + z2 in the form a + bi.

(1)
(c)

Show, using the arguments of z1 and z2, that |z1 + z2| = 5 exactly.

(2)
(d)

Find the argument of z1 + z2 in radians, to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 z1 = 4 cos(π/6) + 4i sin(π/6) = 3.46 + 2.00i✓ 1
A1 z2 = 3 cos(2π/3) + 3i sin(2π/3) = −1.50 + 2.60i✓ 1
Part (b)
A1 z1 + z2 = 1.96 + 4.60i (ECF from (a))✓ 1
Part (c)
M1 difference of arguments = 2π/3 − π/6 = π/2, so the two vectors on the Argand diagram are at right angles✓ 1
R1 by Pythagoras, |z1 + z2| = √(42 + 32) = 5✓ 1
Part (d)
M1 arg = π/6 + tan−1(3 ÷ 4) (or tan−1(4.60 ÷ 1.96))✓ 1
A1 arg = 1.17 rad✓ 1

Exam tip: Sketching the two complex numbers on an Argand diagram shows when a right angle or a symmetry makes the algebra simpler.

290IBMI3H-1-06r35-09Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium8 marks

A port manager plans cargo handling using the system Mkx = r, where Mk = [[1, 2, 1], [2, k, 3], [1, 1, 2]], x = (x, y, z)T and r = (8, 21, 9)T.

(a)

Show that det Mk = k − 3.

(2)
(b)

Write down the value of k for which Mk has no inverse.

(1)
(c)

Show that when k = 3 the system has no solution.

(2)
(d)

When k = 5, find the value of z.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 expansion: 1(2k − 3) − 2(4 − 3) + 1(2 − k)✓ 1
A1 2k − 3 − 2 + 2 − k = k − 3✓ 1
Part (b)
A1 k = 3✓ 1
Part (c)
M1 when k = 3, row 2 = row 1 + row 3 in M3, so the left-hand sides satisfy 2x + 3y + 3z = (x + 2y + z) + (x + y + 2z)✓ 1
R1 but 8 + 9 = 17 ≠ 21, so the equations are inconsistent and there is no solution✓ 1
Part (d)
M1 x = M5−1r (det M5 = 2) or equivalent GDC method✓ 1
A1 x = 1 and y = 2✓ 1
A1 z = 3✓ 1

Exam tip: A zero determinant means there is either no solution or infinitely many; compare the right-hand sides to decide which.

291IBMI3H-1-06r35-10Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium8 marks

During an eruption, the thickness of the ash layer measured at the nth site along a line leading away from the vent forms a geometric sequence. The first site, 1 km from the vent, has a layer 48 mm thick, and each successive site has a layer 85% as thick as the one before.

(a)

Find the thickness of the layer at the fifth site, to 3 significant figures.

(2)
(b)

Find the sum of the thicknesses recorded at the first 10 sites, to 3 significant figures.

(2)
(c)

Find the smallest value of n for which the thickness is less than 2.0 mm.

(3)
(d)

Find the thickness of the layer at this site, to 3 significant figures.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 48 × 0.854✓ 1
A1 25.1 mm✓ 1
Part (b)
M1 48(1 − 0.8510) ÷ (1 − 0.85)✓ 1
A1 257 mm✓ 1
Part (c)
M1 48 × 0.85n−1 < 2✓ 1
M1 n − 1 > ln(2 ÷ 48) ÷ ln(0.85) = 19.55…✓ 1
A1 n = 21✓ 1
Part (d)
A1 48 × 0.8520 = 1.86 mm✓ 1

Common error: When solving an inequality with a base less than 1, take logarithms carefully: dividing by ln 0.85 (a negative number) reverses the inequality.

292IBMI3H-1-06r35-13Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

The energy E (joules) released by a volcanic tremor of magnitude M satisfies log10 E = 5.2 + 1.3M.

(a)

Find the energy released by a tremor of magnitude 4.0, giving your answer in the form a × 10k with 1 ≤ a < 10 and a to 3 significant figures.

(2)
(b)

Find the increase in magnitude that makes the energy released 100 times as great, to 3 significant figures.

(2)
(c)

Find how many times greater the energy of a magnitude 5.5 tremor is than that of a magnitude 4.0 tremor, to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 E = 105.2 + 1.3 × 4.0 = 1010.4✓ 1
A1 E = 2.51 × 1010 J✓ 1
Part (b)
M1 log10 100 = 2 = 1.3 × ΔM✓ 1
A1 ΔM = 1.54✓ 1
Part (c)
M1 difference in log10 E = 1.3 × (5.5 − 4.0) = 1.95✓ 1
M1 ratio = 101.95✓ 1
A1 89.1✓ 1

Exam tip: A difference in log values corresponds to a ratio of the original quantities, so subtract the logs first and then undo the logarithm.

293IBMI3H-1-07r36-04Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

A cyclist rides anticlockwise around a circular velodrome of radius 30 m. Taking the centre of the track as the origin of an Argand diagram (distances in metres), the position of the cyclist at time t seconds is \(z = 30e^{0.15it}\). Use radian mode.

(a)

Find the time taken for the cyclist to complete one lap, correct to 3 significant figures.

(2)
(b)

Find the position of the cyclist when \(t=10\), giving the real and imaginary parts correct to 3 significant figures.

(3)
(c)

Find the first time, after the start, at which the real part of \(z\) equals \(-15\). Give your answer correct to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(0.15T=2\pi\)✓ 1
A1 \(T=41.9\) s✓ 1
Part (b)
M1 \(z=30(\cos 1.5+i\sin 1.5)\)✓ 1
A1 real part \(2.12\)✓ 1
A1 imaginary part \(29.9\)✓ 1
Part (c)
M1 \(30\cos(0.15t)=-15\) so \(0.15t=\frac{2\pi}{3}\)✓ 1
A1 \(t=14.0\) s✓ 1

Common error: In part (c) the first solution of \(\cos\theta=-\tfrac12\) is \(\tfrac{2\pi}{3}\), not \(\tfrac{4\pi}{3}\) (which is the second time the cyclist passes that position).

294IBMI3H-1-07r36-05Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

Three post-holes are found at positions, in metres from a central stone, given by the solutions of \(z^3 = 64i\) in the complex plane.

(a)

Write \(64i\) in the form \(re^{i\theta}\) where \(r>0\) and \(0\le\theta<2\pi\).

(2)
(b)

Find the three solutions of \(z^3=64i\), giving each in the form \(a+bi\) with \(a\) and \(b\) correct to 3 significant figures.

(3)
(c)

Find the distance between any two of the post-holes, correct to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(r=|64i|=64\)✓ 1
A1 \(\theta=\frac{\pi}{2}\), so \(64e^{i\pi/2}\)✓ 1
Part (b)
M1 modulus \(\sqrt[3]{64}=4\) and arguments \(\frac{\pi}{6}+\frac{2\pi k}{3}\), \(k=0,1,2\)✓ 1
A1 two correct solutions from \(3.46+2.00i\), \(-3.46+2.00i\), \(-4i\)✓ 1
A1 all three correct✓ 1
Part (c)
M1 distance between \(3.46+2i\) and \(-3.46+2i\), or \(2\times4\sin\left(\frac{\pi}{3}\right)\)✓ 1
A1 \(6.93\) m✓ 1

Exam tip: The three roots of \(z^3=c\) are equally spaced around a circle, so the arguments differ by \(\frac{2\pi}{3}\) and the post-holes form an equilateral triangle.

295IBMI3H-1-07r36-06Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

A company moves data between two data centres, X and Y, every night. Each night 20% of the data at X moves to Y and 30% of the data at Y moves to X; the rest stays where it is. If \(X_n\) and \(Y_n\) are the amounts of data (in terabytes) at the two sites at the start of night n, then \(\begin{pmatrix}X_{n+1}\\Y_{n+1}\end{pmatrix}=\mathbf{M}\begin{pmatrix}X_n\\Y_n\end{pmatrix}\) where \(\mathbf{M}=\begin{pmatrix}0.8&0.3\\0.2&0.7\end{pmatrix}\).

(a)

Find the eigenvalues of \(\mathbf{M}\).

(3)
(b)

Find an eigenvector of \(\mathbf{M}\) corresponding to the eigenvalue 1.

(2)
(c)

After many nights the data settles into a steady distribution. Use your answer to (b) to find the percentage of the total data held at X in this steady distribution.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 sets \(\det(\mathbf{M}-\lambda\mathbf{I})=0\)✓ 1
A1 \(\lambda^2-1.5\lambda+0.5=0\), or equivalent✓ 1
A1 \(\lambda=1\) and \(\lambda=0.5\)✓ 1
Part (b)
M1 solves \((\mathbf{M}-\mathbf{I})\mathbf{v}=\mathbf{0}\), for example \(-0.2x+0.3y=0\)✓ 1
A1 \(\begin{pmatrix}3\\2\end{pmatrix}\), or any non-zero multiple✓ 1
Part (c)
R1 the steady distribution is proportional to the eigenvector for \(\lambda=1\), since the contribution from \(\lambda=0.5\) tends to zero✓ 1
A1 \(\frac{3}{3+2}=60\%\) (ECF from (b))✓ 1

Common error: An eigenvector is only defined up to a non-zero multiple, so (3, 2) and (1.5, 1) both describe the same steady split; always convert it to a proportion of the total.

296IBMI3H-1-07r36-07Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

A data centre uses three server models, A, B and C. Let a, b and c be the power drawn, in watts, by one server of each model. Rack 1 holds 2 of model A, 3 of model B and 1 of model C and draws 920 W. Rack 2 holds 1 A, 4 B and 2 C and draws 1080 W. Rack 3 holds 3 A, 1 B and 5 C and draws 960 W.

(a)

Write the information as a matrix equation \(\mathbf{N}\begin{pmatrix}a\\b\\c\end{pmatrix}=\mathbf{p}\).

(2)
(b)

Find \(\det\mathbf{N}\) and explain why the system has a unique solution.

(2)
(c)

Hence find \(a\), \(b\) and \(c\).

(2)
(d)

A fourth rack holds 1 server of model A, 6 of model B and 3 of model C. Find the power it draws.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(\mathbf{N}=\begin{pmatrix}2&3&1\\1&4&2\\3&1&5\end{pmatrix}\)✓ 1
A1 \(\mathbf{p}=\begin{pmatrix}920\\1080\\960\end{pmatrix}\)✓ 1
Part (b)
A1 \(\det\mathbf{N}=28\)✓ 1
R1 the determinant is non-zero, so \(\mathbf{N}^{-1}\) exists and the solution is unique✓ 1
Part (c)
M1 attempts \(\mathbf{N}^{-1}\mathbf{p}\) (or solves the system using a calculator)✓ 1
A1 \(a=120\), \(b=200\), \(c=80\)✓ 1
Part (d)
A1 \(1(120)+6(200)+3(80)=1560\) W (ECF from (c))✓ 1

Exam tip: A determinant of zero means the matrix has no inverse, so the system has either no solution or infinitely many; a non-zero determinant guarantees exactly one.

297IBMI3H-1-07r36-08Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

Two alternating voltages in a data-centre power supply are represented by the complex numbers \(w=4e^{i\pi/3}\) and \(v=2e^{-i\pi/6}\), in volts.

(a)

Find \(wv\), giving your answer in the form \(re^{i\theta}\).

(2)
(b)

Find \(\frac{w}{v}\), giving your answer in the form \(re^{i\theta}\).

(2)
(c)

Find \(|w+v|\), correct to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 multiplies the moduli and adds the arguments✓ 1
A1 \(8e^{i\pi/6}\)✓ 1
Part (b)
M1 divides the moduli and subtracts the arguments✓ 1
A1 \(2e^{i\pi/2}\)✓ 1
Part (c)
M1 converts to Cartesian form, \(w=2+3.46i\) and \(v=1.73-i\)✓ 1
M1 adds the parts: \(3.73+2.46i\), then finds the modulus✓ 1
A1 \(|w+v|=\sqrt{20}=4.47\) V✓ 1

Common error: \(|w+v|\) is not \(|w|+|v|\); the modulus of a sum depends on the angle between the two numbers, so add in Cartesian form first.

298IBMI3H-1-07r36-13Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

Archaeologists date a bone fragment using carbon-14. The amount of carbon-14 remaining after t years is modelled by \(N=N_0e^{-kt}\), where \(N_0\) is the initial amount and k is a positive constant. The half-life of carbon-14 is 5730 years.

(a)

Show that \(k=\frac{\ln 2}{5730}\), and find the value of k correct to 3 significant figures.

(2)
(b)

The fragment contains 62% of its original carbon-14. Find the age of the fragment, to the nearest 10 years. Use the exact value \(k=\frac{\ln 2}{5730}\).

(3)
(c)

A second fragment is 9000 years old. Find the percentage of its original carbon-14 remaining, correct to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(\frac12N_0=N_0e^{-5730k}\) so \(5730k=\ln 2\)✓ 1
A1 \(k=1.21\times10^{-4}\)✓ 1
Part (b)
M1 \(0.62=e^{-kt}\)✓ 1
M1 \(t=\frac{\ln(1/0.62)}{k}\)✓ 1
A1 \(3950\) years✓ 1
Part (c)
M1 \(e^{-9000k}\)✓ 1
A1 \(33.7\%\)✓ 1

Exam tip: Keep the exact value of k in your calculator memory rather than the rounded 1.21 × 10⁻⁴, otherwise an age near 4000 years can be out by tens of years.

299IBMI3H-1-06r37-02Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium8 marks

Calculator

A feature wall in a building plan has a fixing point at P, represented by the complex number z = 12 + 5i (coordinates in metres, relative to the origin O). The plan is rotated by 30° anticlockwise about O, taking P to P′ represented by w.

(a)

Write z in the form reiθ, with θ in radians to 3 significant figures.

(3)
(b)

Write down w in the form reiφ, with φ in radians to 3 significant figures.

(2)
(c)

Find the coordinates of P′ to 3 significant figures, using unrounded values from your working.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 r = √(122 + 52)✓ 1
A1 r = 13✓ 1
A1 θ = 0.395✓ 1
Part (b)
M1 adds π/6 (= 0.5236) to the argument (ECF from (a))✓ 1
A1 w = 13e0.918i✓ 1
Part (c)
M1 w = 13(cos(0.9184) + i sin(0.9184)), or (12 + 5i)(cos 30° + i sin 30°)✓ 1
A1 x = 7.89✓ 1
A1 y = 10.3✓ 1

Common error: Mixing degrees and radians: the argument must be in radians when written in the form re^(iθ), so convert 30° to π/6 first.

300IBMI3H-1-06r37-12Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium8 marks

Calculator

A marine laboratory harvests algae from a tank each day. The mass harvested on day 1 is 2.5 kg, and the mass harvested each day is 12% greater than the mass harvested the previous day.

(a)

Find the mass harvested on day 5, to 3 significant figures.

(2)
(b)

Find the smallest number of days after which the total mass harvested exceeds 200 kg.

(4)
(c)

Find the total mass harvested in the first 10 days, to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 2.5 × 1.124✓ 1
A1 3.93 kg✓ 1
Part (b)
M1 uses geometric series sum 2.5(1.12n − 1)/0.12✓ 1
M1 forms 2.5(1.12n − 1)/0.12 > 200, giving 1.12n > 10.6✓ 1
M1 takes logarithms: n > 20.8✓ 1
A1 21 days✓ 1
Part (c)
M1 2.5(1.1210 − 1)/0.12✓ 1
A1 43.9 kg✓ 1

Exam tip: When solving r^n > k with r > 1, take logarithms and keep the inequality direction; round up to the next whole number of days.

301IBMI3H-1-06r39-03Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium8 marks

A wind turbine has three equally spaced blades, each 45 m long. With the origin at the hub, the positions of the three blade tips (in metres, as complex numbers in the plane of rotation) are the solutions of \(z^3=91\,125\,i\).

(a)

Write \(91\,125\,i\) in the form \(re^{i\theta}\).

(2)
(b)

Hence find the three solutions of \(z^3=91\,125\,i\), giving each in the form \(a+bi\) with \(a\) and \(b\) to 3 significant figures.

(4)
(c)

Find the distance between the tips of two adjacent blades, to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(r=91\,125\) (\(=45^3\))✓ 1
A1 \(\theta=\dfrac{\pi}{2}\)✓ 1
Part (b)
M1 takes the cube root of the modulus and divides the argument by 3, adding \(\dfrac{2\pi}{3}\) for each other root✓ 1
A1 \(45e^{i\pi/6}\) and \(45e^{5i\pi/6}\)✓ 1
A1 \(39.0+22.5i\) and \(-39.0+22.5i\)✓ 1
A1 \(-45i\)✓ 1
Part (c)
M1 uses \(|z_1-z_2|\) with the first two solutions (or chord \(2\times45\sin60^\circ\))✓ 1
A1 \(77.9\) m✓ 1

Exam tip: The n solutions of \(z^n=w\) are equally spaced around a circle of radius \(|w|^{1/n}\), separated by an angle of \(\frac{2\pi}{n}\).

302IBMI3H-1-06r39-04Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium6 marks

A library buys three kinds of book: hardbacks (€\(h\) each), paperbacks (€\(p\) each) and audiobooks (€\(a\) each). Three orders were placed:
Order 1: 2 hardbacks, 3 paperbacks, 1 audiobook, total cost €75.
Order 2: 4 hardbacks, 1 paperback, 2 audiobooks, total cost €105.
Order 3: 1 hardback, 2 paperbacks, 3 audiobooks, total cost €72.

(a)

Write the information as a matrix equation \(\mathbf{A}\mathbf{x}=\mathbf{b}\).

(2)
(b)

Use the inverse of \(\mathbf{A}\) to find the price of each type of book.

(3)
(c)

The library plans to buy 8 hardbacks, 15 paperbacks and 10 audiobooks. Find the total cost.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(\mathbf{A}=\begin{pmatrix}2&3&1\\4&1&2\\1&2&3\end{pmatrix}\) and \(\mathbf{x}=\begin{pmatrix}h\\p\\a\end{pmatrix}\)✓ 1
A1 \(\mathbf{b}=\begin{pmatrix}75\\105\\72\end{pmatrix}\)✓ 1
Part (b)
M1 uses \(\mathbf{x}=\mathbf{A}^{-1}\mathbf{b}\)✓ 1
A1 any two of the prices correct✓ 1
A1 hardback €18, paperback €9, audiobook €12✓ 1
Part (c)
A1 \(8\times18+15\times9+10\times12=\) €399✓ 1

Exam tip: Enter the matrix and the right-hand side directly into the calculator and check one equation by hand to catch typing slips.

303IBMI3H-1-06r39-06Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium8 marks

A library keeps stock of two categories of book. Each year the stock vector \(\begin{pmatrix}F\\N\end{pmatrix}\), giving the numbers of fiction and non-fiction books, is updated by purchases and withdrawals according to \(\begin{pmatrix}F_{n+1}\\N_{n+1}\end{pmatrix}=\mathbf{M}\begin{pmatrix}F_n\\N_n\end{pmatrix}\), where \(\mathbf{M}=\begin{pmatrix}0.9&0.2\\0.1&1.0\end{pmatrix}\).

(a)

Find the eigenvalues of \(\mathbf{M}\).

(3)
(b)

Find an eigenvector for each eigenvalue.

(2)
(c)

In year 0 the library holds 200 books in each category. Use an eigenvector to find the number of books in each category after 5 years, to the nearest whole number.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 forms \(\det(\mathbf{M}-\lambda\mathbf{I})=0\)✓ 1
A1 \(\lambda^2-1.9\lambda+0.88=0\)✓ 1
A1 \(\lambda=1.1\) and \(\lambda=0.8\)✓ 1
Part (b)
A1 for \(\lambda=1.1\): \(\begin{pmatrix}1\\1\end{pmatrix}\) (or any multiple)✓ 1
A1 for \(\lambda=0.8\): \(\begin{pmatrix}2\\-1\end{pmatrix}\) (or any multiple)✓ 1
Part (c)
M1 recognises \(\begin{pmatrix}200\\200\end{pmatrix}\) is \(200\) times the eigenvector for 1.1✓ 1
M1 multiplies by \(1.1^5\)✓ 1
A1 322 books in each category✓ 1

Exam tip: If the starting vector is an eigenvector, applying the matrix n times simply multiplies it by the eigenvalue raised to the power n.

304IBMI3H-1-06r39-08Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium6 marks

Two cables in a wind farm have impedances, in ohms, \(z_1=3+4i\) and \(z_2=6-2i\). When the cables are connected in parallel, the combined impedance is \(z=\dfrac{z_1z_2}{z_1+z_2}\).

(a)

Find \(z_1+z_2\) and \(z_1z_2\), giving each in the form \(a+bi\).

(2)
(b)

Hence find \(z\) in the form \(a+bi\), with \(a\) and \(b\) to 3 significant figures.

(3)
(c)

Find the magnitude of the combined impedance, to 3 significant figures.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(z_1+z_2=9+2i\)✓ 1
A1 \(z_1z_2=26+18i\)✓ 1
Part (b)
M1 multiplies numerator and denominator by the conjugate \(9-2i\)✓ 1
A1 \(\dfrac{270+110i}{85}\)✓ 1
A1 \(3.18+1.29i\)✓ 1
Part (c)
A1 \(|z|=\dfrac{\sqrt{270^2+110^2}}{85}=3.43\) Ω✓ 1

Exam tip: To divide complex numbers, multiply the top and bottom by the conjugate of the denominator so the denominator becomes real.

305IBMI3H-1-06r39-10Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium8 marks

At a swimming pool, swimmers choose a slow or fast lane each week. Of the swimmers in the slow lane, 90% choose the slow lane again and 10% move to the fast lane. Of those in the fast lane, 40% move to the slow lane and 60% stay. Let the vector \(\begin{pmatrix}s_n\\f_n\end{pmatrix}\) give the expected numbers in the slow and fast lanes in week \(n\), so \(\begin{pmatrix}s_{n+1}\\f_{n+1}\end{pmatrix}=\mathbf{P}\begin{pmatrix}s_n\\f_n\end{pmatrix}\) with \(\mathbf{P}=\begin{pmatrix}0.9&0.4\\0.1&0.6\end{pmatrix}\). In week 0 there are 100 swimmers, all in the slow lane.

(a)

Show that 1 and 0.5 are eigenvalues of \(\mathbf{P}\).

(2)
(b)

Find an eigenvector for each eigenvalue.

(2)
(c)

Write \(\begin{pmatrix}100\\0\end{pmatrix}\) as a combination of these eigenvectors.

(2)
(d)

Hence find an expression for \(f_n\) and state the number of swimmers in the fast lane in the long term.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(\det(\mathbf{P}-\lambda\mathbf{I})=\lambda^2-1.5\lambda+0.5=0\)✓ 1
A1 \((\lambda-1)(\lambda-0.5)=0\), so \(\lambda=1\) and \(\lambda=0.5\)✓ 1
Part (b)
A1 \(\lambda=1\): \(\begin{pmatrix}4\\1\end{pmatrix}\) (or any multiple)✓ 1
A1 \(\lambda=0.5\): \(\begin{pmatrix}1\\-1\end{pmatrix}\) (or any multiple)✓ 1
Part (c)
M1 solves \(4a+b=100\) and \(a-b=0\)✓ 1
A1 \(a=b=20\)✓ 1
Part (d)
A1 \(f_n=20\times1-20\times0.5^n=20-20(0.5)^n\) (ECF from (c))✓ 1
A1 as \(n\to\infty\), \(0.5^n\to0\), so the long-term number is 20✓ 1

Common error: Eigenvectors can be scaled by any non-zero constant, so choose a scaling that makes the combination of starting vector simple, and check that it reproduces the initial vector.

306IBMI3H-1-06r39-13Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

The number of digital loans per month at a library is modelled by \(N=450\times1.12^{t}\), where \(t\) is the time in years since the service began.

(a)

Find the number of digital loans per month after 5 years, to the nearest whole number.

(2)
(b)

Find the time taken for the loans to reach 2000 per month, giving your answer to 3 significant figures.

(3)
(c)

The model can be written as \(N=450e^{kt}\). Find the value of \(k\) to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(450\times1.12^5\)✓ 1
A1 793✓ 1
Part (b)
M1 sets \(450\times1.12^t=2000\)✓ 1
M1 takes logarithms: \(t=\dfrac{\ln(2000/450)}{\ln1.12}\)✓ 1
A1 \(13.2\) years✓ 1
Part (c)
M1 equates \(e^{k}=1.12\)✓ 1
A1 \(k=\ln1.12=0.113\)✓ 1

Exam tip: Taking logarithms of both sides brings the variable down from the exponent; \(\ln1.12\) is just a number, so keep it unrounded until the end.

307IBMI3H-1-04r40-03Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

A warehouse drone starts at the point represented by the complex number 8 (in metres) on an Argand diagram whose origin is the charging station. At each step, the drone's position is multiplied by w = 0.9eiπ/6, so that after n steps its position is 8wn.

(a)

Find the position of the drone after 5 steps, in the form a + bi with a and b correct to 3 significant figures.

(2)
(b)

Find the least number of steps after which the drone is within 1 m of the charging station.

(2)
(c)

The drone's position is first a positive real number again after m steps. Find m, and find the drone's distance from the charging station at that time, correct to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 8 × 0.95 e5iπ/6✓ 1
A1 -4.09 + 2.36i✓ 1
Part (b)
M1 8 × 0.9n < 1, so n > ln(1/8) ÷ ln 0.9✓ 1
A1 n > 19.74, so 20 steps✓ 1
Part (c)
M1 argument mπ/6 = 2π✓ 1
A1 m = 12✓ 1
A1 8 × 0.912 = 2.26 m✓ 1

Common error: Multiplying by 0.9eiπ/⁶ shrinks the modulus by 0.9 and rotates by π/6 at every step; do not add 0.9 or π/6 once.

308IBMI3H-1-04r40-04Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

An architect places the three corners of a triangular pavilion at the points that represent the solutions of z3 = 216i on an Argand diagram, where each unit is 1 metre.

(a)

Find the three solutions in the form reiθ, where r > 0 and −π < θ ≤ π. Give θ as a multiple of π.

(3)
(b)

The solutions form an equilateral triangle. Find the length of one side of the pavilion, in metres, correct to 3 significant figures.

(2)
(c)

Find the floor area of the pavilion, in m2, correct to 3 significant figures. Use the unrounded side length.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 r3 = 216, so r = 6 and 3θ = π/2 + 2πk✓ 1
A1 6eiπ/6 and 6ei5π/6✓ 1
A1 6e−iπ/2 (accept equivalent forms for all three roots)✓ 1
Part (b)
M1 distance between 6eiπ/6 (≈ 5.196 + 3i) and 6ei5π/6 (≈ −5.196 + 3i), or 6 × 2 sin 60°✓ 1
A1 10.4 m✓ 1
Part (c)
M1 area = ½ × side2 × sin 60° (or (√3/4) × side2)✓ 1
A1 46.8 m2✓ 1

Exam tip: An equation zⁿ = c has n solutions with equal modulus, spaced 2π/n apart in argument: list all n.

309IBMI3H-1-04r40-05Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

A marine biologist mixes three foods for a large aquarium: krill, algae pellets and fish meal. Each kilogram of krill supplies 3 units of protein, 1 unit of fat and 2 units of minerals. Each kilogram of algae pellets supplies 2, 4 and 1 units respectively, and each kilogram of fish meal supplies 1, 2 and 3 units respectively. The daily requirement is 20 units of protein, 20 units of fat and 17 units of minerals. Let k, a and f be the masses, in kg, of krill, algae pellets and fish meal fed each day.

(a)

Write the requirements as a matrix equation of the form AX = B, where X is the column matrix with entries k, a, f.

(2)
(b)

Find A−1.

(1)
(c)

Hence find the daily mass of each food.

(2)
(d)

Krill costs €6.50 per kg, algae pellets €4.20 per kg and fish meal €5.10 per kg. Find the daily cost of the food.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 3 × 3 matrix with rows (3, 2, 1), (1, 4, 2), (2, 1, 3)✓ 1
A1 X = (k, a, f)T and B = (20, 20, 17)T✓ 1
Part (b)
A1 (1/25) × [[10, −5, 0], [1, 7, −5], [−7, 1, 10]] (accept decimal entries 0.4, −0.2, 0, 0.04, 0.28, −0.2, −0.28, 0.04, 0.4)✓ 1
Part (c)
M1 X = A−1B (or use technology to solve the system)✓ 1
A1 krill 4 kg, algae pellets 3 kg, fish meal 2 kg✓ 1
Part (d)
M1 4 × 6.50 + 3 × 4.20 + 2 × 5.10 (accept ECF from (c))✓ 1
A1 €48.80✓ 1

Exam tip: Check a matrix solution by substituting back into at least one original equation: 3(4) + 2(3) + 1(2) = 20.

310IBMI3H-1-04r40-09Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

An architect scales and rotates a rectangular courtyard plan with corners (0, 0), (3, 0), (3, 2) and (0, 2), measured in metres. The transformation is given by the matrix M = [[1.6, −1.2], [1.2, 1.6]], so that a point (x, y) is mapped to M(x, y)T.

(a)

Find the image of the corner (3, 2).

(2)
(b)

Find M−1, and hence find the point on the original plan that is mapped to (8, 4).

(3)
(c)

The transformation is an enlargement with scale factor k followed by an anticlockwise rotation through an angle θ about the origin. Find k and find θ in degrees correct to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 M(3, 2)T✓ 1
A1 (2.4, 6.8)✓ 1
Part (b)
A1 det M = 4, so M−1 = (1/4)[[1.6, 1.2], [−1.2, 1.6]]✓ 1
M1 M−1(8, 4)T✓ 1
A1 (4.4, −0.8)✓ 1
Part (c)
A1 k = 2 (e.g. k2 = det M = 4, or |1.6 + 1.2i| = 2)✓ 1
A1 θ = 36.9° (cos θ = 0.8, sin θ = 0.6)✓ 1

Common error: Since the matrix has a positive determinant, the area of the plan is multiplied by det M = 4 (not by k = 2): areas scale with the square of the scale factor.

311IBMI3H-1-04r40-12Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

The seats in a stadium stand are in rows. The first row has 48 seats and each row has 4 more seats than the row in front of it, so the numbers of seats form an arithmetic sequence.

(a)

Find the number of seats in row 25.

(2)
(b)

Find the total number of seats in the first 20 rows.

(2)
(c)

The stand has 5040 seats in total. Find the number of rows.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 48 + 24 × 4✓ 1
A1 144✓ 1
Part (b)
M1 S20 = 20/2 (2 × 48 + 19 × 4)✓ 1
A1 1720✓ 1
Part (c)
M1 n/2 (2 × 48 + (n − 1) × 4) = 5040✓ 1
A1 2n2 + 46n − 5040 = 0 (or n2 + 23n − 2520 = 0)✓ 1
A1 n = 40 (reject n = −63)✓ 1

Exam tip: When solving a sum equation, a quadratic arises: reject any negative or non-whole root because the number of rows must be a positive integer.

312IBMI3H-1-06r40-03Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium8 marks

The corners of a pentagonal pavilion are modelled by the solutions of \(z^5 = 32\) in the complex plane, with distances in metres.

(a)

Write the five solutions in the form \(2e^{i\phi}\), with \(0 \le \phi < 2\pi\).

(3)
(b)

Find the distance between two adjacent corners, giving your answer to 3 significant figures.

(3)
(c)

Hence find the perimeter of the pavilion, using the unrounded value from part (b), to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(z^5 = 32e^{2\pi ik}\), so \(z = 32^{1/5}e^{2\pi ik/5}\)✓ 1
A1 modulus \(32^{1/5} = 2\)✓ 1
A1 \(\phi = 0,\ \frac{2\pi}{5},\ \frac{4\pi}{5},\ \frac{6\pi}{5},\ \frac{8\pi}{5}\)✓ 1
Part (b)
M1 attempt to use the cosine rule or chord length with angle \(\frac{2\pi}{5}\) at the centre✓ 1
A1 \(\sqrt{2^2+2^2-2(2)(2)\cos\frac{2\pi}{5}}\) or \(4\sin\frac{\pi}{5}\)✓ 1
A1 2.35 m✓ 1
Part (c)
M1 \(5 \times 4\sin\frac{\pi}{5}\)✓ 1
A1 11.8 m✓ 1

Exam tip: The roots of \(z^n = c\) are equally spaced around a circle, separated by an angle of \(\frac{2\pi}{n}\).

313IBMI3H-1-06r40-04Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium8 marks

A logistics firm uses vans (capacity a pallets), trucks (capacity b pallets) and trailers (capacity c pallets). Three routes, each fully loaded, carry the following:
Route X: 2 vans, 1 truck and 1 trailer carry 16 pallets.
Route Y: 1 van, 3 trucks and 2 trailers carry 25 pallets.
Route Z: 3 vans, 1 truck and 4 trailers carry 29 pallets.

(a)

Write the information as a matrix equation \(\mathbf{A}\mathbf{x} = \mathbf{b}\).

(2)
(b)

Find \(\det\mathbf{A}\) and explain why the system has a unique solution.

(2)
(c)

Solve the system to find a, b and c.

(2)
(d)

A depot has 6 vans, 4 trucks and 7 trailers. Find the total number of pallets they can carry when fully loaded.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(\mathbf{A} = \begin{pmatrix}2&1&1\\1&3&2\\3&1&4\end{pmatrix}\) and \(\mathbf{x} = \begin{pmatrix}a\\b\\c\end{pmatrix}\)✓ 1
A1 \(\mathbf{b} = \begin{pmatrix}16\\25\\29\end{pmatrix}\)✓ 1
Part (b)
A1 \(\det\mathbf{A} = 14\)✓ 1
R1 non-zero determinant, so \(\mathbf{A}^{-1}\) exists and the solution is unique✓ 1
Part (c)
M1 \(\mathbf{x} = \mathbf{A}^{-1}\mathbf{b}\)✓ 1
A1 \(a=4,\ b=5,\ c=3\)✓ 1
Part (d)
M1 \(6(4)+4(5)+7(3)\) ECF from (c)✓ 1
A1 65 pallets✓ 1

Exam tip: Check a matrix-inverse solution by substituting it back into at least one of the original equations.

314IBMI3H-1-06r40-06Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

A shipping company moves containers between two depots, A and B. Each night, 30% of the containers at A move to B and 20% of the containers at B move to A. The rest stay where they are. The transition matrix is \(\mathbf{P} = \begin{pmatrix}0.7 & 0.2\\0.3 & 0.8\end{pmatrix}\).

(a)

Find the eigenvalues of \(\mathbf{P}\).

(3)
(b)

Find an eigenvector of \(\mathbf{P}\) corresponding to \(\lambda = 1\).

(2)
(c)

Find the proportion of containers at depot A in the long run.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(\det(\mathbf{P}-\lambda\mathbf{I})=0\)✓ 1
A1 \(\lambda^2 - 1.5\lambda + 0.5 = 0\)✓ 1
A1 \(\lambda = 1\) and \(\lambda = 0.5\)✓ 1
Part (b)
M1 solve \((\mathbf{P}-\mathbf{I})\mathbf{v}=\mathbf{0}\), giving \(-0.3x + 0.2y = 0\)✓ 1
A1 \(\mathbf{v} = \begin{pmatrix}2\\3\end{pmatrix}\) (or any multiple)✓ 1
Part (c)
M1 \(\frac{2}{2+3}\) ECF from (b)✓ 1
A1 0.4✓ 1

Exam tip: The eigenvector for eigenvalue 1 of a transition matrix gives the steady state once its entries are scaled to sum to 1.

315IBMI3H-1-06r40-08Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

Two ocean-wave components are modelled by the phasors \(z = 2e^{i\pi/4}\) and \(w = 3e^{i\pi/6}\), where each modulus is an amplitude in metres.

(a)

Write \(zw\) in the form \(re^{i\theta}\).

(2)
(b)

Find \(z^4\) in the form \(a + bi\).

(2)
(c)

Find the smallest positive integer n for which \(z^n\) is a positive real number, and state the value of \(z^n\).

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 multiply the moduli and add the arguments✓ 1
A1 \(zw = 6e^{i5\pi/12}\)✓ 1
Part (b)
M1 \(z^4 = 2^4 e^{i\pi}\)✓ 1
A1 \(-16 + 0i\)✓ 1
Part (c)
M1 require \(\frac{n\pi}{4} = 2\pi k\) for an integer k✓ 1
A1 \(n = 8\)✓ 1
A1 \(z^8 = 2^8 = 256\)✓ 1

Exam tip: For \(re^{i\theta}\) raised to a power n, the modulus becomes \(r^n\) and the argument becomes \(n\theta\).

316IBMI3H-1-06r40-09Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium8 marks

An architect maps plan coordinates of a canopy by a transformation T, which is a rotation of \(90^\circ\) anticlockwise about the origin followed by an enlargement with scale factor 2 and centre the origin.

(a)

Write down the matrix \(\mathbf{M}\) that represents T.

(2)
(b)

Find the point that T maps to \((6, -4)\).

(3)
(c)

A glass panel has area 3.5 m2. Find the area of its image under T.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 rotation matrix \(\begin{pmatrix}0&-1\\1&0\end{pmatrix}\) combined with the factor 2✓ 1
A1 \(\mathbf{M} = \begin{pmatrix}0&-2\\2&0\end{pmatrix}\)✓ 1
Part (b)
M1 attempt to find \(\mathbf{M}^{-1}\)✓ 1
A1 \(\mathbf{M}^{-1} = \begin{pmatrix}0&0.5\\-0.5&0\end{pmatrix}\)✓ 1
A1 \((-2, -3)\)✓ 1
Part (c)
M1 area scale factor \(|\det\mathbf{M}|\)✓ 1
A1 \(\det\mathbf{M} = 4\)✓ 1
A1 \(4 \times 3.5 = 14\) m2✓ 1

Exam tip: The area of an image is the original area multiplied by the absolute value of the determinant of the transformation matrix.

317IBMI3H-1-06r40-11Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium8 marks

An algae bloom covers 8 m2 of the sea surface on day 1. Each day, the area is 12% larger than on the previous day.

(a)

Write down the common ratio of the geometric sequence of daily areas.

(1)
(b)

Find the area of the bloom on day 10, to 3 significant figures.

(2)
(c)

Find the first day on which the area exceeds 100 m2.

(3)
(d)

Find the sum of the daily areas for the first 10 days, to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 1.12✓ 1
Part (b)
M1 \(8 \times 1.12^{9}\)✓ 1
A1 22.2 m2✓ 1
Part (c)
M1 \(8 \times 1.12^{n-1} > 100\)✓ 1
M1 \(n - 1 > \frac{\ln 12.5}{\ln 1.12} = 22.3\)✓ 1
A1 day 24✓ 1
Part (d)
M1 \(\frac{8\left(1.12^{10}-1\right)}{0.12}\)✓ 1
A1 140 m2 (140.39 before rounding)✓ 1

Common error: For \(u_n = u_1 r^{n-1}\), the exponent is \(n-1\), not n; the 10th term uses \(r^9\).

318IBMI3H-1-06r40-12Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

The light intensity at depth d metres in a lake is modelled by \(I = 100e^{-0.08d}\), where I is the percentage of the surface intensity.

(a)

Find the intensity at a depth of 20 m, to 3 significant figures.

(2)
(b)

Show that the model can be written as \(I = 100b^{d}\), and find the value of b to 3 significant figures.

(2)
(c)

Find the depth at which the intensity falls to 5% of the surface value, to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(100e^{-0.08(20)} = 100e^{-1.6}\)✓ 1
A1 20.2%✓ 1
Part (b)
M1 \(e^{-0.08d} = \left(e^{-0.08}\right)^d\)✓ 1
A1 \(b = 0.923\)✓ 1
Part (c)
M1 \(100e^{-0.08d} = 5\)✓ 1
M1 \(-0.08d = \ln 0.05\), so \(d = \frac{\ln 20}{0.08}\)✓ 1
A1 37.4 m✓ 1

Exam tip: Isolate the exponential term before taking natural logarithms: divide by 100 first, then apply ln.

319IBMI3H-1-07r40-01Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium6 marks

A community theatre pays €12 000 into a fund for new stage lighting. The fund earns interest at a nominal annual rate of 3.6%, compounded quarterly. No further money is added or withdrawn.

(a)

Calculate the value of the fund after 7 years. Give your answer to the nearest cent (2 decimal places).

(3)
(b)

Find how many years it takes for the fund to reach €18 000. Give your answer to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 use of compound interest formula with \(r=3.6\), \(k=4\) and \(n=7\)✓ 1
A1 correct substitution \(12000\left(1+\frac{3.6}{400}\right)^{28}\)✓ 1
A1 €15 421.76✓ 1
Part (b)
M1 equation \(12000(1.009)^{4t}=18000\) or equivalent with a calculator solver or graph✓ 1
(M1) correct method to solve for the exponent, e.g. logarithms or intersection of graphs, giving 45.25… quarters✓ 1
A1 \(t=11.3\) years✓ 1

Exam tip: Check the compounding period first: quarterly compounding means \(k=4\) and \(4n\) periods, not \(n\).

320IBMI3H-1-07r40-02Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium6 marks

A shipping company fits a new hull coating to one of its tankers. The coating reduces fuel use, so the fuel saved in the first year is 84 tonnes. Because the coating wears, the saving each year is 95% of the saving in the previous year.

(a)

Find the fuel saved in year 10. Give your answer to 3 significant figures.

(2)
(b)

Find the total fuel saved over the first 10 years. Give your answer to 3 significant figures.

(2)
(c)

The company claims that, however long the coating lasts, the total fuel saved can never reach 1700 tonnes. Determine whether the claim is correct.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 use of \(u_n=u_1r^{n-1}\) with \(u_1=84\), \(r=0.95\), \(n=10\)✓ 1
A1 52.9 tonnes✓ 1
Part (b)
M1 use of \(S_n=\frac{u_1(1-r^n)}{1-r}\)✓ 1
A1 674 tonnes✓ 1
Part (c)
M1 sum to infinity \(S_\infty=\frac{84}{1-0.95}\)✓ 1
R1 \(S_\infty=1680<1700\), so the claim is correct✓ 1

Common error: Using \(r=0.05\) (the percentage lost) instead of \(r=0.95\) (the fraction kept) in a geometric sequence.

321IBMI3H-1-07r40-08Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium6 marks

A prop on a revolving theatre stage is at the position \(p=3+2\mathrm{i}\) metres relative to the centre of the stage, which is the origin. The stage turns anticlockwise through \(24^\circ\) about its centre. The new position of the prop is \(w=p\,e^{\mathrm{i}\theta}\), where \(\theta\) is the angle turned, in radians.

(a)

Find \(w\) in the form \(a+b\mathrm{i}\), with \(a\) and \(b\) to 3 significant figures.

(3)
(b)

The stage makes repeated turns of \(24^\circ\). Find the smallest number of turns after which the prop is back at its starting position.

(2)
(c)

Find the distance of the prop from the centre of the stage. Give your answer to 3 significant figures.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 convert \(24^\circ=\frac{2\pi}{15}\) rad or use \(\cos24^\circ+\mathrm{i}\sin24^\circ\)✓ 1
M1 multiply \((3+2\mathrm{i})(\cos24^\circ+\mathrm{i}\sin24^\circ)\)✓ 1
A1 \(w=1.93+3.05\mathrm{i}\)✓ 1
Part (b)
M1 \(24n\) is a multiple of \(360\), or \(e^{\mathrm{i}n\theta}=1\)✓ 1
A1 \(n=15\)✓ 1
Part (c)
A1 \(|p|=\sqrt{13}=3.61\) m (the distance is unchanged by rotation)✓ 1

Exam tip: Multiplying by \(e^{\mathrm{i}\theta}\) rotates a point about the origin through \(\theta\) without changing its modulus.

322IBMI3H-1-07r40-09Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

Three seismometers are placed around a volcano at the points on an Argand diagram (units: km) that satisfy \(z^3=64\mathrm{i}\), where the volcano is at the origin.

(a)

Write \(64\mathrm{i}\) in the form \(re^{\mathrm{i}\theta}\), with \(r>0\) and \(0\le\theta<2\pi\).

(1)
(b)

Find the three solutions of \(z^3=64\mathrm{i}\), giving each in the form \(a+b\mathrm{i}\) with \(a\) and \(b\) to 3 significant figures.

(4)
(c)

Find the distance between two neighbouring seismometers. Give your answer to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(64e^{\mathrm{i}\pi/2}\)✓ 1
Part (b)
A1 modulus of each solution is \(\sqrt[3]{64}=4\)✓ 1
A1 one argument is \(\frac{\pi}{6}\)✓ 1
M1 add \(\frac{2\pi}{3}\) (or \(\frac{2k\pi}{3}\)) to obtain the other arguments \(\frac{5\pi}{6}\) and \(\frac{3\pi}{2}\)✓ 1
A1 \(3.46+2\mathrm{i}\), \(-3.46+2\mathrm{i}\), \(-4\mathrm{i}\)✓ 1
Part (c)
M1 e.g. distance between \(3.46+2\mathrm{i}\) and \(-3.46+2\mathrm{i}\), or chord \(2\times4\sin60^\circ\)✓ 1
A1 6.93 km✓ 1

Exam tip: The \(n\)th roots of a complex number are equally spaced around a circle, separated by \(\frac{2\pi}{n}\) in argument.

323IBMI3H-1-07r40-10Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium6 marks

A theatre sells tickets in three zones: stalls, circle and balcony, at prices of €s, €c and €b respectively. On three nights the numbers of tickets sold and the total takings are shown.
Night 1: 120 stalls, 80 circle, 40 balcony; takings €8680.
Night 2: 100 stalls, 90 circle, 60 balcony; takings €8460.
Night 3: 150 stalls, 60 circle, 30 balcony; takings €9210.

(a)

Write this information as a matrix equation \(\mathbf{A}\mathbf{x}=\mathbf{t}\), where \(\mathbf{x}=\begin{pmatrix}s\\c\\b\end{pmatrix}\).

(2)
(b)

Hence find the price of each type of ticket.

(3)
(c)

On night 4, the theatre sells 130 stalls, 70 circle and 50 balcony tickets. Find the takings.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(\mathbf{A}=\begin{pmatrix}120&80&40\\100&90&60\\150&60&30\end{pmatrix}\)✓ 1
A1 \(\mathbf{t}=\begin{pmatrix}8680\\8460\\9210\end{pmatrix}\)✓ 1
Part (b)
M1 \(\mathbf{x}=\mathbf{A}^{-1}\mathbf{t}\)✓ 1
M1 evaluate on a calculator✓ 1
A1 stalls €45, circle €32, balcony €18✓ 1
Part (c)
A1 €8990 (ECF from (b))✓ 1

Exam tip: Check a matrix solution by substituting it back into at least one of the original equations.

324IBMI3H-1-07r40-12Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium6 marks

Each month a volcano is classed as Quiet (Q) or in Unrest (U). The probabilities for next month's class are modelled by the transition matrix \(\mathbf{T}=\begin{pmatrix}0.8&0.3\\0.2&0.7\end{pmatrix}\), where the first column gives the probabilities after a Quiet month and the second column those after an Unrest month.

(a)

Find the eigenvalues of \(\mathbf{T}\).

(2)
(b)

Find an eigenvector of \(\mathbf{T}\) corresponding to the eigenvalue 1.

(2)
(c)

Hence find the long-term proportion of months in which the volcano is in Unrest.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(\det(\mathbf{T}-\lambda\mathbf{I})=\lambda^2-1.5\lambda+0.5=0\)✓ 1
A1 \(\lambda=1\) and \(\lambda=0.5\)✓ 1
Part (b)
M1 solve \((\mathbf{T}-\mathbf{I})\mathbf{v}=\mathbf{0}\), e.g. \(-0.2q+0.3u=0\)✓ 1
A1 \(\mathbf{v}=\begin{pmatrix}3\\2\end{pmatrix}\) or any non-zero multiple✓ 1
Part (c)
M1 scale the eigenvector so the entries sum to 1: \(\frac{1}{5}\begin{pmatrix}3\\2\end{pmatrix}\)✓ 1
A1 \(0.4\)✓ 1

Exam tip: The long-term state of a transition matrix is the eigenvector for eigenvalue 1, scaled so its entries add to 1.

325IBMI3H-1-04r41-02Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium6 marks

Calculator

The vibration of a long water pipe is modelled by a system whose characteristic equation is z2 − 6z + 34 = 0, where z is a complex number.

(a)

Solve the equation, giving the roots in the form a + bi.

(3)
(b)

Write the root with positive imaginary part in the form r(cos θ + i sin θ), where r and θ are given correct to 3 significant figures and −π < θ ≤ π.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 quadratic formula (or polynomial solver) with discriminant 36 − 136 = −100✓ 1
A1 z = 3 + 5i✓ 1
A1 z = 3 − 5i✓ 1
Part (b)
M1 r = √(32 + 52)✓ 1
A1 r = 5.83✓ 1
A1 θ = 1.03✓ 1

Common error: For real coefficients, complex roots come in conjugate pairs; write both roots, not just one.

326IBMI3H-1-04r41-03Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

Calculator

In a game, a sprite's position on the complex plane is repeatedly multiplied by the complex number w = √3 + i, so that each step rotates and enlarges the position.

(a)

Write w in the form reiθ.

(2)
(b)

The sprite starts at position 1. Find its exact position after 8 steps, in the form a + bi.

(3)
(c)

Find the smallest positive number of steps after which the sprite is again on the positive real axis.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 r = 2✓ 1
A1 θ = π/6✓ 1
Part (b)
M1 w8 = 28ei·8π/6 (de Moivre)✓ 1
A1 256(cos 4π/3 + i sin 4π/3)✓ 1
A1 −128 − 128√3 i✓ 1
Part (c)
M1 nπ/6 = 2π (argument a multiple of 2π)✓ 1
A1 n = 12✓ 1

Exam tip: To use de Moivre's theorem, convert to modulus–argument form first: raise the modulus to the power and multiply the argument by the power.

327IBMI3H-1-04r41-06Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium6 marks

Calculator

Three water pumps A, B and C deliver water at constant rates of x, y and z megalitres (ML) per hour. In three tests, the pumps were run for different times. Test 1: A, B and C each ran for 1 hour, delivering 11 ML in total. Test 2: A ran for 2 hours, B for 1 hour and C for 3 hours, delivering 19 ML in total. Test 3: A ran for 3 hours, B for 2 hours and C for 1 hour, delivering 24 ML in total.

(a)

Write the information as a matrix equation.

(1)
(b)

Find the determinant of the coefficient matrix and state what this shows about the solution.

(1)
(c)

In a fourth test, replace Test 3 by: A runs for 3 hours, B for 2 hours and C for k hours. Find the value of k for which the system no longer has a unique solution.

(2)
(d)

Using the original three tests, find the delivery rate of each pump.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(\begin{pmatrix}1&1&1\\2&1&3\\3&2&1\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}11\\19\\24\end{pmatrix}\)✓ 1
Part (b)
A1 determinant = 3, which is non-zero, so the system has a unique solution✓ 1
Part (c)
M1 determinant of [[1,1,1],[2,1,3],[3,2,k]] = 4 − k set equal to 0✓ 1
A1 k = 4✓ 1
Part (d)
M1 solves the system using the inverse matrix or technology✓ 1
A1 x = 4, y = 5, z = 2 (ML per hour)✓ 1

Exam tip: A zero determinant means the equations are dependent (here the new test would be the sum of the first two), so it adds no new information.

328IBMI3H-1-04r41-07Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium6 marks

Calculator

Players of an online game choose each week between Game A and Game B. Each week, 60% of players of Game A stay with Game A and the rest switch to Game B; 70% of players of Game B stay with Game B and the rest switch to Game A. The transition matrix is M = \(\begin{pmatrix}0.6&0.3\\0.4&0.7\end{pmatrix}\).

(a)

Show that the eigenvalues of M are 1 and 0.3.

(3)
(b)

Find an eigenvector of M corresponding to the eigenvalue 1.

(1)
(c)

Hence find the long-term proportion of players who choose Game A. Give your answer correct to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 det(M − λI) = 0✓ 1
M1 (0.6 − λ)(0.7 − λ) − 0.12 = 0, i.e. λ2 − 1.3λ + 0.3 = 0✓ 1
A1 λ = 1 and λ = 0.3 (shown from factorising or solving)✓ 1
Part (b)
A1 any multiple of (3, 4)T, from −0.4x + 0.3y = 0✓ 1
Part (c)
M1 3/(3 + 4) (normalising so the proportions sum to 1)✓ 1
A1 0.429✓ 1

Common error: The long-term proportions come from the eigenvector for eigenvalue 1 scaled so that its entries sum to 1, not from the eigenvector as first found.

329IBMI3H-1-04r41-08Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium6 marks

Calculator

A game sprite is the triangle with vertices O(0, 0), P(4, 0) and Q(0, 3). Each frame, every point of the plane is transformed by the matrix T = \(\begin{pmatrix}2&-1\\1&3\end{pmatrix}\).

(a)

Find the coordinates of the images P′ and Q′ of P and Q.

(2)
(b)

Find the area of triangle OP′Q′.

(2)
(c)

Find the coordinates of the point that is mapped to (7, 14) by T.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 P′ = (8, 4)✓ 1
A1 Q′ = (−3, 9)✓ 1
Part (b)
M1 area scale factor = det T = 7 (or uses coordinates of the image)✓ 1
A1 7 × 6 = 42 square units✓ 1
Part (c)
M1 \(\mathbf{T}^{-1}\begin{pmatrix}7\\14\end{pmatrix}\) with \(\mathbf{T}^{-1}=\frac{1}{7}\begin{pmatrix}3&1\\-1&2\end{pmatrix}\) (or solving 2x − y = 7, x + 3y = 14)✓ 1
A1 (5, 3)✓ 1

Exam tip: The area of an image equals the area of the original multiplied by the absolute value of the determinant.

330IBMI3H-1-04r41-14Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium6 marks

Calculator

A fund of €60 000 earns interest at a nominal annual rate of 3.6%, compounded monthly. At the end of each month, after interest has been added, €1500 is withdrawn. The final withdrawal may be smaller than €1500.

(a)

Find the number of withdrawals needed to empty the fund.

(3)
(b)

Find the amount of the final withdrawal. Give your answer to the nearest cent.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 TVM solver for N: I% = 3.6, PV = 60000, PMT = −1500, FV = 0, P/Y = C/Y = 12✓ 1
A1 N = 42.7 (42.675…)✓ 1
A1 43 withdrawals (42 of €1500, then a smaller final one)✓ 1
Part (b)
M1 balance after 42 withdrawals = 60000(1.003)42 − 1500((1.003)42 − 1)/0.003 = €1009.98✓ 1
M1 final withdrawal = balance × 1.003 (interest added before the last withdrawal)✓ 1
A1 €1013.01✓ 1

Exam tip: When a loan or fund ends with a smaller final payment, find the balance after the last full payment and then add one month of interest.

331IBMI3H-1-04r43-02Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium6 marks

Two alternating voltages are represented by the complex numbers V1 = 12eiπ⁄6 and V2 = 5e−iπ⁄3, where the modulus is the amplitude in volts and the argument is the phase in radians. The combined voltage is represented by V = V1 + V2.

(a)

Write V1 and V2 in Cartesian form, with each part given to 3 significant figures.

(2)
(b)

Hence find V in Cartesian form, with each part given to 3 significant figures.

(1)
(c)

Find the amplitude of the combined voltage, and its phase to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 V1 = 12 cos(π⁄6) + 12 sin(π⁄6)i = 10.4 + 6.00i✓ 1
A1 V2 = 5 cos(π⁄3) − 5 sin(π⁄3)i = 2.50 − 4.33i✓ 1
Part (b)
A1 V = 12.9 + 1.67i✓ 1
Part (c)
M1 attempts the modulus of V (using unrounded values)✓ 1
A1 amplitude = 13.0 V (exactly 13)✓ 1
A1 phase = arctan(1.6699… ÷ 12.8923…) = 0.129 radians✓ 1

Common error: Keep unrounded values from part (a) when you add: using the rounded Cartesian parts can shift the third significant figure of the argument.

332IBMI3H-1-04r43-03Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium6 marks

A game physics engine models the displacement, x metres, of a swinging door by a solution of the form x = eRe(λ)t cos(Im(λ)t), where λ satisfies the equation λ2 + 4λ + 13 = 0 and t is the time in seconds.

(a)

Solve λ2 + 4λ + 13 = 0, giving your answers in the form a + bi.

(2)
(b)

Using the root with positive imaginary part, write down the model for x. Find the period of the oscillation, to 3 significant figures.

(2)
(c)

The size of the swings is governed by the factor e−2t. Find the time taken for this factor to fall to 5%. Give your answer to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 uses the quadratic formula or completes the square: λ = (−4 ± √(16 − 52)) ÷ 2✓ 1
A1 λ = −2 + 3i and λ = −2 − 3i✓ 1
Part (b)
A1 x = e−2t cos(3t)✓ 1
A1 period = 2π ÷ 3 = 2.09 seconds✓ 1
Part (c)
M1 sets e−2t = 0.05✓ 1
A1 t = ln 20⁄2 = 1.50 seconds✓ 1

Exam tip: For complex roots of a quadratic with real coefficients, the real part controls growth or decay and the imaginary part controls the frequency.

333IBMI3H-1-04r43-04Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium6 marks

Five pumping stations are placed around a water-treatment plant. Taking the plant as the origin of an Argand diagram with distances in kilometres, the stations are at the five solutions of z5 = 32.

(a)

Write the five solutions in the form reiθ, with −π < θ ≤ π.

(2)
(b)

Find the Cartesian coordinates, to 3 significant figures, of the station with the smallest positive argument.

(2)
(c)

Find the distance between two neighbouring stations. Give your answer to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 r = 321/5 = 2 and arguments spaced by 2π⁄5✓ 1
A1 2, 2ei2π/5, 2ei4π/5, 2e−i4π/5, 2e−i2π/5✓ 1
Part (b)
M1 2 cos(2π⁄5) and 2 sin(2π⁄5)✓ 1
A1 (0.618, 1.90) km✓ 1
Part (c)
M1 chord of a circle of radius 2 subtending 2π⁄5 at the centre: 2 × 2 sin(π⁄5) (or cosine rule, or distance formula)✓ 1
A1 2.35 km✓ 1

Exam tip: The nth roots of a complex number lie on a circle and are equally spaced in argument by 2π ÷ n; draw a quick sketch to keep track of the arguments.

334IBMI3H-1-04r43-07Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium6 marks

A pharmacist mixes three stock solutions A, B and C. Each millilitre of A, B and C contains the following masses, in mg, of three compounds:
Compound 1: 2 (in A), 3 (in B), 1 (in C)
Compound 2: 1 (in A), 2 (in B), 4 (in C)
Compound 3: 3 (in A), 1 (in B), 2 (in C)
A prescription requires 25 mg of compound 1, 22 mg of compound 2 and 21 mg of compound 3. Let x, y and z be the volumes in mL of A, B and C used.

(a)

Write down a matrix equation that represents this situation.

(2)
(b)

Hence find x, y and z.

(2)
(c)

A second prescription requires 14 mg of compound 1, 25 mg of compound 2 and 20 mg of compound 3. Find the total volume of stock solution needed.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 coefficient matrix \(\begin{pmatrix}2&3&1\\1&2&4\\3&1&2\end{pmatrix}\) with column vector \((x, y, z)\)✓ 1
A1 right-hand side \(\begin{pmatrix}25\\22\\21\end{pmatrix}\)✓ 1
Part (b)
M1 uses the inverse matrix (or the GDC equation solver)✓ 1
A1 x = 4, y = 5, z = 2✓ 1
Part (c)
M1 solves with the new right-hand side: x = 3, y = 1, z = 5✓ 1
A1 total volume = 9 mL✓ 1

Exam tip: Once the coefficient matrix is stored in the GDC, a new right-hand side only needs one more multiplication by the inverse: do not solve the whole system again by hand.

335IBMI3H-1-04r43-09Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium6 marks

Each day, a player of an online game is either active (A) or inactive (I). An active player is still active the next day with probability 0.9. An inactive player becomes active the next day with probability 0.3. The proportions of active and inactive players on one day are multiplied by the transition matrix \(\mathbf{T}=\begin{pmatrix}0.9&0.3\\0.1&0.7\end{pmatrix}\) to give the proportions on the next day.

(a)

Show that the eigenvalues of T are 1 and 0.6.

(2)
(b)

Find the long-term proportions of active and inactive players.

(2)
(c)

The game has 8000 players. Find the number of active players in the long term.

(1)
(d)

Explain how the eigenvalue 0.6 shows that the proportions settle down to their long-term values.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \((0.9-\lambda)(0.7-\lambda)-0.03=0\), which gives λ2 − 1.6λ + 0.6 = 0✓ 1
A1 (λ − 1)(λ − 0.6) = 0, so λ = 1 and λ = 0.6✓ 1
Part (b)
M1 solves (T − I)v = 0: −0.1a + 0.3i = 0 and scales so that the proportions add to 1✓ 1
A1 0.75 active and 0.25 inactive✓ 1
Part (c)
A1 6000✓ 1
Part (d)
R1 the part of the starting vector along the other eigenvector is multiplied by 0.6 each day, so it shrinks towards zero✓ 1

Common error: The long-term vector must be scaled so its entries add up to 1 (proportions): an eigenvector such as (3, 1) is only a direction.

336IBMI3H-1-04r43-10Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

Three connected water tanks A, B and C hold 600, 300 and 100 m3 of water respectively. Each hour, water is moved between tanks so that the new volumes (in m3) are found by multiplying the column vector of volumes by \(\mathbf{P}=\begin{pmatrix}0.7&0.2&0.1\\0.2&0.6&0.3\\0.1&0.2&0.6\end{pmatrix}\).

(a)

Explain why every column of P adds up to 1.

(1)
(b)

Find the volume of water in each tank after one hour.

(2)
(c)

Find the volume of water in each tank after 5 hours, to 1 decimal place.

(2)
(d)

The process continues for a very long time. Find the volume of water in tank B in the long term. Give your answer to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
R1 all the water in a tank is accounted for: the fractions staying in the tank and moving to the other tanks add up to 1 (no water is lost)✓ 1
Part (b)
M1 multiplies P by the column vector (600, 300, 100)✓ 1
A1 A: 490, B: 330, C: 180 m3✓ 1
Part (c)
M1 uses P5 applied to (600, 300, 100) (or repeats the multiplication from the result of part (b))✓ 1
A1 A: 359.5, B: 373.8, C: 266.7 m3✓ 1
Part (d)
M1 finds the eigenvector for eigenvalue 1 (proportional to (10, 11, 8)) and scales it to a total of 1000 m3✓ 1
A1 379 m3✓ 1

Exam tip: In a long-term question the total amount is conserved: scale the eigenvector for eigenvalue 1 so that its entries add up to the original total.

337IBMI3H-1-05r43-02Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium5 marks

A fashion label has two shops, Nord and Sud. Matrix A gives the numbers of shirts, trousers and jackets ordered by each shop (rows: Nord, Sud). Matrix B gives the metres of cotton and wool needed for one of each garment (rows: shirt, trousers, jacket).
\(\mathbf{A}=\begin{pmatrix}20&15&10\\30&10&5\end{pmatrix}\), \(\mathbf{B}=\begin{pmatrix}1.5&0.5\\2.0&1.2\\2.5&2.0\end{pmatrix}\)

(a)

Find the matrix product \(\mathbf{AB}\).

(2)
(b)

State what the entry 48 in the product represents.

(1)
(c)

Cotton costs €4.20 per metre and wool costs €9.50 per metre. Find the total cost of the fabric for Sud's order.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt at matrix multiplication (row × column)✓ 1
A1 \(\begin{pmatrix}85&48\\77.5&37\end{pmatrix}\)✓ 1
Part (b)
A1 metres of wool needed for Nord's order✓ 1
Part (c)
M1 attempt 77.5 × 4.20 + 37 × 9.50 (or multiply the second row of AB by the cost vector)✓ 1
A1 €677.00✓ 1

Exam tip: Check the matrices can be multiplied: the number of columns in the first must equal the number of rows in the second.

338IBMI3H-1-05r43-03Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium8 marks

A fish farm counts juvenile and adult fish in one tank each year (in thousands). Each year every adult produces on average 2 juveniles that survive to the next count, 50% of juveniles survive to become adults, and 50% of adults survive. No fish are harvested. Let the counts in year n be \(\binom{J_n}{A_n}\).

(a)

Write down the matrix L such that \(\binom{J_{n+1}}{A_{n+1}}=\mathbf{L}\binom{J_n}{A_n}\).

(1)
(b)

Find the eigenvalues of L, to 3 significant figures.

(3)
(c)

Interpret the larger eigenvalue in the context of the fish farm.

(1)
(d)

Find an eigenvector for the larger eigenvalue, and hence the long-term proportion of the fish that are juveniles, to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(\mathbf{L}=\begin{pmatrix}0&2\\0.5&0.5\end{pmatrix}\)✓ 1
Part (b)
M1 attempt to solve det(L − λI) = 0✓ 1
A1 λ2 − 0.5λ − 1 = 0✓ 1
A1 λ = 1.28 and λ = −0.781✓ 1
Part (c)
A1 in the long term the population grows by about 28% each year (it is multiplied by 1.28 each year)✓ 1
Part (d)
M1 attempt to solve (L − λI)v = 0 for the larger eigenvalue✓ 1
A1 J : A = 1 : 0.640 (from −λJ + 2A = 0)✓ 1
A1 proportion = 1/(1 + 0.640) = 0.610✓ 1

Common error: The long-term ratio comes from the eigenvector of the dominant (largest) eigenvalue, not from the initial counts.

339IBMI3H-1-05r43-04Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

Two motors drive a robot arm. When motor 1 runs at x A and motor 2 runs at y A, the torques at the two joints are (4x + 3y) N m and (2x + 5y) N m. The torques are given by \(\binom{T_1}{T_2}=\mathbf{M}\binom{x}{y}\) where \(\mathbf{M}=\begin{pmatrix}4&3\\2&5\end{pmatrix}\).

(a)

Find det M and \(\mathbf{M}^{-1}\).

(3)
(b)

The joints need torques of 17 N m and 19 N m. Use M−1 to find the currents x and y.

(2)
(c)

A redesigned arm has matrix \(\begin{pmatrix}4&3\\2&k\end{pmatrix}\). Find the value of k for which the currents are not uniquely determined by the torques.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 det M = 14✓ 1
M1 attempt to use the inverse formula (swap the diagonal, change signs of the other two, divide by det)✓ 1
A1 \(\mathbf{M}^{-1}=\frac{1}{14}\begin{pmatrix}5&-3\\-2&4\end{pmatrix}\)✓ 1
Part (b)
M1 attempt to multiply M−1 by the column (17, 19)✓ 1
A1 x = 2 A, y = 3 A✓ 1
Part (c)
M1 attempt to set the determinant 4k − 6 = 0✓ 1
A1 k = 1.5✓ 1

Exam tip: A system with a square matrix has a unique solution exactly when the determinant is non-zero.

340IBMI3H-1-05r43-05Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium6 marks

A designer places five studs at equal spacing around a circular collar. In an Argand diagram (units of cm, centre at the origin), the studs are at the solutions of \(z^5=32\).

(a)

Find the five solutions in the form reiθ, with −π < θ ≤ π or 0 ≤ θ < 2π.

(3)
(b)

Find the distance between two adjacent studs, to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 write 32 = 32e2πik (general argument) or equivalent✓ 1
M1 take fifth roots: modulus 2 and arguments 2πk/5✓ 1
A1 2, 2e0.4πi, 2e0.8πi, 2e1.2πi, 2e1.6πi (equivalent forms accepted)✓ 1
Part (b)
M1 attempt to find |2e0.4πi − 2| or use an isosceles triangle with apex angle 72°✓ 1
M1 correct method, for example 2 × 2 sin 36° or cosine rule✓ 1
A1 2.35 cm✓ 1

Common error: Remember to include all five roots; the angles between neighbouring roots are 2π/5 apart.

341IBMI3H-1-05r43-10Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium6 marks

A fish farm mixes three feeds. Feed A is 40% protein, 10% fat and 5% fibre by mass. Feed B is 30% protein, 20% fat and 10% fibre. Feed C is 20% protein, 30% fat and 25% fibre. The farm needs a mixture with a total of 47 kg of protein, 28 kg of fat and 18 kg of fibre. Let the masses of A, B and C used be a, b and c kilograms.

(a)

Write down three equations for a, b and c.

(2)
(b)

Solve the equations.

(2)
(c)

Feed A costs €1.20 per kg, feed B costs €0.95 per kg and feed C costs €0.80 per kg. Find the total cost of the mixture.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to form equations from protein, fat and fibre✓ 1
A1 0.40a + 0.30b + 0.20c = 47; 0.10a + 0.20b + 0.30c = 28; 0.05a + 0.10b + 0.25c = 18✓ 1
Part (b)
M1 use of a calculator solver or matrix inverse✓ 1
A1 a = 60, b = 50, c = 40✓ 1
Part (c)
M1 60 × 1.20 + 50 × 0.95 + 40 × 0.80 (ECF from (b))✓ 1
A1 €151.50✓ 1

Exam tip: Check each answer by substituting back into all three original equations.

342IBMI3H-1-05r43-12Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium8 marks

A printed fabric is stretched by the transformation with matrix \(\mathbf{M}=\begin{pmatrix}3&1\\0&2\end{pmatrix}\) (lengths in cm).

(a)

Find the eigenvalues of M.

(2)
(b)

Find an eigenvector for each eigenvalue.

(3)
(c)

A seam of length 10 cm lies along the direction (1, −1) before stretching. State its length after the transformation.

(1)
(d)

A motif has area 25 cm2 before stretching. Find its area afterwards.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to solve (3 − λ)(2 − λ) = 0✓ 1
A1 λ = 3 and λ = 2✓ 1
Part (b)
M1 attempt to solve (M − λI)v = 0✓ 1
A1 λ = 3: (1, 0) or any multiple✓ 1
A1 λ = 2: (1, −1) or any multiple✓ 1
Part (c)
A1 20 cm (the length is multiplied by the eigenvalue 2)✓ 1
Part (d)
M1 area scale factor = det M = 6✓ 1
A1 150 cm2✓ 1

Common error: An eigenvector keeps its direction under the transformation, but it is stretched by the eigenvalue; the area scale factor is the determinant, not an eigenvalue.

343IBMI3H-1-07r44-02Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

On an architect's plan, a decorative column stands at the point representing the complex number P = 6 + 2i, where distances are in metres from the centre of a circular atrium. The design is modified by applying the transformation z ↦ 1.2(cos 35° + i sin 35°)z, so that every point on the plan is moved.

(a)

Write P in the form r(cos θ + i sin θ), with θ in degrees, giving r and θ to 3 significant figures.

(2)
(b)

Hence find the new position of the column, in the form a + bi, with a and b given to 3 significant figures.

(3)
(c)

The architect instead wants to rotate the original column about the centre, with no change in distance from the centre, until it lies on the positive imaginary axis. Find the required anticlockwise rotation, in degrees, to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 r = √40 = 6.32 m✓ 1
A1 θ = arctan(2/6) = 18.4°✓ 1
Part (b)
M1 new modulus = 1.2 × 6.32… = 7.589… and new argument = 18.43… + 35 = 53.43…° (or equivalent multiplication)✓ 1
A1 correct real part 4.52✓ 1
A1 correct imaginary part 6.10✓ 1
Part (c)
M1 attempt to use 90° − θ✓ 1
A1 71.6°✓ 1

Common error: Multiplying two complex numbers in polar form multiplies the moduli but adds the arguments; do not multiply the arguments.

344IBMI3H-1-07r44-04Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium8 marks

Three supporting pillars of a pavilion stand at the points representing the solutions of z3 = 27(cos 120° + i sin 120°) on a plan with distances in metres from the centre of a circular floor.

(a)

Use de Moivre's theorem to find the three solutions in the form r(cos θ + i sin θ), with 0° ≤ θ < 360°.

(3)
(b)

Find the coordinates of the pillar in the second quadrant, with each coordinate given to 3 significant figures.

(2)
(c)

Find the distance between two adjacent pillars, in metres, to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to take the cube root of the modulus and divide the argument by 3, so r = 3 and θ = 40°✓ 1
A1 3(cos 40° + i sin 40°)✓ 1
A1 the other two solutions have θ = 160° and θ = 280° (adding 120° each time)✓ 1
Part (b)
M1 attempt to use 3 cos 160° and 3 sin 160°✓ 1
A1 (−2.82, 1.03)✓ 1
Part (c)
R1 adjacent pillars subtend an angle of 120° at the centre (equilateral triangle)✓ 1
M1 attempt to use the cosine rule (or chord length 2 × 3 × sin 60°)✓ 1
A1 5.20 m✓ 1

Common error: When finding roots with de Moivre's theorem, do not forget the other roots: they are equally spaced round a circle, so add 360°/n to the argument each time.

345IBMI3H-1-07r44-06Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

A logistics firm loads three types of crate onto delivery vans. The masses of one crate of each type are x kg, y kg and z kg. Van 1 carries 2 crates of type X, 1 of type Y and 3 of type Z, with total crate mass 70 kg. Van 2 carries 1 crate of type X, 4 of type Y and 2 of type Z, with total crate mass 72 kg. Van 3 carries 3 crates of type X, 2 of type Y and 1 of type Z, with total crate mass 71 kg.

(a)

Write the information as a matrix equation of the form Av = b, where v is the column vector (x, y, z)T.

(2)
(b)

Use technology to find det A and state what this tells you about the solution of the system.

(2)
(c)

Find the mass of one crate of each type.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 A = rows (2, 1, 3), (1, 4, 2), (3, 2, 1)✓ 1
A1 b = (70, 72, 71)T and the product written in the correct order✓ 1
Part (b)
A1 det A = −25✓ 1
R1 the determinant is non-zero, so A is invertible and there is a unique solution✓ 1
Part (c)
M1 attempt to use v = A−1b (or equivalent technology method)✓ 1
A1 x = 14 kg✓ 1
A1 y = 9 kg and z = 11 kg✓ 1

Exam tip: When solving Av = b, multiply by the inverse on the left: v = A⁻¹b. Reversing the order gives a different (and usually undefined) product.

346IBMI3H-1-07r44-11Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

An architecture firm takes out a loan of €240 000 to buy a studio. The annual interest rate is 3.6%, compounded monthly. The loan is repaid in equal monthly payments made at the end of each month over 25 years.

(a)

Use a financial solver to find the monthly payment, giving your answer to the nearest cent.

(2)
(b)

Using the payment from part (a) rounded to the nearest cent, find the total interest paid over the 25 years.

(2)
(c)

The firm decides to pay €1500 each month instead. Use a financial solver to find the value of N for this payment, to 1 decimal place, and hence state the number of monthly payments needed to repay the loan.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 correct use of solver or formula with N = 300, I% = 3.6, PV = 240 000, FV = 0, P/Y = C/Y = 12✓ 1
A1 €1214.41✓ 1
Part (b)
M1 300 × 1214.41 − 240 000✓ 1
A1 €124 323.00✓ 1
Part (c)
M1 attempt to use solver with PMT = −1500 (or equivalent formula) to find N✓ 1
A1 N = 218.3✓ 1
A1 219 payments are needed (the last being smaller)✓ 1

Common error: Number of payments must be a whole number: a solver value such as 218.3 means 219 payments are needed, not 218.

347IBMI3H-1-07r44-12Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium8 marks

A harmful algal bloom is spreading across a bay. On day 1 it covers an area of 4.0 m², and its area grows by 15% each day, so the daily areas form a geometric sequence.

(a)

Find the area covered on day 8, in m², to 3 significant figures.

(2)
(b)

Find the sum of the daily areas for the first 10 days, in m², to 3 significant figures.

(2)
(c)

Find the first day on which the area covered exceeds 50 m².

(4)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to use un = 4.0 × 1.15n−1 with n = 8✓ 1
A1 10.6 m²✓ 1
Part (b)
M1 attempt to use the geometric series formula with a = 4.0, r = 1.15 and n = 10✓ 1
A1 81.2 m²✓ 1
Part (c)
M1 4.0 × 1.15n−1 > 50 (or equality)✓ 1
M1 attempt to solve using logarithms or technology, e.g. n − 1 > ln 12.5 / ln 1.15✓ 1
A1 n − 1 > 18.07…, so n > 19.07✓ 1
A1 day 20✓ 1

Common error: The nth term of a geometric sequence is arⁿ⁻¹, not arⁿ: the first term has no growth factor applied.

348IBMI3H-1-04r46-03Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium6 marks

The steps of a spiral staircase are placed using complex numbers. Step \(n\) has its outer corner at \(P_n = z^n\) metres from the central pillar on an Argand diagram of the floor plan, where \(z = 1.08\,e^{i\pi/12}\) and \(n = 0, 1, 2, \dots\)

(a)

Find the position of \(P_5\) in the form \(a + bi\), with \(a\) and \(b\) to 3 significant figures.

(3)
(b)

Find the smallest positive integer \(n\) for which \(P_n\) lies on the negative real axis.

(2)
(c)

Find the distance of this step from the central pillar, in metres, to 3 significant figures.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(P_5 = 1.08^5\,e^{5i\pi/12}\)✓ 1
A1 modulus 1.469… and argument 75°✓ 1
A1 \(0.380 + 1.42i\)✓ 1
Part (b)
M1 sets \(n\pi/12 = \pi\) (plus \(2k\pi\) if used)✓ 1
A1 \(n = 12\)✓ 1
Part (c)
A1 \(1.08^{12} = 2.52\) m✓ 1

Exam tip: In exponential form, powers multiply the modulus by repeated multiplication and the argument by \(n\): \((re^{i\theta})^n = r^n e^{in\theta}\).

349IBMI3H-1-04r46-06Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

The numbers of juvenile fish \(J_n\) and adult fish \(A_n\) in a reef population in year \(n\) are modelled by \(\begin{pmatrix}J_{n+1}\\A_{n+1}\end{pmatrix} = L\begin{pmatrix}J_n\\A_n\end{pmatrix}\), where \(L = \begin{pmatrix}0.2&1.4\\0.5&0.5\end{pmatrix}\).

(a)

Show that the eigenvalues of \(L\) satisfy \(\lambda^2 - 0.7\lambda - 0.6 = 0\), and find the eigenvalues.

(3)
(b)

Find an eigenvector corresponding to the positive eigenvalue.

(2)
(c)

Interpret the positive eigenvalue in context, and find the long-term percentage of the fish population that are juveniles. Give your answer to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(\det(L - \lambda I) = (0.2-\lambda)(0.5-\lambda) - 0.7 = 0\)✓ 1
A1 \(\lambda^2 - 0.7\lambda - 0.6 = 0\) (AG)✓ 1
A1 \(\lambda = 1.2\) and \(\lambda = -0.5\)✓ 1
Part (b)
M1 solves \((L - 1.2I)\mathbf{v} = \mathbf{0}\), e.g. \(-x + 1.4y = 0\)✓ 1
A1 \(\mathbf{v} = \begin{pmatrix}1.4\\1\end{pmatrix}\) (or any non-zero multiple)✓ 1
Part (c)
R1 the population changes by a factor of 1.2 each year, i.e. grows 20% a year in the long run✓ 1
A1 \(1.4 \div 2.4 = 58.3\%\)✓ 1

Common error: An eigenvector gives only a ratio of juveniles to adults, so convert it to a percentage by dividing by the total of its components.

350IBMI3H-1-04r46-08Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium8 marks

The vibration of a seabed sensor is modelled using the equation \(z^2 - 4z + 13 = 0\), where \(z\) is a complex number.

(a)

Solve the equation, giving your answers in the form \(a + bi\).

(2)
(b)

Write the root with positive imaginary part in the form \(r(\cos\theta + i\sin\theta)\), giving \(r\) and \(\theta\) to 3 significant figures.

(3)
(c)

Find \((2 + 3i)^4\) in the form \(a + bi\).

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 uses the quadratic formula or completes the square, \((z-2)^2 = -9\)✓ 1
A1 \(z = 2 \pm 3i\)✓ 1
Part (b)
A1 \(r = \sqrt{13} = 3.61\)✓ 1
M1 \(\theta = \tan^{-1}(3/2)\)✓ 1
A1 \(\theta = 0.983\)✓ 1
Part (c)
M1 \((2+3i)^2 = -5 + 12i\)✓ 1
M1 squares again, \((-5+12i)^2 = 25 - 120i - 144\)✓ 1
A1 \(-119 - 120i\)✓ 1

Exam tip: Roots of a quadratic with real coefficients come as a conjugate pair, so checking that \(b^2 - 4ac < 0\) first confirms complex roots.

351IBMI3H-1-04r46-09Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium6 marks

A distribution company holds stock of three products P, Q and R in three warehouses. The matrix \(S\) gives the number of units in each warehouse; its rows are warehouses W1, W2, W3 and its columns are products P, Q, R:
\(S = \begin{pmatrix}120&80&50\\60&150&90\\100&40&200\end{pmatrix}\).
The cost per unit is €4.50 for P, €12.00 for Q and €7.25 for R.

(a)

Use matrix multiplication to find the total number of units of each product held across all three warehouses.

(2)
(b)

Use matrix multiplication to find the total value of stock in each warehouse.

(3)
(c)

All unit costs rise by 8%. Find the new value of stock in W2.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(\begin{pmatrix}1&1&1\end{pmatrix}S\)✓ 1
A1 \((280,\ 270,\ 340)\)✓ 1
Part (b)
M1 \(S\begin{pmatrix}4.5\\12\\7.25\end{pmatrix}\)✓ 1
A1 W1: €1862.50 and W2: €2722.50✓ 1
A1 W3: €2380.00✓ 1
Part (c)
A1 \(2722.50 \times 1.08 = €2940.30\)✓ 1

Common error: Check the order of multiplication: the number of columns of the first matrix must equal the number of rows of the second.

352IBMI3H-1-04r46-10Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium6 marks

In an architect's plan, a decorative node is placed at the point represented by the complex number \(z\) on an Argand diagram of the floor, where \(2z + 3z^* = 10 + 5i\) and \(z^*\) is the complex conjugate of \(z\).

(a)

Find \(z\), giving your answer in the form \(a + bi\).

(4)
(b)

Find the distance of the node from the origin, to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 writes \(z = x + iy\) and \(z^* = x - iy\)✓ 1
M1 substitutes to obtain \(5x - iy = 10 + 5i\)✓ 1
A1 equates real and imaginary parts: \(x = 2\), \(y = -5\)✓ 1
A1 \(z = 2 - 5i\)✓ 1
Part (b)
M1 \(\sqrt{2^2 + (-5)^2}\)✓ 1
A1 5.39✓ 1

Common error: The conjugate changes only the sign of the imaginary part, so \(2z + 3z^*\) gives \(5x - iy\), not \(5x + iy\).

353IBMI3H-1-04r46-13Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium6 marks

An open-air amphitheatre is being designed with rows of seats. The first row has 18 seats, and each row after the first has 3 more seats than the row before.

(a)

Find the number of seats in row 20.

(2)
(b)

The designers need at least 1500 seats in total. Find the smallest number of rows that achieves this.

(3)
(c)

Find the total number of seats when the amphitheatre has this number of rows.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(u_{20} = 18 + 19 \times 3\)✓ 1
A1 75✓ 1
Part (b)
M1 \(S_n = \tfrac{n}{2}(2 \times 18 + (n-1) \times 3) \ge 1500\)✓ 1
M1 solves, e.g. the positive root is about 26.6, or finds \(S_{26} = 1443\) and \(S_{27} = 1539\)✓ 1
A1 27 rows✓ 1
Part (c)
A1 1539✓ 1

Common error: The answer to a question like part (b) must be a whole number, and rounding 26.6 up is correct only because 26 rows gives fewer than 1500 seats.

354IBMI3H-1-05r46-03Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

Calculator

A data centre runs jobs on two clusters, A and B. Each hour, 20% of the jobs on A move to B and 30% of the jobs on B move to A; all other jobs stay where they are. The proportions of jobs on (A, B) at the start of an hour are written as a column vector and the next hour's proportions are found by multiplying by \(\mathbf{M} = \begin{pmatrix} 0.8 & 0.3 \\ 0.2 & 0.7 \end{pmatrix}\).

(a)

Find the eigenvalues of \(\mathbf{M}\).

(3)
(b)

Find an eigenvector of \(\mathbf{M}\) corresponding to the eigenvalue 1.

(2)
(c)

Hence find the long-term percentage of jobs on cluster A.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 set \(\det(\mathbf{M} - \lambda\mathbf{I}) = 0\)✓ 1
A1 \((0.8-\lambda)(0.7-\lambda) - 0.06 = 0\), i.e. \(\lambda^2 - 1.5\lambda + 0.5 = 0\)✓ 1
A1 \(\lambda = 1\) and \(\lambda = 0.5\)✓ 1
Part (b)
M1 solve \((\mathbf{M} - \mathbf{I})\mathbf{v} = \mathbf{0}\), e.g. \(-0.2x + 0.3y = 0\)✓ 1
A1 any multiple of \(\begin{pmatrix} 3 \\ 2 \end{pmatrix}\)✓ 1
Part (c)
M1 scale the eigenvector so that the entries sum to 1, i.e. \(\frac{3}{3+2}\) (ECF from (b))✓ 1
A1 60%✓ 1

Exam tip: The long-term state of a column-stochastic matrix is the eigenvector for eigenvalue 1, scaled so its entries add to 1.

355IBMI3H-1-05r46-04Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium6 marks

Calculator

The centre of a bicycle wheel is at the origin of the complex plane, with distances in metres. The tyre valve is at the point \(z = 0.34\,e^{i\pi/5}\).

(a)

Write \(z\) in the form \(a + bi\), giving \(a\) and \(b\) to 3 significant figures.

(2)
(b)

The wheel then turns anticlockwise through \(\frac{2\pi}{3}\) radians about its centre. Write the new position of the valve in the form \(re^{i\theta}\).

(2)
(c)

Find the horizontal displacement of the valve from the centre after the turn, correct to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(a = 0.34\cos\frac{\pi}{5} = 0.275\)✓ 1
A1 \(b = 0.34\sin\frac{\pi}{5} = 0.200\)✓ 1
Part (b)
M1 multiply by \(e^{2\pi i/3}\) or add \(\frac{2\pi}{3}\) to the argument✓ 1
A1 \(0.34\,e^{13\pi i/15}\)✓ 1
Part (c)
M1 \(0.34\cos\frac{13\pi}{15}\) (ECF from (b))✓ 1
A1 \(-0.311\) m (the valve is to the left of the centre)✓ 1

Exam tip: Multiplying by \(e^{i\phi}\) rotates a point anticlockwise by \(\phi\) about the origin without changing its distance from it.

356IBMI3H-1-05r46-07Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium5 marks

Calculator

A city has three bike-share stations, A, B and C. At the end of each day the expected distribution of bikes is found by multiplying the column vector of bikes at (A, B, C) by \(\mathbf{T} = \begin{pmatrix} 0.7 & 0.2 & 0.1 \\ 0.2 & 0.6 & 0.3 \\ 0.1 & 0.2 & 0.6 \end{pmatrix}\). At the start of Monday there are 240 bikes at A, 180 at B and 120 at C.

(a)

Explain why each column of \(\mathbf{T}\) sums to 1.

(1)
(b)

Find the expected number of bikes at each station at the end of Monday.

(2)
(c)

Find the expected number of bikes at station C at the end of Friday (after five days), to the nearest whole bike.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
R1 every bike starting at a station finishes the day at one of the three stations, so the proportions sum to 1✓ 1
Part (b)
M1 \(\mathbf{T}\begin{pmatrix} 240 \\ 180 \\ 120 \end{pmatrix}\)✓ 1
A1 216 at A, 192 at B and 132 at C✓ 1
Part (c)
M1 \(\mathbf{T}^5\begin{pmatrix} 240 \\ 180 \\ 120 \end{pmatrix}\)✓ 1
A1 147✓ 1

Exam tip: Use the matrix power directly; the total number of bikes (540) is unchanged each day, which gives a quick check on your answer.

357IBMI3H-1-05r46-09Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium8 marks

Calculator

A cycling club pays €250 at the end of every month into an account that pays a nominal annual interest rate of 3.6%, compounded monthly. The account starts with a balance of zero.

(a)

Find the balance after 4 years, giving your answer to the nearest cent.

(3)
(b)

Find the number of complete months needed for the balance to first reach at least €15 000.

(3)
(c)

Find the balance at the end of that month, to the nearest cent.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 use of a finance solver or annuity formula with N = 48, I% = 3.6, PMT = 250, P/Y = C/Y = 12, PV = 0✓ 1
A1 correct values entered (rate 0.3% per month)✓ 1
A1 €12 886.26✓ 1
Part (b)
M1 solve \(250\times\frac{1.003^n - 1}{0.003} \ge 15000\) or trial of values of \(n\)✓ 1
A1 \(n\) is between 55 and 56 so the balance first reaches €15 000 at the end of month 56✓ 1
A1 56 months✓ 1
Part (c)
M1 use \(N = 56\) in the solver or formula (ECF from (b))✓ 1
A1 €15 219.93✓ 1

Exam tip: Always state the values you enter for N, I%, PV, PMT, FV, P/Y and C/Y; with a monthly compounding period the per-month rate is the nominal rate divided by 12.

358IBMI3H-1-05r46-10Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium5 marks

Calculator

An archaeology laboratory uses a radioactive tracer to study pottery glaze. The activity of a sample is modelled by \(A = A_0 e^{-kt}\), where \(A\) is measured in counts per minute and \(t\) is in years. The activity falls from 850 to 620 counts per minute over 18 years.

(a)

Find the value of \(k\), correct to 3 significant figures.

(3)
(b)

Hence find the half-life of the tracer, using the unrounded value of \(k\) and giving your answer to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(620 = 850e^{-18k}\)✓ 1
M1 take logarithms, e.g. \(k = \frac{1}{18}\ln\frac{850}{620}\)✓ 1
A1 \(k = 0.0175\)✓ 1
Part (b)
M1 \(\frac{1}{2} = e^{-kt}\) so \(t = \frac{\ln 2}{k}\)✓ 1
A1 39.5 years✓ 1

Exam tip: Halving time depends only on \(k\): it does not depend on the starting activity.

359IBMI3H-1-05r47-05Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

Volcanic ash is released into two atmospheric layers, A (lower) and B (upper). Each hour, 70% of the ash in layer A stays in A and 30% moves to B. Each hour, 80% of the ash in layer B stays in B and 20% moves to A. The amounts after n hours are given by \(\begin{pmatrix}a_n\\b_n\end{pmatrix}=\mathbf{M}^n\begin{pmatrix}a_0\\b_0\end{pmatrix}\), where \(\mathbf{M}=\begin{pmatrix}0.7&0.2\\0.3&0.8\end{pmatrix}\).

(a)

Find the eigenvalues of \(\mathbf{M}\).

(3)
(b)

Find an eigenvector of \(\mathbf{M}\) corresponding to the eigenvalue 1.

(2)
(c)

A total of 4000 kg of ash is released. Use your answer to part (b) to find the mass of ash in layer A in the long term.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to solve \(\det(\mathbf{M}-\lambda\mathbf{I})=0\)✓ 1
A1 \(\lambda^2-1.5\lambda+0.5=0\)✓ 1
A1 \(\lambda=1\) and \(\lambda=0.5\)✓ 1
Part (b)
M1 attempt to solve \((\mathbf{M}-\mathbf{I})\mathbf{v}=\mathbf{0}\), e.g. \(-0.3x+0.2y=0\)✓ 1
A1 \(\begin{pmatrix}2\\3\end{pmatrix}\) or any non-zero multiple✓ 1
Part (c)
M1 use proportion \(\frac{2}{2+3}\) of the total✓ 1
A1 \(1600\) kg✓ 1

Exam tip: An eigenvector for the eigenvalue 1 of a transition matrix describes the long-term proportions, so scale it to the total amount available.

360IBMI3H-1-05r47-07Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium8 marks

On a theatre stage, a spotlight is aimed at a sequence of points \(P_n\) whose positions, in metres from the centre, are given by the complex numbers \(P_n=z^n\), where \(z=1.5\,e^{i\pi/8}\) and \(n=1,2,3,\dots\)

(a)

Use de Moivre's theorem to show that \(P_4\) lies on the imaginary axis, and state its exact distance from the centre.

(3)
(b)

Find the smallest value of n for which \(P_n\) is more than 20 m from the centre.

(3)
(c)

Write \(P_8\) in the form a + bi, giving your answer to 4 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(z^4=1.5^4e^{i\pi/2}\) or \(1.5^4\left(\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}\right)\)✓ 1
A1 \(z^4=5.0625\,i\), which has real part 0, so \(P_4\) is on the imaginary axis✓ 1
A1 distance \(=5.0625\) m✓ 1
Part (b)
M1 set up \(1.5^n>20\)✓ 1
M1 use of logarithms or a table of values✓ 1
A1 \(n=8\)✓ 1
Part (c)
M1 \(P_8=1.5^8e^{i\pi}\)✓ 1
A1 \(-25.63\) (or \(-25.63+0i\))✓ 1

Exam tip: In de Moivre's theorem the modulus is raised to the power n while the argument is multiplied by n.

361IBMI3H-1-05r47-09Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

The energy E joules released by a volcanic earthquake of magnitude M is modelled by \(\log_{10}E=4.8+1.5M\).

(a)

Find the energy released by an earthquake of magnitude 5.2. Give your answer in the form \(a\times10^k\) where \(1\le a<10\) and \(k\in\mathbb{Z}\), with a to 3 significant figures.

(2)
(b)

Show that, when the magnitude increases by 1, the energy released is multiplied by a constant factor, and find this factor to 3 significant figures.

(2)
(c)

A second earthquake releases 50 times as much energy as the earthquake in part (a). Find the magnitude of the second earthquake, to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(\log_{10}E=4.8+1.5\times5.2=12.6\)✓ 1
A1 \(E=3.98\times10^{12}\) J✓ 1
Part (b)
M1 \(\log_{10}E_2-\log_{10}E_1=1.5\) or \(\frac{E_2}{E_1}=10^{1.5}\)✓ 1
A1 factor \(=31.6\) (independent of the starting magnitude)✓ 1
Part (c)
M1 \(\log_{10}(50E_1)=\log_{10}50+\log_{10}E_1\) or equivalent✓ 1
M1 \(1.5M=\log_{10}50+12.6-4.8\)✓ 1
A1 \(M=6.33\)✓ 1

Exam tip: Multiplying the energy by a factor adds the logarithm of that factor to \(\log_{10}E\), so work with the logarithmic equation rather than the large numbers.

362IBMI3H-1-05r47-14Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium6 marks

The vibration of a theatre fly-system is described by the cubic equation \(z^3+az^2+bz+c=0\), where a, b and c are real. One root is \(z=3\) and another is \(z=1+2i\).

(a)

Write down a third root, giving a reason for your answer.

(2)
(b)

Find the values of a, b and c.

(3)
(c)

The vibration grows without limit if any root has positive real part. State whether the vibration grows without limit, with a reason.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(1-2i\)✓ 1
R1 complex roots of a polynomial with real coefficients occur in conjugate pairs✓ 1
Part (b)
M1 quadratic factor \(z^2-2z+5\) from the sum 2 and product 5 of the conjugate roots✓ 1
M1 expand \((z-3)(z^2-2z+5)\)✓ 1
A1 \(a=-5\), \(b=11\), \(c=-15\)✓ 1
Part (c)
R1 yes, since all three roots (3, 1+2i and 1−2i) have positive real part✓ 1

Common error: If \(1+2i\) is a root of a real polynomial, its conjugate must also be a root. Do not forget the third factor when expanding.

363IBMI3H-1-05r48-03Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium6 marks

Calculator

The signal from a rotating sensor on a wind-turbine blade is modelled by the complex number \(z = 2\left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right)\).

(a)

Use de Moivre's theorem to write \(z^5\) in the form \(r(\cos\theta + i\sin\theta)\).

(2)
(b)

Hence write \(z^5\) in the form \(a + bi\), giving \(a\) and \(b\) correct to 3 significant figures.

(2)
(c)

Find the smallest positive integer \(n\) for which \(z^n\) is a positive real number.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to use \(r^5\) and \(5\theta\)✓ 1
A1 \(32\left(\cos\frac{5\pi}{6} + i\sin\frac{5\pi}{6}\right)\)✓ 1
Part (b)
M1 evaluate \(32\cos\frac{5\pi}{6}\) and \(32\sin\frac{5\pi}{6}\)✓ 1
A1 \(-27.7 + 16.0i\)✓ 1
Part (c)
M1 argument of \(z^n\), namely \(\frac{n\pi}{6}\), equated to a multiple of \(2\pi\)✓ 1
A1 \(n = 12\)✓ 1

Exam tip: A complex number is a positive real number exactly when its argument is a multiple of \(2\pi\); a multiple of \(\pi\) alone also allows negative values.

364IBMI3H-1-05r48-04Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium5 marks

Calculator

The three phase voltages in a wind-farm transformer are the solutions of \(z^3 = 8i\), where \(z\) is a complex number.

(a)

Write \(8i\) in the form \(re^{i\theta}\), where \(r > 0\) and \(-\pi < \theta \le \pi\).

(1)
(b)

Hence find the three solutions of \(z^3 = 8i\), giving each in the form \(re^{i\theta}\) with \(-\pi < \theta \le \pi\).

(4)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(8e^{i\pi/2}\)✓ 1
Part (b)
M1 attempt to take the cube root of the modulus and one third of the argument, giving \(r = 2\), \(\theta = \frac{\pi}{6}\)✓ 1
M1 attempt to add \(\frac{2\pi}{3}\) (or \(2\pi k\) before dividing by 3) to find further arguments✓ 1
A1 two correct solutions from \(2e^{i\pi/6}\), \(2e^{i5\pi/6}\), \(2e^{-i\pi/2}\)✓ 1
A1 all three solutions correct and in the required range✓ 1

Common error: When finding roots, give every argument in the required interval; a value such as \(\frac{3\pi}{2}\) must be rewritten as \(-\frac{\pi}{2}\).

365IBMI3H-1-05r48-05Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

Calculator

A market stall sells small and large baskets. The matrix \(M = \begin{pmatrix} 3 & 2 \\ 5 & 4 \end{pmatrix}\) records the baskets sold: row 1 is Saturday (3 small, 2 large) and row 2 is Sunday (5 small, 4 large). The takings were €34 on Saturday and €62 on Sunday. Let €\(x\) be the price of a small basket and €\(y\) the price of a large basket, so that \(M\binom{x}{y} = \binom{34}{62}\).

(a)

Find \(\det M\), and state why \(M^{-1}\) exists.

(2)
(b)

Find \(M^{-1}\).

(2)
(c)

Use \(M^{-1}\) to find the price of a small basket and the price of a large basket.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to find \(3\times4 - 2\times5\)✓ 1
A1 \(\det M = 2\), which is non-zero, so \(M^{-1}\) exists✓ 1
Part (b)
M1 attempt to swap the leading-diagonal entries and change the signs of the others✓ 1
A1 \(M^{-1} = \frac{1}{2}\begin{pmatrix} 4 & -2 \\ -5 & 3 \end{pmatrix}\)✓ 1
Part (c)
M1 attempt to evaluate \(M^{-1}\binom{34}{62}\)✓ 1
A1 small basket \(x = €6\)✓ 1
A1 large basket \(y = €8\)✓ 1

Exam tip: Pre-multiply by the inverse: \(M^{-1}\) must go on the left of the column vector, not the right.

366IBMI3H-1-05r48-10Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium6 marks

Calculator

A café buys an espresso machine for €12 000 using a loan. The annual nominal interest rate is 6%, compounded monthly. The loan is repaid by equal payments at the end of each month for 4 years.

(a)

Find the monthly payment, correct to the nearest cent.

(2)
(b)

Find the total interest paid over the 4 years, using the unrounded monthly payment. Give your answer to the nearest euro.

(2)
(c)

Find the amount still owed immediately after the 24th payment. Give your answer to the nearest euro.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 use of a finance solver (or the annuity formula) with \(N = 48\), \(I\% = 6\), \(PV = 12000\), \(FV = 0\), 12 payments and 12 compounding periods per year✓ 1
A1 €281.82✓ 1
Part (b)
M1 attempt to find \(48 \times 281.82\ldots - 12000\)✓ 1
A1 €1527✓ 1
Part (c)
M1 use of a finance solver with \(N = 24\) (or \(12000(1.005)^{24}\) minus the accumulated payments)✓ 1
A1 €6359✓ 1

Exam tip: State the values you enter in a finance solver (N, I%, PV, PMT, FV, P/Y, C/Y) so that a slip in one setting can still earn the method mark.

367IBMI3H-1-05r48-11Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium8 marks

Calculator

A solar array produces 420 MWh of electricity in its first year. Because of ageing, the output in each later year is 0.5% less than in the previous year.

(a)

Write down the common ratio of this geometric sequence.

(1)
(b)

Find the output in year 10, correct to 3 significant figures.

(2)
(c)

Find the total output over the first 25 years, correct to 3 significant figures.

(2)
(d)

Find the first year in which the output is below 385 MWh.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(0.995\)✓ 1
Part (b)
M1 \(420 \times 0.995^{9}\)✓ 1
A1 \(401\) MWh✓ 1
Part (c)
M1 \(\frac{420(1 - 0.995^{25})}{1 - 0.995}\)✓ 1
A1 \(9890\) MWh✓ 1
Part (d)
M1 \(420 \times 0.995^{n-1} < 385\)✓ 1
M1 attempt to use logarithms, for example \(n - 1 > \frac{\ln(385/420)}{\ln 0.995}\)✓ 1
A1 \(n = 19\)✓ 1

Common error: The \(n\)th term of a geometric sequence is \(u_1 r^{n-1}\), not \(u_1 r^{n}\); check the exponent against year 1.

368IBMI3H-1-05r48-12Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium6 marks

Calculator

The number of cases of an infection is modelled by \(N = 120e^{kt}\), where \(t\) is the number of days after 1 March and \(k\) is a constant. On 15 March (\(t = 14\)) there were 840 cases.

(a)

Find the value of \(k\), correct to 3 significant figures.

(2)
(b)

Find the time taken for the number of cases to double, using the unrounded value of \(k\). Give your answer to 3 significant figures.

(2)
(c)

An extra clinic will open when the number of cases reaches 10 000. Find the value of \(t\) at which this first happens, correct to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(840 = 120e^{14k}\), so \(k = \frac{\ln 7}{14}\)✓ 1
A1 \(k = 0.139\)✓ 1
Part (b)
M1 \(e^{kt} = 2\), so \(t = \frac{\ln 2}{k}\)✓ 1
A1 \(4.99\) days✓ 1
Part (c)
M1 \(10000 = 120e^{kt}\), so \(t = \frac{\ln(10000/120)}{k}\)✓ 1
A1 \(t = 31.8\)✓ 1

Exam tip: Keep the unrounded value of \(k\) in your calculator memory for later parts, then round only the final answer.

369IBMI3H-1-05r48-13Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Medium7 marks

Calculator

The motion of a wave-energy buoy involves the equation \(s^2 + 4s + 13 = 0\), where \(s\) is a complex number.

(a)

Solve the equation, giving your answers in the form \(a + bi\).

(2)
(b)

For the root \(s = -2 + 3i\), find the modulus and the argument, in radians, each correct to 3 significant figures.

(3)
(c)

The amplitude of the buoy's motion decays in proportion to \(e^{-2t}\), where \(t\) is in seconds. Find the time for the amplitude to fall to 1% of its initial value, correct to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to use the quadratic formula or to complete the square, for example \((s+2)^2 = -9\)✓ 1
A1 \(s = -2 \pm 3i\)✓ 1
Part (b)
A1 modulus \(\sqrt{13} = 3.61\)✓ 1
M1 attempt to find the argument as \(\pi - \tan^{-1}\left(\frac{3}{2}\right)\) (second quadrant)✓ 1
A1 argument \(= 2.16\) rad✓ 1
Part (c)
M1 \(e^{-2t} = 0.01\)✓ 1
A1 \(t = 2.30\) s✓ 1

Common error: A calculator gives \(\tan^{-1}\left(\frac{3}{-2}\right)\) in the fourth quadrant; sketch the complex number to place the argument in the correct quadrant.

370IBMI3H-1-04r10-14Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Hard7 marks

A cyclist pays £150 at the end of each month into a savings account that pays a nominal annual interest rate of r%, compounded monthly. After 5 years the account holds £9945, to the nearest pound.

(a)

State the values to enter in a TVM solver to find r.

(3)
(b)

Find the value of r, to 2 significant figures.

(2)
(c)

Find the total interest earned over the 5 years.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
N = 60 and PMT = −150✓ 1
PV = 0 and FV = 9945✓ 1
P/Y = C/Y = 12 with payments at the end of each period✓ 1
Part (b)
use of solver to find I%✓ 1
r = 4.0✓ 1
Part (c)
total paid in = 150 × 60 = £9000✓ 1
interest = 9945 − 9000 = £945✓ 1

Exam tip: Interest earned is the final balance minus the total of the deposits, not the final balance on its own.

371IBMI3H-1-03r11-13Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Hard8 marks

Three signal beacons are placed at the points in the complex plane (units: km) that satisfy \(p(z) = z^3 - 5z^2 + 17z - 13 = 0\).

(a)

Show that \(z = 1\) is a solution.

(1)
(b)

Find the other two solutions.

(3)
(c)

Find the area of the triangle formed by the three beacons.

(2)
(d)

Write the beacon with positive imaginary part as \(re^{i\theta}\), giving \(r\) and \(\theta\) (in radians) to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
R1 substituting \(z = 1\) gives \(1 - 5 + 17 - 13 = 0\), so \(z = 1\) is a root✓ 1
Part (b)
M1 factorise: \(p(z) = (z - 1)(z^2 - 4z + 13)\)✓ 1
M1 solve \(z^2 - 4z + 13 = 0\) (formula or completing the square)✓ 1
A1 \(z = 2 + 3i\) and \(z = 2 - 3i\)✓ 1
Part (c)
M1 beacons at \((1,0)\), \((2,3)\) and \((2,-3)\): base 6 on the line \(x = 2\) and perpendicular height \(2 - 1 = 1\)✓ 1
A1 area \(= \frac{1}{2} \times 6 \times 1 = 3\) km2✓ 1
Part (d)
A1 \(r = \sqrt{13} = 3.61\)✓ 1
A1 \(\theta = \tan^{-1}\frac{3}{2} = 0.983\)✓ 1

Common error: Writing only one of the complex conjugate roots. A polynomial with real coefficients has its non-real roots in conjugate pairs.

372IBMI3H-1-03r11-14Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Hard7 marks

In a game, positions on the map \((x, y)\) are transformed by the matrix \(\mathbf{A} = \begin{pmatrix}k&2\\3&1\end{pmatrix}\), where \(k\) is a constant. It is known that 4 is an eigenvalue of \(\mathbf{A}\).

(a)

Find the value of \(k\).

(3)
(b)

Find the other eigenvalue of \(\mathbf{A}\).

(2)
(c)

Find an eigenvector for the eigenvalue 4 and describe the effect of the transformation on points on the corresponding line through the origin.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(\det(\mathbf{A} - 4\mathbf{I}) = 0\)✓ 1
M1 \((k - 4)(1 - 4) - (2)(3) = 0\)✓ 1
A1 \(k = 2\)✓ 1
Part (b)
M1 \(\lambda^2 - 3\lambda - 4 = 0\) (or trace \(= 3\))✓ 1
A1 \(\lambda = -1\)✓ 1
Part (c)
A1 eigenvector \(\begin{pmatrix}1\\1\end{pmatrix}\) (any multiple), line \(y = x\)✓ 1
R1 points on \(y = x\) stay on the line and move 4 times further from the origin✓ 1

Exam tip: If \(\lambda\) is an eigenvalue then \(\det(\mathbf{A}-\lambda\mathbf{I}) = 0\). When an eigenvalue is given, this gives an equation for the unknown in the matrix.

373IBMI3H-1-03r13-05Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Hard6 marks

Three thrusters on a probe are positioned at the points in the Argand diagram (units: metres) that satisfy z3 = 8i.

(a)

Write 8i in the form reiθ.

(1)
(b)

Find the three solutions of z3 = 8i, giving each in the form reiθ with −π < θ ≤ π.

(3)
(c)

The three thrusters form an equilateral triangle. Find its area, giving your answer to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 8i = 8eπi/2✓ 1
Part (b)
M1 modulus 81/3 = 2 and argument (π/2 + 2kπ)/3✓ 1
A1 2eπi/6 and 2e5πi/6✓ 1
A1 2e−πi/2✓ 1
Part (c)
M1 attempt to use area = ½r2sin120° three times, or ½ × side × side × sin60° with side 2√3✓ 1
A1 area = 3√3 = 5.20 m2 (5.196…)✓ 1

Common error: When taking an nth root, add multiples of 2π to the argument before dividing by n; the roots are equally spaced by 2π/n around a circle.

374IBMI3H-1-03r13-09Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Hard8 marks

In a warehouse of stored green coffee, a beetle population has juveniles J and adults A. Each year, every adult produces 1.5 juveniles, 40% of juveniles become adults, and 60% of adults survive. Hence Jn+1 = 1.5An and An+1 = 0.4Jn + 0.6An.

(a)

Write down the matrix M such that (Jn+1, An+1) = M(Jn, An).

(1)
(b)

Find the eigenvalues of M, giving each to 3 significant figures.

(3)
(c)

Use the dominant eigenvalue to state the long-term percentage growth in the population each year, giving your answer to 2 significant figures.

(2)
(d)

In the long term, find the ratio J : A in the form k : 1, giving k to 3 significant figures. Use the unrounded eigenvalue.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 M = [[0, 1.5],[0.4, 0.6]]✓ 1
Part (b)
M1 set det(M − λI) = 0, giving λ2 − 0.6λ − 0.6 = 0✓ 1
A1 λ = 1.13 (1.1307…)✓ 1
A1 λ = −0.531 (−0.5307…)✓ 1
Part (c)
R1 the largest eigenvalue determines long-term growth✓ 1
A1 growth is about 13% per year✓ 1
Part (d)
M1 use −λJ + 1.5A = 0 (or equivalent) with the dominant λ✓ 1
A1 k = 1.5/1.1307… = 1.33✓ 1

Exam tip: The long-term behaviour follows the eigenvector of the dominant (largest modulus) eigenvalue; only the ratio of its entries matters.

375IBMI3H-1-03r13-14Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Hard7 marks

The vibration of a bridge deck is governed by the characteristic equation λ2 + 6λ + 25 = 0.

(a)

Solve the equation, giving the roots in the form a + bi.

(2)
(b)

For the root with positive imaginary part, find its modulus and its argument. Give the argument in radians to 3 significant figures, with −π < arg ≤ π.

(3)
(c)

Find the principal argument of λ5 for this root (−π < arg ≤ π), to 3 significant figures. Use the unrounded argument from (b).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 use of the quadratic formula or completing the square: (λ + 3)2 = −16✓ 1
A1 λ = −3 + 4i and λ = −3 − 4i✓ 1
Part (b)
A1 modulus = 5✓ 1
M1 argument = π − arctan(4/3) (second quadrant)✓ 1
A1 argument = 2.21 rad (2.2143…)✓ 1
Part (c)
M1 argument of λ5 is 5 × 2.2143… = 11.07…✓ 1
A1 subtract 4π to obtain −1.49 rad (−1.4949…)✓ 1

Common error: The argument of a power is the argument multiplied by the power, but then reduce it by multiples of 2π to give the principal value.

376IBMI3H-1-04r14-06Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Hard8 marks

A pest population on a farm has juveniles \(j_n\) and adults \(a_n\) in generation \(n\), with \(\begin{pmatrix}j_{n+1}\\a_{n+1}\end{pmatrix}=M\begin{pmatrix}j_n\\a_n\end{pmatrix}\), where \(M=\begin{pmatrix}0&3\\0.4&0.6\end{pmatrix}\).

(a)

Show that the eigenvalues of \(M\) satisfy \(\lambda^2-0.6\lambda-1.2=0\).

(2)
(b)

Find the dominant eigenvalue, to 4 significant figures.

(2)
(c)

Interpret the dominant eigenvalue in terms of the long-term growth of the population.

(1)
(d)

Using the unrounded dominant eigenvalue \(\lambda\), find the long-term percentage of the population that are juveniles, to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 sets \(\det(M-\lambda I)=0\): \((0-\lambda)(0.6-\lambda)-3\times0.4=0\)✓ 1
A1 \(\lambda^2-0.6\lambda-1.2=0\)✓ 1
Part (b)
M1 quadratic formula \(\lambda=\frac{0.6\pm\sqrt{5.16}}{2}\)✓ 1
A1 1.436✓ 1
Part (c)
R1 in the long run the population grows by a factor of about 1.436 per generation, i.e. about 43.6% per generation✓ 1
Part (d)
M1 eigenvector condition \(3a=\lambda j\) from the first row✓ 1
A1 \(j:a=3:\lambda\) so proportion \(\frac{3}{3+\lambda}\)✓ 1
A1 67.6 %✓ 1

Exam tip: The long-term proportions come from the eigenvector of the dominant eigenvalue, scaled so that its entries sum to 1.

377IBMI3H-1-04r15-05Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Hard8 marks

A robot follows a spiral path. Its position at the end of move n is the complex number \(w_n = \left(1.2e^{0.5i}\right)^n\), where \(n = 1, 2, 3, \ldots\) and distances are in metres from the starting point at the origin.

(a)

Show that \(w_n = 1.2^n e^{0.5ni}\) and state the distance of the robot from the origin after move n.

(2)
(b)

Find the smallest value of n for which the robot is more than 10 m from the origin.

(3)
(c)

For this value of n, find the position of the robot in the form \(a + bi\), with a and b to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 use of \((re^{i\theta})^n = r^ne^{in\theta}\)✓ 1
A1 distance \(= 1.2^n\) m✓ 1
Part (b)
M1 \(1.2^n > 10\)✓ 1
M1 \(n > \dfrac{\ln 10}{\ln 1.2} = 12.6\ldots\) (or table of values on GDC)✓ 1
A1 \(n = 13\)✓ 1
Part (c)
M1 \(1.2^{13}(\cos 6.5 + i\sin 6.5)\) (angle in radians, ECF from (b))✓ 1
A1 \(a = 10.4\)✓ 1
A1 \(b = 2.30\)✓ 1

Exam tip: State the angle \(0.5n\) in radians and keep your calculator in radian mode; a degree-mode answer here would be wildly wrong.

378IBMI3H-1-04r15-12Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Hard8 marks

A fishery model divides a fish stock into three age classes: young, medium and old. Let \(\mathbf{x}_n\) be the column vector of the numbers (in thousands) in each class after n years. Then \(\mathbf{x}_{n+1} = \mathbf{L}\mathbf{x}_n\) where \(\mathbf{L} = \begin{pmatrix}0&3&5\\0.4&0&0\\0&0.6&0\end{pmatrix}\) and \(\mathbf{x}_0 = \begin{pmatrix}200\\120\\80\end{pmatrix}\).

(a)

State what the entry 0.6 in L represents.

(1)
(b)

Find the numbers in each age class after 2 years.

(2)
(c)

Find the total number of fish after 5 years, in thousands, to 3 significant figures.

(2)
(d)

Use technology to find the real eigenvalue of L to 4 significant figures. Hence state the long-term annual percentage growth of the stock, to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 60% of the medium-aged fish survive to become old fish the next year✓ 1
Part (b)
M1 \(\mathbf{L}^2\mathbf{x}_0\) or two successive products✓ 1
A1 young 600, medium 304, old 48 (thousand)✓ 1
Part (c)
M1 \(\mathbf{L}^5\mathbf{x}_0 = (2102.4,\ 652.8,\ 276.48)^T\) (technology)✓ 1
A1 3030 thousand (accept \(3.03\times10^3\))✓ 1
Part (d)
M1 use of technology to find the eigenvalues of L✓ 1
A1 real eigenvalue 1.428✓ 1
A1 long-term growth of about 42.8% per year✓ 1

Common error: Eigenvalues of a model like this can include complex pairs; the long-term growth rate comes from the real eigenvalue with the largest modulus.

379IBMI3H-1-05r19-04Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Hard8 marks

A fisheries scientist models the stock of a fish species in a lake using two groups, juveniles (\(J\)) and adults (\(A\)). Each year, 40% of juveniles remain juveniles, 60% of juveniles become adults, each adult produces 1.5 new juveniles, and 30% of adults survive to the next year. The stock one year later is given by
\(\begin{pmatrix}J_{n+1}\\A_{n+1}\end{pmatrix}=\mathbf{M}\begin{pmatrix}J_{n}\\A_{n}\end{pmatrix},\quad \mathbf{M}=\begin{pmatrix}0.4&1.5\\0.6&0.3\end{pmatrix}.\)

(a)

Find the eigenvalues of \(\mathbf{M}\).

(3)
(b)

Explain what the largest eigenvalue tells the scientist about the long-term behaviour of the total stock.

(1)
(c)

Find an eigenvector of \(\mathbf{M}\) corresponding to the largest eigenvalue.

(2)
(d)

Hence find the long-term percentage of the stock that is juvenile.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to solve \(\det(\mathbf{M}-\lambda\mathbf{I})=0\)✓ 1
A1 \(\lambda^2-0.7\lambda-0.78=0\)✓ 1
A1 \(\lambda=1.3\) and \(\lambda=-0.6\)✓ 1
Part (b)
R1 in the long term the stock grows by a factor of about 1.3 each year, an increase of about 30% per year✓ 1
Part (c)
M1 \((0.4-1.3)J+1.5A=0\), so \(-0.9J+1.5A=0\)✓ 1
A1 any multiple of \(\begin{pmatrix}5\\3\end{pmatrix}\)✓ 1
Part (d)
M1 \(\frac{5}{5+3}\) (ECF from (c))✓ 1
A1 62.5%✓ 1

Exam tip: An eigenvector is only defined up to a scalar multiple, so any non-zero multiple of the correct vector earns full credit; for proportions, only the ratio of its entries matters.

380IBMI3H-1-05r20-13Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Hard6 marks

Calculator

In a crafting game a player brews \(x\), \(y\) and \(z\) units of three potions. The total strength, speed and stealth gained are
strength \(=x+2y+kz\), speed \(=2x+y+z\), stealth \(=x-y+2z\),
where \(k\) is a constant set by a rare herb. This is written as \(\mathbf{A}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}\text{strength}\\\text{speed}\\\text{stealth}\end{pmatrix}\) with \(\mathbf{A}=\begin{pmatrix}1&2&k\\2&1&1\\1&-1&2\end{pmatrix}\).

(a)

Show that \(\det\mathbf{A}=-3-3k\).

(2)
(b)

Find the value of \(k\) for which \(\mathbf{A}\) has no inverse.

(1)
(c)

Explain in context why, for this value of \(k\), the player cannot choose the three stat totals freely.

(1)
(d)

Now \(k=2\). The player wants strength 10, speed 8 and stealth 7. Find \(x\), \(y\) and \(z\).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to expand the determinant, e.g. \(1(2+1)-2(4-1)+k(-2-1)\)✓ 1
A1 \(3-6-3k=-3-3k\) (AG)✓ 1
Part (b)
A1 \(-3-3k=0\) so \(k=-1\)✓ 1
Part (c)
R1 the three equations are not independent (the matrix is singular), so for some targets there is no solution and for others there are infinitely many✓ 1
Part (d)
M1 \(\begin{pmatrix}x\\y\\z\end{pmatrix}=\mathbf{A}^{-1}\begin{pmatrix}10\\8\\7\end{pmatrix}\) or use of technology to solve the system✓ 1
A1 \(x=2\), \(y=1\), \(z=3\)✓ 1

Exam tip: A matrix has no inverse exactly when its determinant is zero. A system with a singular matrix has either no solution or infinitely many, never exactly one.

381IBMI3H-1-04r24-05Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Hard7 marks

The characteristic equation of the signal filter in an acoustic tracking buoy is \(z^3+pz^2+qz+15=0\), where \(p\) and \(q\) are real constants. One root of the equation is \(1+2i\).

(a)

Write down a second root of the equation.

(1)
(b)

Find the third root.

(3)
(c)

Find the values of \(p\) and \(q\).

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(1-2i\) (the complex conjugate)✓ 1
Part (b)
M1 the product of the three roots is \(-15\)✓ 1
A1 \((1+2i)(1-2i)=5\)✓ 1
A1 third root \(=-3\)✓ 1
Part (c)
M1 use the sum of roots \(=-p\) (or expand \((z^2-2z+5)(z+3)\))✓ 1
A1 \(p=1\)✓ 1
A1 \(q=5+(-3)(2)=-1\) using the sum of the products of pairs✓ 1

Exam tip: For a polynomial with real coefficients, complex roots come in conjugate pairs; this gives you a second root for free.

382IBMI3H-1-04r24-14Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Hard8 marks

The surface area covered by an algal species A in a bay is modelled by \(A=12e^{0.18t}\) m2, where \(t\) is the time in days after monitoring begins. A second species B covers an area \(B=300e^{-0.05t}\) m2.

(a)

Find the doubling time of species A, to 3 significant figures.

(3)
(b)

Find the time at which species A covers 500 m2, to 3 significant figures.

(2)
(c)

Find the time at which the two species cover equal areas, to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(e^{0.18T}=2\)✓ 1
A1 \(T=\dfrac{\ln2}{0.18}\)✓ 1
A1 3.85 days✓ 1
Part (b)
M1 \(12e^{0.18t}=500\), so \(t=\dfrac{\ln(500/12)}{0.18}\)✓ 1
A1 20.7 days✓ 1
Part (c)
M1 \(12e^{0.18t}=300e^{-0.05t}\)✓ 1
M1 \(e^{0.23t}=25\) and take logarithms✓ 1
A1 \(t=\dfrac{\ln25}{0.23}=14.0\) days✓ 1

Exam tip: To solve an equation with exponentials on both sides, collect them into a single exponential, then take natural logarithms.

383IBMI3H-1-05r32-11Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Hard7 marks

A climber ascends 420 m in the first hour of a trek. In each following hour she gains 30 m less height than in the previous hour, so the heights gained each hour form an arithmetic sequence.

(a)

Find the height gained in the 5th hour.

(2)
(b)

Find the total height gained in the first 8 hours.

(2)
(c)

The next camp is 2800 m above the start. Find the least number of complete hours needed to reach it, and the total height gained after that number of hours.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(420+4\times(-30)\)✓ 1
A1 \(300\) m✓ 1
Part (b)
M1 \(\frac{8}{2}\left(2\times420+7\times(-30)\right)\)✓ 1
A1 \(2520\) m✓ 1
Part (c)
M1 sets up \(S_n\ge2800\) (or evaluates \(S_9\) and \(S_{10}\))✓ 1
A1 \(n=10\) hours (\(S_9=2700<2800\))✓ 1
A1 total \(2850\) m✓ 1

Exam tip: When asked for the least whole number of terms, check the neighbouring values (here \(S_9\) and \(S_{10}\)) rather than just rounding a solution of the equation.

384IBMI3H-1-08r33-04Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Hard8 marks

A model for a volcanic tremor has the characteristic equation z3 − 7z2 + 25z − 39 = 0, which has real coefficients. One root is 2 + 3i.

(a)

Write down a second root and justify your answer.

(2)
(b)

Find the third root.

(3)
(c)

Express (2 + 3i)5 in the form reiθ with −π < θ ≤ π, giving r and θ to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 2 − 3i✓ 1
R1 complex roots of a polynomial with real coefficients occur in conjugate pairs✓ 1
Part (b)
M1 sum of the roots = 7 (or factor out z2 − 4z + 13)✓ 1
M1 third root = 7 − (2 + 3i) − (2 − 3i) or division of the cubic✓ 1
A1 3✓ 1
Part (c)
M1 modulus √13 and argument arctan(3/2) = 0.983 for 2 + 3i✓ 1
A1 r = 135/2 = 609✓ 1
A1 θ = 5 × 0.9828 − 2π = −1.37 (the Cartesian form is 122 − 597i)✓ 1

Common error: When de Moivre's theorem gives an argument outside the required interval, subtract 2π until it lies in (−π, π], then check the quadrant against the Cartesian form.

385IBMI3H-1-08r33-07Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Hard8 marks

A ship carries x type S, y type M and z type L containers. The ship carries 120 containers in total. The masses of a type S, type M and type L container are 14 t, 10 t and 6 t respectively, and the total mass carried is 1160 t. The shipping fees per container are €800, €600 and €450 respectively, and the total fees are €72000.

(a)

Write the information as a matrix equation Av = b, where v = (x, y, z) as a column vector.

(2)
(b)

Use the inverse of A to find the number of containers of each type.

(3)
(c)

A different design of type L container has mass m tonnes instead of 6 t (the other data are unchanged, so the third column of the mass row becomes m). Show that the determinant of the new matrix is 200m − 1400, and find the value of m for which the numbers of containers cannot be found uniquely.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 A with rows (1, 1, 1), (14, 10, 6), (800, 600, 450)✓ 1
A1 v = (x, y, z)T and b = (120, 1160, 72000)T✓ 1
Part (b)
M1 v = A−1b✓ 1
A1 x = 30, y = 50✓ 1
A1 z = 40✓ 1
Part (c)
M1 expansion of the determinant, e.g. 1(4500 − 600m) − 1(6300 − 800m) + 1(8400 − 8000)✓ 1
A1 200m − 1400 (AG)✓ 1
A1 matrix is singular when m = 7✓ 1

Exam tip: A matrix has no inverse exactly when its determinant is zero; then the system has either no solution or infinitely many, so the unknowns cannot be found uniquely.

386IBMI3H-1-08r33-10Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Hard8 marks

The concentrations of sulfur dioxide and carbon dioxide in a volcanic crater, in suitable units, are modelled hourly by xn+1 = Mxn, where xn = (SO2, CO2) as a column vector and M has rows (0.9, 0.4) and (0.3, 0.8). The initial concentrations are x0 = (11, 3).

(a)

Show that (4, 3) is an eigenvector of M and state the corresponding eigenvalue.

(2)
(b)

Find the other eigenvalue and a corresponding eigenvector.

(3)
(c)

Write x0 as a combination of the two eigenvectors, and hence find the SO2 concentration after 6 hours, to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 M(4, 3)T = (4.8, 3.6)T✓ 1
A1 = 1.2 × (4, 3)T, so eigenvalue 1.2✓ 1
Part (b)
M1 use trace: 0.9 + 0.8 = 1.7 = 1.2 + λ✓ 1
A1 λ = 0.5✓ 1
A1 eigenvector (1, −1) (any multiple)✓ 1
Part (c)
M1 (11, 3) = 2(4, 3) + 3(1, −1)✓ 1
M1 x6 = 2 × 1.26(4, 3) + 3 × 0.56(1, −1)✓ 1
A1 23.9 (23.93...)✓ 1

Exam tip: Because M^n applied to an eigenvector just multiplies it by the eigenvalue to the power n, write the starting vector in eigenvector form before taking powers.

387IBMI3H-1-08r34-11Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Hard8 marks

A population of lava lizards on a volcanic island is split into three age classes: hatchlings, juveniles and adults. Each year, every juvenile produces on average 3 hatchlings and every adult produces on average 2 hatchlings. 50% of hatchlings survive to become juveniles, and 40% of juveniles survive to become adults; adults do not survive more than a year. The Leslie matrix is \(L = \begin{pmatrix}0&3&2\\0.5&0&0\\0&0.4&0\end{pmatrix}\).

(a)

Show that the eigenvalues of L satisfy \(\lambda^3 = 1.5\lambda + 0.4\).

(3)
(b)

Find the positive eigenvalue of L, to four significant figures.

(1)
(c)

Explain what this eigenvalue tells you about the long-term population.

(1)
(d)

In the long term, the numbers in the three classes are in the ratio \(1 : \frac{0.5}{\lambda} : \frac{0.2}{\lambda^2}\). Find the long-term percentage of the population that are hatchlings, to three significant figures. Use \(\lambda = 1.3410013\).

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(\det(L - \lambda I) = 0\) expanded✓ 1
A1 \(-\lambda(\lambda^2) - 3(-0.5\lambda) + 2(0.2) = 0\) (cofactor terms correct)✓ 1
A1 \(\lambda^3 = 1.5\lambda + 0.4\) (AG)✓ 1
Part (b)
A1 \(\lambda = 1.341\) (1.34100…)✓ 1
Part (c)
R1 the population grows by a factor of about 1.34 each year, i.e. increases by about 34% per year✓ 1
Part (d)
M1 total \(1 + \frac{0.5}{\lambda} + \frac{0.2}{\lambda^2}\) with \(\lambda\) substituted✓ 1
A1 hatchling proportion \(\frac{1}{1 + 0.5/\lambda + 0.2/\lambda^2} = 0.6738\)✓ 1
A1 67.4%✓ 1

Exam tip: In a Leslie model the dominant eigenvalue is the long-term yearly growth factor and its eigenvector gives the stable age proportions.

388IBMI3H-1-06r35-05Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Hard8 marks

A volcano is classified at the start of each year as Quiet (Q) or Restless (R). If it is Quiet, it is Quiet next year with probability 0.9. If it is Restless, it is Quiet next year with probability 0.3. The transition matrix is T = [[0.9, 0.3], [0.1, 0.7]], acting on the state vector (q, r)T.

(a)

Show that the eigenvalues of T are 1 and 0.6.

(3)
(b)

Find an eigenvector of T corresponding to λ = 1, and hence find the long-term probability that the volcano is Restless.

(3)
(c)

The volcano is Quiet now. Find the probability that it is Restless in three years' time.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 det(T − λI) = (0.9 − λ)(0.7 − λ) − 0.03 = 0✓ 1
A1 λ2 − 1.6λ + 0.6 = 0✓ 1
A1 (λ − 1)(λ − 0.6) = 0, so λ = 1 or 0.6✓ 1
Part (b)
M1 (T − I)v = 0 gives −0.1q + 0.3r = 0, so v = (3, 1)T✓ 1
M1 scale so that q + r = 1: (0.75, 0.25)T✓ 1
A1 long-term probability of Restless = 0.25✓ 1
Part (c)
M1 T3(1, 0)T (or 0.25 − 0.25 × 0.63 from the eigenvalues)✓ 1
A1 0.196✓ 1

Exam tip: The eigenvector for λ = 1 only gives probabilities once it is scaled so that its entries add to 1.

389IBMI3H-1-06r35-06Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Hard8 marks

A seabird colony on a volcanic island has Jn juveniles and An adults in year n. Each adult produces on average 1.2 juveniles per year, half of the juveniles survive to become adults, and 80% of adults survive each year. Hence Jn+1 = 1.2An and An+1 = 0.5Jn + 0.8An, which can be written (Jn+1, An+1)T = L(Jn, An)T.

(a)

Write down L and show that its characteristic equation is λ2 − 0.8λ − 0.6 = 0.

(2)
(b)

Find the dominant eigenvalue of L, to 3 significant figures.

(2)
(c)

Interpret this eigenvalue in the context of the colony.

(1)
(d)

Find the long-term proportion of the colony that are juveniles, giving your answer to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 L = [[0, 1.2], [0.5, 0.8]]✓ 1
A1 det(L − λI) = −λ(0.8 − λ) − 0.6 = λ2 − 0.8λ − 0.6 = 0✓ 1
Part (b)
M1 solve the quadratic, λ = (0.8 ± √3.04) ÷ 2✓ 1
A1 λ = 1.27✓ 1
Part (c)
A1 in the long term the colony grows by about 27% each year (the population is multiplied by about 1.27)✓ 1
Part (d)
M1 from the first row of (L − λI)v = 0: 1.2A = λJ, so J ÷ A = 1.2 ÷ 1.2718… = 0.9436…✓ 1
M1 proportion = J ÷ (J + A) = 0.9436 ÷ 1.9436✓ 1
A1 0.485 (48.5%)✓ 1

Exam tip: The dominant eigenvalue is the long-term growth factor, and its eigenvector gives the long-term proportions in each class.

390IBMI3H-1-06r35-08Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Hard8 marks

Containers move weekly between three ports A, B and C. Each week, the proportion of containers at a port that move to each port is given by the transition matrix
T = [[0.7, 0.2, 0.1], [0.2, 0.6, 0.3], [0.1, 0.2, 0.6]],
acting on the column vector (number at A, number at B, number at C)T. At the start of week 0 there are 400 containers at A, 400 at B and 200 at C.

(a)

Explain why each column of T sums to 1.

(1)
(b)

Find the number of containers at each port after 3 weeks, giving each answer to the nearest whole number.

(3)
(c)

The total number of containers stays at 1000. Find, to the nearest whole number, the number of containers at port A in the long term.

(4)
Show mark scheme
Marking pointMarkNotes
Part (a)
R1 every container at a port this week is at exactly one of the three ports next week, so the proportions leaving a given port add to 1✓ 1
Part (b)
M1 T3(400, 400, 200)T✓ 1
A1 357 at A and 376 at B✓ 1
A1 267 at C✓ 1
Part (c)
M1 solve Ts = s with sA + sB + sC = 1 (eigenvector for eigenvalue 1)✓ 1
A1 s = (10/29, 11/29, 8/29)T✓ 1
M1 1000 × 10/29✓ 1
A1 345 containers✓ 1

Exam tip: The long-term distribution is the eigenvector for eigenvalue 1, scaled so that its entries sum to 1 (or to the total number of containers).

391IBMI3H-1-06r40-05Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Hard8 marks

The population of a fish species is split into juveniles, J, and adults, A. Each year, every adult produces on average 1.5 juveniles, 40% of juveniles survive to become adults, and 80% of adults survive. The populations satisfy \(\begin{pmatrix}J_{n+1}\\A_{n+1}\end{pmatrix} = \mathbf{M}\begin{pmatrix}J_n\\A_n\end{pmatrix}\), where n is the number of years. Initially \(J_0 = 200\) and \(A_0 = 300\).

(a)

Write down the matrix \(\mathbf{M}\).

(2)
(b)

Find the number of juveniles and adults after two years.

(3)
(c)

Find the largest eigenvalue of \(\mathbf{M}\), to 3 significant figures, and interpret it for the long-term population.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 first row \((0,\ 1.5)\)✓ 1
A1 second row \((0.4,\ 0.8)\)✓ 1
Part (b)
M1 attempt at \(\mathbf{M}^2\begin{pmatrix}200\\300\end{pmatrix}\) or two successive products✓ 1
A1 after one year: 450 juveniles, 320 adults✓ 1
A1 after two years: 480 juveniles, 436 adults✓ 1
Part (c)
M1 \(\det(\mathbf{M} - \lambda\mathbf{I}) = \lambda^2 - 0.8\lambda - 0.6 = 0\)✓ 1
A1 \(\lambda = 1.27\) (other eigenvalue \(-0.472\))✓ 1
R1 in the long run the population grows by about 27% each year✓ 1

Exam tip: In a population model, the eigenvalue with the largest magnitude gives the long-term growth factor per time step.

392IBMI3H-1-06r40-14Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Hard8 marks

A conservation team models sea turtles as juveniles, J, and adults, A, using \(\begin{pmatrix}J_{n+1}\\A_{n+1}\end{pmatrix} = \mathbf{M}\begin{pmatrix}J_n\\A_n\end{pmatrix}\) with \(\mathbf{M} = \begin{pmatrix}0 & f\\0.5 & 0.6\end{pmatrix}\), where f is the mean number of juveniles produced per adult each year and n is in years.

(a)

Show that the eigenvalues of \(\mathbf{M}\) satisfy \(\lambda^2 - 0.6\lambda - 0.5f = 0\).

(2)
(b)

The population is stable when it stays unchanged from year to year. Find the value of f for which this is possible.

(3)
(c)

For this value of f, the stable population has 900 turtles in total. Find the number of juveniles.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(\det(\mathbf{M}-\lambda\mathbf{I}) = (-\lambda)(0.6-\lambda) - 0.5f\)✓ 1
A1 expand to \(\lambda^2 - 0.6\lambda - 0.5f = 0\) (AG)✓ 1
Part (b)
R1 a stable non-zero population requires \(\mathbf{M}\mathbf{v}=\mathbf{v}\), so \(\lambda = 1\) is an eigenvalue✓ 1
M1 substitute \(\lambda = 1\): \(1 - 0.6 - 0.5f = 0\)✓ 1
A1 \(f = 0.8\)✓ 1
Part (c)
M1 solve \((\mathbf{M}-\mathbf{I})\mathbf{v}=\mathbf{0}\): \(-J + 0.8A = 0\)✓ 1
A1 ratio \(J:A = 4:5\)✓ 1
A1 \(J = 400\)✓ 1

Exam tip: For a stable population, look for the eigenvector with eigenvalue 1, then scale it to the known total.

393IBMI3H-1-04r43-14Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Hard7 marks

Two drugs decay in a patient's blood exponentially. Drug X starts at 240 mg with a half-life of 5 hours. Drug Y starts at 100 mg with a half-life of 12 hours.

(a)

Write down expressions for the mass of each drug, X and Y in mg, after t hours.

(2)
(b)

Find the time at which the two masses are equal. Give your answer in hours to 3 significant figures.

(3)
(c)

Find the common mass of the drugs at this time, to 3 significant figures. Use the unrounded value of t.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 X = 240 × (1⁄2)t/5✓ 1
A1 Y = 100 × (1⁄2)t/12✓ 1
Part (b)
M1 equates 240 × 2−t/5 = 100 × 2−t/12✓ 1
M1 rearranges to 27t/60 = 2.4 and takes logarithms (or uses the GDC)✓ 1
A1 t = 10.8 hours✓ 1
Part (c)
M1 substitutes into one of the expressions✓ 1
A1 53.5 mg✓ 1

Common error: When a half-life is T, the exponent is t ÷ T with base ¹⁄₂; do not write the base as 0.5 raised to t × T.

394IBMI3H-1-05r43-11Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Hard8 marks

A warehouse robot moves in a spiral. Its position in the complex plane (metres, origin at the charging point) after n steps is \(z_n = 2\left(1.1\,\mathrm{e}^{\mathrm{i}\pi/9}\right)^n\).

(a)

Find the position after 4 steps in the form a + bi, with a and b to 3 significant figures.

(3)
(b)

Find the smallest number of steps after which the robot is more than 10 m from the origin.

(2)
(c)

The robot first returns to the positive real axis after N steps. Find N and the distance of the robot from the origin at that moment, to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to use z4 = 2 × 1.14e4πi/9✓ 1
A1 modulus 2.93 and argument 4π/9 (80°)✓ 1
A1 0.508 + 2.88i✓ 1
Part (b)
M1 set up 2 × 1.1n > 10, giving n > 16.89…✓ 1
A1 n = 17✓ 1
Part (c)
M1 set Nπ/9 = 2π✓ 1
A1 N = 18✓ 1
A1 11.1 m✓ 1

Common error: The argument of the product is the sum of the arguments, so after n steps it is nπ/9; the robot is on the positive real axis when this is a multiple of 2π, not π.

395IBMI3H-1-05r43-13Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 1Hard8 marks

The motion of a robot arm is controlled by a system whose characteristic equation is s2 + bs + 34 = 0, where b is an adjustable positive constant with b < 2√34.

(a)

Show that the roots are s = −b/2 ± i√(34 − b2/4).

(2)
(b)

For b = 6, write the root with positive imaginary part in the form reiθ, giving r and θ to 3 significant figures.

(3)
(c)

The adjustable constant is now changed so that the root with positive imaginary part has argument 2.2 radians. Show that the modulus of this root is the same for every b, and find b to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to use the quadratic formula✓ 1
A1 discriminant b2 − 136 < 0, so s = (−b ± i√(136 − b2))/2 = −b/2 ± i√(34 − b2/4) (AG)✓ 1
Part (b)
A1 root is −3 + 5i✓ 1
A1 r = √34 = 5.83✓ 1
A1 θ = π − tan−1(5/3) = 2.11 rad✓ 1
Part (c)
R1 modulus2 = b2/4 + (34 − b2/4) = 34, independent of b✓ 1
M1 real part: −b/2 = √34 cos 2.2✓ 1
A1 b = 6.86✓ 1

Common error: Use the real part of the root, not the whole root, when linking the argument to b; the argument lies in the second quadrant, so the cosine is negative.

396IBMI-030-04-r1-03Complex numbers and matrices Paper 2Medium12 marks

A city bike-share scheme has 300 bikes and three docking stations, A, B and C. Each bike is used once per day and is left at a station overnight. Data suggest the following daily behaviour.
• Of bikes starting at A: 60% return to A, 30% go to B, 10% go to C.
• Of bikes starting at B: 20% go to A, 70% return to B, 10% go to C.
• Of bikes starting at C: 30% go to A, 20% go to B, 50% return to C.
On Monday morning there are 100 bikes at each station.

(a)

Write down the transition matrix \(T\), where the entry in row \(i\), column \(j\) is the probability that a bike starting the day at station \(j\) ends the day at station \(i\).

(2)
(b)

Find the expected number of bikes at each station after three days.

(3)
(c)

Find the long-term expected number of bikes at each station.

(5)
(d)

Each station has 120 docking points. With reference to your answer to part (c), comment on whether this is suitable in the long term.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A2 \(T = \begin{pmatrix}0.6 & 0.2 & 0.3\\0.3 & 0.7 & 0.2\\0.1 & 0.1 & 0.5\end{pmatrix}\)✓ 2
(Award A1 only if the matrix is transposed or one column is incorrect.)—
Part (b)
M1 calculates \(T^3\begin{pmatrix}100\\100\\100\end{pmatrix}\)✓ 1
A1 A: 110 (110.2), B: 137 (136.6)✓ 1
A1 C: 53 (53.2)✓ 1
Part (c)
M1 sets up \(T\mathbf{v} = \mathbf{v}\), or finds the eigenvector for eigenvalue 1, or evaluates \(T^n\) for large \(n\)✓ 1
M1 uses the condition \(a + b + c = 300\) (or proportions summing to 1)✓ 1
A1 A: 108 (\(300\times\frac{13}{36}\))✓ 1
A1 C: 50 (\(300\times\frac{6}{36}\))✓ 1
A1 B: 142 (\(300\times\frac{17}{36}\))✓ 1
Part (d)
R1 in the long term about 142 bikes are expected at B, which is more than its 120 docking points✓ 1
A1 so B needs more docks (or bikes must be moved from B), while C has spare capacity (about 50 bikes for 120 docks)✓ 1

Exam tip: Check that every column of your transition matrix sums to 1 before doing any calculation. A transposed matrix gives completely different predictions, and every later part will be wrong.

397IBMI-030-04-r2-05Complex numbers and matrices Paper 2Medium11 marks

A drone's horizontal position, relative to its landing pad at the origin, is modelled by a complex number \(z\), with distances in metres. The drone starts at \(z_0 = 40\). At each step of a landing manoeuvre, its position is multiplied by \(w = 0.9e^{i\pi/6}\), so that \(z_n = 40w^n\).

(a)

Write \(w\) in the form \(a + bi\).

(2)
(b)

Find \(z_4\), giving your answer in the form \(a + bi\).

(3)
(c)

Find the first step at which the drone is less than 10 m from the landing pad.

(3)
(d)

For the step found in the previous part, find the drone's distance from the pad and the argument of its position in the interval \(-\pi < \theta \le \pi\).

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
\(0.9\cos\frac{\pi}{6} + 0.9i\sin\frac{\pi}{6}\)✓ 1
\(w = 0.779 + 0.45i\)✓ 1
Part (b)
\(z_4 = 40(0.9)^4 e^{i4\pi/6}\)✓ 1
\(= 26.244e^{i2\pi/3}\)✓ 1
\(z_4 = -13.1 + 22.7i\)✓ 1
Part (c)
Uses \(|z_n| = 40(0.9)^n\)✓ 1
Solves \(40(0.9)^n < 10\), giving \(n > 13.2\)✓ 1
Step 14✓ 1
Part (d)
Distance \(= 40(0.9)^{14} = 9.15\) m✓ 1
Argument \(= \frac{14\pi}{6} = \frac{7\pi}{3}\)✓ 1
Reduces by \(2\pi\): \(\frac{\pi}{3}\) (1.05 rad)✓ 1

Exam tip: Multiplying by \(re^{i\theta}\) scales the distance by \(r\) and rotates the point by \(\theta\). Work with modulus and argument separately rather than expanding powers in Cartesian form.

398IBMI3H-1-03r7-03Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Medium9 marks

A farmer needs to apply exactly 110 kg of nitrogen and 130 kg of phosphorus to a field using two fertiliser blends.
Blend X is 20% nitrogen and 10% phosphorus by mass. Blend Y is 10% nitrogen and 30% phosphorus by mass.
The farmer uses \(x\) kg of blend X and \(y\) kg of blend Y.

(a)

Write the requirements as a matrix equation of the form \(\mathbf{M}\begin{pmatrix}x\\y\end{pmatrix}=\mathbf{b}\).

(2)
(b)

Find \(\mathbf{M}^{-1}\).

(2)
(c)

Hence find the mass of each blend the farmer must use.

(3)
(d)

Blend X costs £0.85 per kg and blend Y costs £1.40 per kg. Find the total cost of the fertiliser.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 matrix with entries 0.2, 0.1, 0.1, 0.3 arranged consistently with the unknowns✓ 1
A1 \(\begin{pmatrix}0.2&0.1\\0.1&0.3\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}110\\130\end{pmatrix}\)✓ 1
Part (b)
M1 attempt to find the determinant, \(0.2\times0.3-0.1\times0.1=0.05\)✓ 1
A1 \(\mathbf{M}^{-1}=\begin{pmatrix}6&-2\\-2&4\end{pmatrix}\)✓ 1
Part (c)
M1 attempt to calculate \(\mathbf{M}^{-1}\mathbf{b}\) (ECF from (b))✓ 1
A1 \(x=400\) kg✓ 1
A1 \(y=300\) kg✓ 1
Part (d)
M1 \(0.85\times400+1.40\times300\) (ECF from (c))✓ 1
A1 £760.00✓ 1

Exam tip: Matrix multiplication is not commutative: write the inverse on the left of the vector, \(\mathbf{M}^{-1}\mathbf{b}\), when solving \(\mathbf{M}\mathbf{x}=\mathbf{b}\).

399IBMI3H-1-03r7-04Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Medium11 marks

A jet burns fuel at a constant rate in each phase of a flight: \(c\) kg per minute in the climb, \(r\) kg per minute in the cruise and \(d\) kg per minute in the descent. Three flights were recorded.

Flight A: climb 20 min, cruise 100 min, descent 25 min, fuel used 6000 kg.
Flight B: climb 25 min, cruise 160 min, descent 30 min, fuel used 8815 kg.
Flight C: climb 18 min, cruise 60 min, descent 20 min, fuel used 4230 kg.

(a)

Write the information as a matrix equation \(\mathbf{T}\begin{pmatrix}c\\r\\d\end{pmatrix}=\mathbf{f}\).

(2)
(b)

Use your matrix equation to find \(c\), \(r\) and \(d\).

(4)
(c)

A fourth flight has a 22 minute climb, a 27 minute descent and a cruise of \(t\) minutes. Find the fuel used, in kg, when \(t=130\).

(2)
(d)

The tanks hold 8000 kg, and the last 10% of the capacity must remain unused as reserve. Find the greatest possible cruise time of the fourth flight, in minutes, correct to 1 decimal place.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 a 3×3 matrix of times with rows in the order climb, cruise, descent✓ 1
A1 \(\begin{pmatrix}20&100&25\\25&160&30\\18&60&20\end{pmatrix}\begin{pmatrix}c\\r\\d\end{pmatrix}=\begin{pmatrix}6000\\8815\\4230\end{pmatrix}\)✓ 1
Part (b)
M1 attempt to solve the system, for example using \(\mathbf{T}^{-1}\mathbf{f}\)✓ 1
A1 \(c=95\)✓ 1
A1 \(r=38\)✓ 1
A1 \(d=12\)✓ 1
Part (c)
M1 \(22\times95+130\times38+27\times12\)✓ 1
A1 7354 kg✓ 1
Part (d)
M1 usable fuel \(=0.9\times8000=7200\) kg✓ 1
M1 \(22\times95+38t+27\times12=7200\) (ECF from (b))✓ 1
A1 \(t=125.9\) minutes✓ 1

Common error: Write the rows and columns of a matrix equation in the same order as the unknowns, otherwise the product is not defined as intended.

400IBMI3H-1-03r7-07Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Medium9 marks

Three navigation beacons are placed so that their positions relative to a control tower at the origin of the complex plane, in km, are the three solutions of \(z^3=64i\).

(a)

Write \(64i\) in the form \(re^{i\theta}\) with \(r>0\).

(2)
(b)

Find the three solutions of \(z^3=64i\), giving each in the form \(re^{i\theta}\) with \(-\pi<\theta\le\pi\).

(3)
(c)

Find the distance between two neighbouring beacons, correct to 3 significant figures.

(2)
(d)

The beacons form an equilateral triangle. Find its area, correct to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to find the modulus and argument✓ 1
A1 \(64e^{\frac{\pi}{2}i}\)✓ 1
Part (b)
M1 attempt to take the cube root of the modulus and divide the argument by 3, allowing for \(+2k\pi\)✓ 1
A1 \(4e^{\frac{\pi}{6}i}\) and \(4e^{\frac{5\pi}{6}i}\)✓ 1
A1 \(4e^{-\frac{\pi}{2}i}\)✓ 1
Part (c)
M1 attempt to use the angle \(\frac{2\pi}{3}\) between neighbouring beacons, for example \(2\times4\sin\frac{\pi}{3}\)✓ 1
A1 6.93 km✓ 1
Part (d)
M1 \(\frac12 s^2\sin\frac{\pi}{3}\) with \(s=4\sqrt3\) (or equivalent)✓ 1
A1 20.8 km2✓ 1

Common error: The argument of a solution of \(z^n=w\) is \(\frac{\arg w+2k\pi}{n}\); forgetting the \(2k\pi\) leaves out all but one of the solutions.

401IBMI3H-1-03r7-12Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Medium9 marks

During a harvest, a farmer cuts 12.5 tonnes of grain on the first day. On each later day the amount cut is 6% less than on the previous day, so the daily amounts form a geometric sequence.

(a)

Find the amount cut on day 5, correct to 3 significant figures.

(2)
(b)

Find the total amount cut in the first 10 days, correct to 3 significant figures.

(2)
(c)

Find the least number of days after which the total amount cut exceeds 150 tonnes.

(3)
(d)

Find the greatest total amount that could ever be cut if the harvest continued indefinitely, correct to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(u_5=12.5\times0.94^4\)✓ 1
A1 9.76 tonnes✓ 1
Part (b)
M1 \(S_{10}=\frac{12.5(1-0.94^{10})}{1-0.94}\)✓ 1
A1 96.1 tonnes✓ 1
Part (c)
M1 \(\frac{12.5(1-0.94^n)}{0.06}>150\)✓ 1
M1 \(0.94^n<0.28\) and attempt to solve, using logarithms or a table (n = 20.57…)✓ 1
A1 21 days✓ 1
Part (d)
M1 \(S_\infty=\frac{12.5}{1-0.94}\)✓ 1
A1 208 tonnes✓ 1

Common error: When solving \(0.94^n<c\) the inequality sign reverses on dividing by the negative number \(\ln0.94\); round \(n\) up to a whole number of days.

402IBMI3H-1-04r17-09Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Medium9 marks

A cycle shop orders tyres, helmets and lights from a supplier. Let \(t\), \(h\) and \(l\) be the prices in euros of one tyre, one helmet and one light. Three orders were placed:
Order A: 4 tyres, 2 helmets and 6 lights cost €274.
Order B: 6 tyres, 1 helmet and 2 lights cost €237.
Order C: 3 tyres, 3 helmets and 5 lights cost €279.

(a)

Write the information as a matrix equation of the form \(\mathbf{A}\mathbf{x}=\mathbf{b}\).

(2)
(b)

Find the determinant of \(\mathbf{A}\) and state what it shows about the prices.

(2)
(c)

Use \(\mathbf{x}=\mathbf{A}^{-1}\mathbf{b}\) to find the price of one tyre, one helmet and one light.

(3)
(d)

The shop plans an order of 5 tyres, 4 helmets and 10 lights. Find its cost.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(\begin{pmatrix}4&2&6\\6&1&2\\3&3&5\end{pmatrix}\begin{pmatrix}t\\h\\l\end{pmatrix}\)✓ 1
A1 \(=\begin{pmatrix}274\\237\\279\end{pmatrix}\)✓ 1
Part (b)
A1 \(\det\mathbf{A}=38\)✓ 1
R1 it is non-zero, so \(\mathbf{A}^{-1}\) exists and the prices are uniquely determined✓ 1
Part (c)
M1 \(\mathbf{x}=\mathbf{A}^{-1}\mathbf{b}\) (or equivalent GDC solve)✓ 1
A1 tyre €28✓ 1
A1 helmet €45 and light €12✓ 1
Part (d)
M1 \(5\times28+4\times45+10\times12\)✓ 1
A1 €440✓ 1

Exam tip: After solving a system with the inverse matrix, check the answers by substituting into one of the original equations (for example Order B: \(6\times28+45+2\times12=237\)).

403IBMI3H-1-04r17-10Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Medium9 marks

A laboratory matches the iron and silica content of a pottery sample by mixing \(x\) kg of clay A and \(y\) kg of clay B. For a particular choice of clay B, the requirements are $\(2x+3y=14,\qquad kx+6y=16,\)$ where $k$ is a constant.

(a)

Write the system as \(\mathbf{M}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}14\\16\end{pmatrix}\) and find the value of \(k\) for which \(\mathbf{M}\) has no inverse.

(3)
(b)

For \(k=1\), find \(\mathbf{M}^{-1}\).

(3)
(c)

Hence find \(x\) and \(y\) when \(k=1\).

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(\mathbf{M}=\begin{pmatrix}2&3\\k&6\end{pmatrix}\)✓ 1
M1 \(\det\mathbf{M}=12-3k=0\)✓ 1
A1 \(k=4\)✓ 1
Part (b)
A1 \(\det\mathbf{M}=9\)✓ 1
M1 swaps the leading-diagonal entries and changes the signs of the others✓ 1
A1 \(\mathbf{M}^{-1}=\frac19\begin{pmatrix}6&-3\\-1&2\end{pmatrix}\)✓ 1
Part (c)
M1 \(\begin{pmatrix}x\\y\end{pmatrix}=\mathbf{M}^{-1}\begin{pmatrix}14\\16\end{pmatrix}\)✓ 1
A1 \(x=\frac{6\times14-3\times16}{9}=4\)✓ 1
A1 \(y=\frac{-14+32}{9}=2\)✓ 1

Common error: The inverse of a 2×2 matrix has the factor \(\frac{1}{\det\mathbf{M}}\) multiplying the whole adjusted matrix; leaving it out is a frequent slip.

404IBMI3H-1-04r18-07Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Medium11 marks

The numbers of brood bees B and worker bees W, both in thousands, in a hive at the start of week n are modelled by
\(\begin{pmatrix}B_{n+1}\\W_{n+1}\end{pmatrix}=M\begin{pmatrix}B_n\\W_n\end{pmatrix}\), where \(M=\begin{pmatrix}0.3&1.2\\0.5&0.7\end{pmatrix}\).

(a)

Find the eigenvalues of \(M\).

(3)
(b)

Find an eigenvector of \(M\) corresponding to the larger eigenvalue.

(2)
(c)

Interpret the larger eigenvalue and its eigenvector for the long-term behaviour of the hive.

(2)
(d)

An eigenvector for \(\lambda=-0.3\) is \(\begin{pmatrix}2\\-1\end{pmatrix}\). At week 0, \(\begin{pmatrix}B_0\\W_0\end{pmatrix}=\begin{pmatrix}1.4\\0.9\end{pmatrix}\). Express this as a combination of the two eigenvectors and hence find the number of worker bees, in thousands, at week 10 to 3 significant figures.

(4)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(\det(M-\lambda I)=0\)✓ 1
A1 \(\lambda^2-\lambda-0.39=0\)✓ 1
A1 \(\lambda=1.3\) and \(\lambda=-0.3\)✓ 1
Part (b)
M1 solves \((M-1.3I)\mathbf{v}=\mathbf{0}\), e.g. \(-1.0x+1.2y=0\)✓ 1
A1 \(\mathbf{v}=\begin{pmatrix}6\\5\end{pmatrix}\) (or any non-zero multiple)✓ 1
Part (c)
R1 long-term, each population grows by a factor of about 1.3 per week (30% growth)✓ 1
R1 the ratio of brood to workers tends to \(6:5\)✓ 1
Part (d)
M1 solves \(\begin{pmatrix}1.4\\0.9\end{pmatrix}=p\begin{pmatrix}6\\5\end{pmatrix}+q\begin{pmatrix}2\\-1\end{pmatrix}\)✓ 1
A1 \(p=0.2,\ q=0.1\)✓ 1
M1 \(\mathbf{v}_{10}=0.2(1.3)^{10}\begin{pmatrix}6\\5\end{pmatrix}+0.1(-0.3)^{10}\begin{pmatrix}2\\-1\end{pmatrix}\)✓ 1
A1 \(W_{10}=13.8\) thousand✓ 1

Exam tip: The component along the eigenvector with \(|\lambda|<1\) dies away. After many steps the state is dominated by the term with the largest \(|\lambda|\).

405IBMI3H-1-05r19-08Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Medium9 marks

A fashion house buys three fabrics: cotton, silk and wool, priced at €\(x\), €\(y\) and €\(z\) per metre respectively. Three garments use these amounts of fabric:
Dress: 3 m cotton, 2 m silk, 1 m wool, with a material cost of €87.
Coat: 2 m cotton, 1 m silk, 4 m wool, with a material cost of €100.
Scarf: 1 m cotton, 1 m silk, 2 m wool, with a material cost of €62.

(a)

Write the information as a matrix equation \(\mathbf{A}\mathbf{v}=\mathbf{b}\).

(2)
(b)

Find \(\det\mathbf{A}\) and explain what it tells you about the solution of the system.

(2)
(c)

Find the price per metre of each fabric.

(3)
(d)

A new jacket needs 2.5 m of cotton, 1.5 m of silk and 3 m of wool. Find its material cost.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(\mathbf{A}=\begin{pmatrix}3&2&1\\2&1&4\\1&1&2\end{pmatrix}\) with \(\mathbf{v}=\begin{pmatrix}x\\y\\z\end{pmatrix}\)✓ 1
A1 \(\mathbf{b}=\begin{pmatrix}87\\100\\62\end{pmatrix}\)✓ 1
Part (b)
A1 \(\det\mathbf{A}=-5\)✓ 1
R1 the determinant is non-zero, so the matrix is invertible and the system has a unique solution✓ 1
Part (c)
M1 \(\mathbf{v}=\mathbf{A}^{-1}\mathbf{b}\) or a GDC simultaneous-equation solver✓ 1
A1 cotton €8 and silk €24 per metre✓ 1
A1 wool €15 per metre✓ 1
Part (d)
M1 \(2.5\times8+1.5\times24+3\times15\) ECF from (c)✓ 1
A1 €101✓ 1

Exam tip: Check a solved system by substituting back into one equation you did not use to find the answers, for example \(3\times8+2\times24+15=87\).

406IBMI3H-1-05r19-14Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Medium9 marks

A designer decorates a dress with rows of sequins. The numbers of sequins in the rows form an arithmetic sequence. Row 1 has 12 sequins and row 20 has 126 sequins.

(a)

Find the common difference.

(2)
(b)

Find the total number of sequins in the first 20 rows.

(2)
(c)

Find the least number of rows needed for the total number of sequins to exceed 5000.

(3)
(d)

Sequins cost €0.032 each. Find the cost of the sequins for the first 40 rows, in €.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(12+19d=126\)✓ 1
A1 \(d=6\)✓ 1
Part (b)
M1 \(S_{20}=\frac{20}{2}(12+126)\)✓ 1
A1 1380✓ 1
Part (c)
M1 \(S_n=\frac{n}{2}(24+6(n-1))=3n^2+9n\)✓ 1
M1 solve \(3n^2+9n>5000\) (GDC), \(n>39.35\) (or \(S_{39}=4914\), \(S_{40}=5160\))✓ 1
A1 \(n=40\)✓ 1
Part (d)
M1 \(5160\times0.032\) ECF from (c)✓ 1
A1 €165.12✓ 1

Common error: Be sure to test the integers either side of a root: the answer is the least whole number of rows, so round up from 39.35, not to the nearest.

407IBMI3H-1-04r21-12Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Medium10 marks

A patient takes a 200 mg dose of a drug every 8 hours. Between doses, 60% of the drug in the body is cleared, so 40% remains. Let An be the amount of drug in the body, in mg, immediately after the nth dose.

(a)

Find A2 and A3.

(2)
(b)

Show that An = 200(1 − 0.4n)/0.6, and hence find A5 to 2 decimal places.

(3)
(c)

Find the smallest number of doses after which the amount of drug immediately after a dose exceeds 330 mg.

(3)
(d)

Find the value that An approaches as n becomes very large. Give your answer to 4 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 A2 = 200 + 0.4 × 200 = 280✓ 1
A1 A3 = 200 + 0.4 × 280 = 312✓ 1
Part (b)
M1 recognise a geometric series with first term 200 and common ratio 0.4✓ 1
A1 use of the sum formula, An = 200(1 − 0.4n)/(1 − 0.4)✓ 1
A1 A5 = 329.92 mg✓ 1
Part (c)
M1 200(1 − 0.4n)/0.6 > 330✓ 1
M1 0.4n < 0.01 and use of logarithms, or a table or graph on the calculator✓ 1
A1 n > 5.03, so n = 6✓ 1
Part (d)
M1 sum to infinity 200/(1 − 0.4)✓ 1
A1 333.3 mg✓ 1

Common error: Dividing both sides of an inequality by a negative number, or by log 0.4, without reversing the inequality sign. Since log 0.4 is negative, the sign must flip.

408IBMI3H-1-04r22-04Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Medium12 marks

A robotic arm positions mirror segments in a telescope. On an Argand diagram, the position of the arm tip after n moves is given by \(z_n = 3w^n\) where \(w = 1.12e^{i\pi/9}\) and \(n = 0, 1, 2, \ldots\) Distances are in metres.

(a)

Find \(z_3\) in the form a + bi, with a and b given to 3 significant figures.

(3)
(b)

Write down expressions for \(|z_n|\) and \(\arg z_n\) in terms of n.

(2)
(c)

Find the smallest value of n for which the arm tip is more than 10 m from the origin.

(3)
(d)

Find the smallest positive value of n for which the arm tip is on the positive real axis.

(2)
(e)

For the value of n found in (d), find the distance of the arm tip from the origin, to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(z_3 = 3 \times 1.12^3 e^{i\pi/3}\) (de Moivre)✓ 1
A1 \(a = 2.11\)✓ 1
A1 \(b = 3.65\)✓ 1
Part (b)
A1 \(|z_n| = 3 \times 1.12^{n}\)✓ 1
A1 \(\arg z_n = \dfrac{n\pi}{9}\)✓ 1
Part (c)
M1 \(3 \times 1.12^{n} > 10\)✓ 1
A1 \(n > \dfrac{\ln(10/3)}{\ln 1.12} = 10.62\)✓ 1
A1 \(n = 11\)✓ 1
Part (d)
M1 \(\dfrac{n\pi}{9} = 2k\pi\) for a positive integer k✓ 1
A1 \(n = 18\)✓ 1
Part (e)
M1 \(|z_{18}| = 3 \times 1.12^{18}\)✓ 1
A1 23.1 m✓ 1

Common error: Do not mix up the modulus and the argument in de Moivre's theorem: the modulus is raised to the power n, while the argument is multiplied by n.

409IBMI3H-1-04r22-07Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Medium10 marks

Commuters in a city choose each day between the train and the bus. After each day, 80% of train users and 30% of bus users choose the train the next day; the rest choose the bus. The total number of commuters stays at 10 000. This is modelled by \(\begin{pmatrix} T_{n+1} \\ B_{n+1} \end{pmatrix} = M\begin{pmatrix} T_{n} \\ B_{n} \end{pmatrix}\) where \(T_n\) and \(B_n\) are the numbers choosing train and bus on day n. On day 0, 3000 choose the train and 7000 choose the bus.

(a)

Write down the matrix M and find the numbers choosing the train and the bus on day 2.

(3)
(b)

Find the eigenvalues of M.

(3)
(c)

Find an eigenvector of M corresponding to the eigenvalue 1.

(2)
(d)

After many days the numbers settle to a steady state, which is proportional to the eigenvector in (c). Find the number of commuters who choose the train in the steady state.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(M = \begin{pmatrix} 0.8 & 0.3 \\ 0.2 & 0.7 \end{pmatrix}\)✓ 1
M1 \(M^2\begin{pmatrix} 3000 \\ 7000 \end{pmatrix}\) (or two successive multiplications; day 1 is 4500 and 5500)✓ 1
A1 train 5250 and bus 4750✓ 1
Part (b)
M1 \(\det(M - \lambda I) = 0\), i.e. \((0.8 - \lambda)(0.7 - \lambda) - 0.06 = 0\)✓ 1
A1 \(\lambda^2 - 1.5\lambda + 0.5 = 0\)✓ 1
A1 \(\lambda = 1\) and \(\lambda = 0.5\)✓ 1
Part (c)
M1 solves \((M - I)\mathbf{v} = \mathbf{0}\), e.g. \(-0.2a + 0.3b = 0\)✓ 1
A1 any multiple of \(\begin{pmatrix} 3 \\ 2 \end{pmatrix}\)✓ 1
Part (d)
M1 scales the eigenvector so that the total is 10 000, i.e. \(\dfrac{3}{3 + 2} \times 10000\)✓ 1
A1 6000✓ 1

Exam tip: A steady state of a transition matrix is an eigenvector for the eigenvalue 1; scale it so the total equals the known population.

410IBMI3H-1-04r22-09Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Medium13 marks

A coach models weekly changes in a player's stamina index S and power index P by \(S_{n+1} = 4S_n + 2P_n\) and \(P_{n+1} = S_n + 3P_n\), where n is the number of weeks. This can be written \(\begin{pmatrix} S_{n+1} \\ P_{n+1} \end{pmatrix} = A\begin{pmatrix} S_n \\ P_n \end{pmatrix}\) with \(A = \begin{pmatrix} 4 & 2 \\ 1 & 3 \end{pmatrix}\). At the start, \(S_0 = 3\) and \(P_0 = 0\).

(a)

Find the eigenvalues of A.

(3)
(b)

Find an eigenvector for each eigenvalue.

(3)
(c)

Write \(\begin{pmatrix} 3 \\ 0 \end{pmatrix}\) as a linear combination of the two eigenvectors.

(2)
(d)

Hence find \(S_6\) and \(P_6\).

(3)
(e)

Find \(P_6\) as a percentage of \(S_6\), to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(\det(A - \lambda I) = (4 - \lambda)(3 - \lambda) - 2 = 0\)✓ 1
A1 \(\lambda^2 - 7\lambda + 10 = 0\)✓ 1
A1 \(\lambda = 5\) and \(\lambda = 2\)✓ 1
Part (b)
M1 solves \((A - \lambda I)\mathbf{v} = \mathbf{0}\) for one eigenvalue✓ 1
A1 \(\lambda = 5\): multiples of \(\begin{pmatrix} 2 \\ 1 \end{pmatrix}\)✓ 1
A1 \(\lambda = 2\): multiples of \(\begin{pmatrix} 1 \\ -1 \end{pmatrix}\)✓ 1
Part (c)
M1 \(\begin{pmatrix} 3 \\ 0 \end{pmatrix} = a\begin{pmatrix} 2 \\ 1 \end{pmatrix} + b\begin{pmatrix} 1 \\ -1 \end{pmatrix}\), so \(2a + b = 3\) and \(a - b = 0\)✓ 1
A1 \(a = b = 1\)✓ 1
Part (d)
M1 \(A^6\begin{pmatrix} 3 \\ 0 \end{pmatrix} = 5^6\begin{pmatrix} 2 \\ 1 \end{pmatrix} + 2^6\begin{pmatrix} 1 \\ -1 \end{pmatrix}\)✓ 1
A1 \(S_6 = 2 \times 15625 + 64 = 31314\)✓ 1
A1 \(P_6 = 15625 - 64 = 15561\)✓ 1
Part (e)
M1 \(\dfrac{15561}{31314} \times 100\)✓ 1
A1 49.7%✓ 1

Common error: The eigenvalue goes to the power n for each eigenvector component separately; you cannot raise the matrix entries individually to the power n.

411IBMI3H-1-04r22-11Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Medium9 marks

A ball is dropped from a height of 3.0 m onto a hard floor. After each bounce it rises to 80% of the height from which it last fell. Heights are measured above the floor, and air resistance is ignored.

(a)

Find the height the ball reaches after the 4th bounce, to 3 significant figures.

(2)
(b)

Find the number of the first bounce after which the ball rises to a height of less than 0.2 m.

(3)
(c)

Find the total distance travelled by the ball from the moment it is released until it hits the floor for the 8th time. Give your answer to 3 significant figures.

(4)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(3.0 \times 0.8^{4}\)✓ 1
A1 1.23 m✓ 1
Part (b)
M1 \(3.0 \times 0.8^{n} < 0.2\)✓ 1
A1 \(n > \dfrac{\ln(0.2/3)}{\ln 0.8} = 12.14\)✓ 1
A1 13th bounce✓ 1
Part (c)
M1 distance = 3.0 + 2(3.0 × 0.8) + 2(3.0 × 0.82) + … (initial drop plus an up-and-down distance after each bounce)✓ 1
M1 recognises a geometric series for the up-and-down distances with first term 4.8 and common ratio 0.8, with 7 terms✓ 1
M1 \(3.0 + \dfrac{4.8(1 - 0.8^{7})}{1 - 0.8}\)✓ 1
A1 22.0 m✓ 1

Common error: After each bounce the ball travels the rebound height twice (up and down), so the geometric series for the distance has a factor of 2 apart from the first drop.

412IBMI3H-1-04r22-13Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Medium9 marks

A swimmer plans to train for \(u_k = 3.5 + 0.4k\) hours in week k, for \(k = 1, 2, 3, \ldots\)

(a)

Write down the first term and the common difference of this arithmetic sequence.

(2)
(b)

Find the value of \(\displaystyle\sum_{k=1}^{20}(3.5 + 0.4k)\) and state what it represents.

(2)
(c)

Find the least number of weeks needed for the total training time to reach at least 500 hours.

(3)
(d)

Find the total training time after this number of weeks, to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 first term 3.9✓ 1
A1 common difference 0.4✓ 1
Part (b)
M1 \(\dfrac{20}{2}(2 \times 3.9 + 19 \times 0.4)\) (or \(20 \times 3.5 + 0.4 \times 210\))✓ 1
A1 154 hours, the total training time in the first 20 weeks✓ 1
Part (c)
M1 \(\dfrac{n}{2}(7.8 + 0.4(n-1)) \ge 500\), i.e. \(0.2n^2 + 3.7n \ge 500\)✓ 1
A1 \(n \ge 41.6\)✓ 1
A1 42 weeks (total after 41 weeks is 487.9 hours)✓ 1
Part (d)
M1 \(S_{42} = \dfrac{42}{2}(2 \times 3.9 + 41 \times 0.4)\)✓ 1
A1 508 hours✓ 1

Exam tip: For sigma notation, the first term is found by substituting the lower limit, so \(k = 1\) gives 3.9, not 3.5.

413IBMI3H-1-05r22-02Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Medium11 marks

A mountaineering club buys three kinds of equipment from a supplier: coils of rope (£r each), sets of ice screws (£s each) and headlamps (£h each). The club places three orders.
Order 1: 3 coils, 2 sets, 4 headlamps, total £459.
Order 2: 2 coils, 5 sets, 1 headlamp, total £410.
Order 3: 4 coils, 1 set, 3 headlamps, total £472.

(a)

Write the information as a matrix equation \(M\mathbf{x}=\mathbf{b}\), where \(\mathbf{x}=\begin{pmatrix}r\\s\\h\end{pmatrix}\).

(2)
(b)

Use your calculator to find the price of each item.

(3)
(c)

Find det M and explain what this tells you about the system.

(2)
(d)

A fourth order is for 5 coils, 3 sets and 6 headlamps. Find its total cost.

(2)
(e)

The supplier then raises every price by 8 %. Find the new cost of Order 1.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(M=\begin{pmatrix}3&2&4\\2&5&1\\4&1&3\end{pmatrix}\)✓ 1
A1 \(\mathbf{b}=\begin{pmatrix}459\\410\\472\end{pmatrix}\)✓ 1
Part (b)
M1 attempt to use \(\mathbf{x}=M^{-1}\mathbf{b}\) (or equivalent solver)✓ 1
A1 two prices correct✓ 1
A1 rope £85, ice screws £42, headlamp £30✓ 1
Part (c)
A1 det M = −34✓ 1
R1 non-zero determinant so M is invertible, hence a unique solution for the prices✓ 1
Part (d)
M1 5(85) + 3(42) + 6(30)✓ 1
A1 £731✓ 1
Part (e)
M1 459 × 1.08✓ 1
A1 £495.72✓ 1

Exam tip: Substituting your prices back into one of the original orders is a quick check on a calculator-solved system.

414IBMI3H-1-05r22-09Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Medium10 marks

In a radio tuner, a series circuit has resistance R = 50 Ω, inductance L = 0.020 H and capacitance C = 2.5 × 10−6 F. At angular frequency ω (rad s−1) its impedance is the complex number
\(Z=R+i\left(\omega L-\frac{1}{\omega C}\right)\) ohms.

(a)

Find Z in the form a + bi when ω = 3000.

(3)
(b)

Find the modulus and argument of Z at this frequency. Give the modulus in ohms to 3 significant figures and the argument in radians to 3 significant figures.

(3)
(c)

The peak voltage is 12 V. Using the unrounded value of |Z|, find the peak current in amperes, to 3 significant figures.

(2)
(d)

The circuit is at resonance when Z is real. Find the resonant angular frequency, to the nearest rad s−1.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(\omega L=60\) and \(\frac{1}{\omega C}=133.3\ldots\)✓ 1
A1 imaginary part \(-73.3\)✓ 1
A1 \(Z=50-73.3i\)✓ 1
Part (b)
M1 \(|Z|=\sqrt{50^2+73.33^2}\) or \(\tan^{-1}\left(\frac{-73.33}{50}\right)\)✓ 1
A1 \(|Z|=88.8\) Ω✓ 1
A1 argument = −0.972 rad✓ 1
Part (c)
M1 \(I=\frac{12}{|Z|}\)✓ 1
A1 0.135 A✓ 1
Part (d)
M1 \(\omega L=\frac{1}{\omega C}\), so \(\omega=\frac{1}{\sqrt{LC}}\)✓ 1
A1 4472 rad s−1✓ 1

Exam tip: The argument of a complex number with a negative imaginary part and positive real part lies in the fourth quadrant, so it is negative.

415IBMI3H-1-05r22-12Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Medium11 marks

Six antenna elements are placed at positions in the complex plane (units: metres) given by the solutions of \(z^6=64\).

(a)

Find the six solutions in the form \(re^{i\theta}\), with \(0\le\theta<2\pi\).

(4)
(b)

Write the position with the smallest positive argument in Cartesian form.

(2)
(c)

Write down the sum of the six positions and give a reason.

(2)
(d)

Find the distance between the elements with k = 0 and k = 2, to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(r=64^{1/6}\)✓ 1
A1 r = 2✓ 1
M1 arguments \(\theta=\frac{2\pi k}{6}\), k = 0, …, 5✓ 1
A1 \(2e^{i0},\,2e^{i\pi/3},\,2e^{i2\pi/3},\,2e^{i\pi},\,2e^{i4\pi/3},\,2e^{i5\pi/3}\)✓ 1
Part (b)
M1 \(2\cos\frac{\pi}{3}+2i\sin\frac{\pi}{3}\)✓ 1
A1 \(1+\sqrt{3}\,i\)✓ 1
Part (c)
A1 0✓ 1
R1 the points are equally spaced on a circle centred at the origin (or the coefficient of \(z^5\) is zero), so they cancel in pairs✓ 1
Part (d)
M1 attempt at \(|2e^{i2\pi/3}-2|\)✓ 1
A1 \(\sqrt{(-3)^2+(\sqrt{3})^2}=\sqrt{12}\)✓ 1
A1 3.46 m✓ 1

Common error: The n solutions of \(z^n=c\) are spread at equal angles of \(\frac{2\pi}{n}\); do not stop after finding the first one.

416IBMI3H-1-08r27-06Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Medium9 marks

A warehouse uses two types of robot: pickers (p) and carriers (c). The expected fleet sizes at the start of month n follow
pn+1 = 1.1pn + 0.3cn, cn+1 = 0.2pn + 0.6cn,
that is, xn+1 = Mxn where xn = (pn, cn)T and M = [[1.1, 0.3], [0.2, 0.6]] (rows listed in turn). Initially there are 95 pickers and 20 carriers.

(a)

Show that the eigenvalues of M are 1.2 and 0.5.

(3)
(b)

Find an eigenvector for each eigenvalue.

(2)
(c)

Write the initial vector (95, 20)T as a combination of your eigenvectors. Hence find an expression for pn and use it to find the expected number of pickers after 6 months, to the nearest whole number.

(4)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 det(M − λI) = 0 expanded✓ 1
A1 λ2 − 1.7λ + 0.6 = 0✓ 1
A1 (λ − 1.2)(λ − 0.5) = 0, so λ = 1.2 or 0.5 (AG)✓ 1
Part (b)
A1 λ = 1.2: (3, 1) (or any multiple)✓ 1
A1 λ = 0.5: (1, −2) (or any multiple)✓ 1
Part (c)
M1 (95, 20)T = a(3, 1)T + b(1, −2)T solved✓ 1
A1 a = 30, b = 5✓ 1
A1 pn = 90(1.2)n + 5(0.5)n✓ 1
A1 p6 = 268.817 ≈ 269✓ 1

Common error: When using xn = aλ₁ⁿv₁ + bλ₂ⁿv₂, remember the constants multiply the whole eigenvector, so the first component carries a factor 3 × 30 = 90, not 30.

417IBMI3H-1-08r27-11Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Medium9 marks

The vibration of a robot gripper after a sudden movement has displacement x(t) = e−2tcos(3t) cm, where t is in seconds. This arises from the characteristic equation m2 + 4m + 13 = 0.

(a)

Solve m2 + 4m + 13 = 0.

(2)
(b)

Write the root with positive imaginary part in the form reiθ, with r and θ to 3 significant figures.

(3)
(c)

Find the period of the oscillation cos(3t) in seconds, to 3 significant figures.

(2)
(d)

The amplitude of the vibration is bounded by e−2t. Find the time after which this bound is below 0.01 cm, to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 use of the quadratic formula or completing the square, with discriminant 16 − 52 = −36✓ 1
A1 m = −2 ± 3i✓ 1
Part (b)
M1 r = √((−2)2 + 32)✓ 1
A1 r = 3.61✓ 1
A1 θ = π − tan−1(3 ÷ 2) = 2.16 (second quadrant)✓ 1
Part (c)
M1 period = 2π ÷ 3✓ 1
A1 2.09 s✓ 1
Part (d)
M1 e−2t < 0.01, so t > ln(100) ÷ 2✓ 1
A1 2.30 s✓ 1

Common error: The argument of a complex number with negative real part and positive imaginary part lies in the second quadrant; a calculator's tan⁻¹ alone gives a negative angle, which is wrong here.

418IBMI3H-1-06r35-03Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Medium9 marks

A shipping line carries three types of container, A, B and C, with masses a, b and c tonnes respectively. Three loadings are weighed:
2 of A, 3 of B and 1 of C: 46 tonnes
1 of A, 2 of B and 4 of C: 54 tonnes
3 of A, 1 of B and 2 of C: 52 tonnes

(a)

Write the information as a matrix equation of the form Mx = r.

(2)
(b)

Find det M and state what its value shows about the solution of the system.

(2)
(c)

Find the mass of each type of container.

(3)
(d)

A fourth loading has 6 of A, 4 of B and 5 of C. Find its total mass.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 M = rows (2, 3, 1), (1, 2, 4), (3, 1, 2) and x = (a, b, c)T✓ 1
A1 r = (46, 54, 52)T✓ 1
Part (b)
A1 det M = 25✓ 1
R1 det M ≠ 0, so M−1 exists and there is a unique solution✓ 1
Part (c)
M1 x = M−1r (or equivalent GDC method)✓ 1
A1 a = 10 and b = 6✓ 1
A1 c = 8✓ 1
Part (d)
M1 6 × 10 + 4 × 6 + 5 × 8 (ECF from (c))✓ 1
A1 124 tonnes✓ 1

Exam tip: Write the unknowns in the same order in every equation before forming the matrix, and check your solution in one original equation.

419IBMI3H-1-06r35-07Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Medium9 marks

A tilt sensor on a lava dome is modelled by the sequence of complex numbers zn = 8wn, n = 0, 1, 2, …, where w = 0.9ei π/8. The modulus of zn is the size of the tilt reading in mm.

(a)

Find z2 in the form reiθ and in the form a + bi, with a and b to 3 significant figures.

(3)
(b)

Find the smallest value of n for which the tilt reading is less than 0.5 mm.

(3)
(c)

Find the smallest positive integer n for which zn is a negative real number, and find the tilt reading |zn| for this value of n.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 z2 = 8 × 0.92 ei 2π/8✓ 1
A1 z2 = 6.48ei π/4✓ 1
A1 z2 = 4.58 + 4.58i✓ 1
Part (b)
M1 8 × 0.9n < 0.5✓ 1
M1 n > ln(0.0625) ÷ ln(0.9) = 26.3…✓ 1
A1 n = 27✓ 1
Part (c)
M1 nπ ÷ 8 = π (the argument must be an odd multiple of π)✓ 1
A1 n = 8✓ 1
A1 |z8| = 8 × 0.98 = 3.44 mm✓ 1

Exam tip: To use powers in exponential form, raise the modulus to the power and multiply the argument by the power.

420IBMI3H-1-06r37-03Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Medium9 marks

Calculator

The columns of a circular pavilion stand at the six solutions of z6 = 86 in the complex plane, where one unit represents one metre and the origin is the centre of the pavilion.

(a)

Write down all six solutions in the form reiθ.

(3)
(b)

Find the Cartesian coordinates of the column with argument 2π/3, to 3 significant figures.

(2)
(c)

The columns at 8 and 8eiπ/3 are adjacent. Find the distance between them, and hence the perimeter of the hexagon joining the six columns.

(4)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 r = 8 for every solution✓ 1
M1 argument spacing 2π/6 = π/3✓ 1
A1 8eikπ/3 for k = 0, 1, 2, 3, 4, 5✓ 1
Part (b)
M1 8(cos(2π/3) + i sin(2π/3))✓ 1
A1 (−4, 6.93)✓ 1
Part (c)
M1 8eiπ/3 = 4 + 6.93i✓ 1
M1 |8 − (4 + 6.93i)| = √(42 + 6.932)✓ 1
A1 8 m✓ 1
A1 perimeter 48 m✓ 1

Exam tip: The n solutions of z^n = r^n are equally spaced around a circle of radius r, with angle 2π/n between neighbours.

421IBMI3H-1-06r37-04Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Medium9 marks

Calculator

At a harbour, the tidal height above mean sea level is modelled as the sum of two tidal components. Each component is represented by a complex number whose modulus is its amplitude (metres) and whose argument is its phase (radians). The components are z1 = 1.8e0.6i and z2 = 1.1e2.1i. The combined component is z = z1 + z2.

(a)

Write z1 and z2 in the form x + yi, to 3 significant figures.

(3)
(b)

Find z in the form x + yi, to 3 significant figures, using unrounded values of z1 and z2.

(2)
(c)

Find the amplitude of the combined component, to 3 significant figures.

(2)
(d)

Find the phase of the combined component, to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 uses r(cos θ + i sin θ) with radians✓ 1
A1 z1 = 1.49 + 1.02i✓ 1
A1 z2 = −0.555 + 0.950i✓ 1
Part (b)
M1 adds real parts and imaginary parts✓ 1
A1 0.930 + 1.97i✓ 1
Part (c)
M1 |z| = √(0.93032 + 1.96592)✓ 1
A1 2.17 m✓ 1
Part (d)
M1 arg z = tan−1(1.9659/0.9303), with z in the first quadrant✓ 1
A1 1.13 radians✓ 1

Common error: Amplitudes of components with different phases do not simply add: 1.8 + 1.1 = 2.9, but the combined amplitude is smaller (2.17).

422IBMI3H-1-06r37-06Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Medium9 marks

Calculator

A logistics company charges €x, €y and €z per pallet for delivery to zones 1, 2 and 3. Customer 1 sends 2, 1 and 3 pallets to zones 1, 2 and 3 and pays €66. Customer 2 sends 1, 3 and 2 pallets and pays €75. Customer 3 sends 3, 2 and 1 pallets and pays €75.

(a)

Write the information as a matrix equation Av = c.

(2)
(b)

Find det A and state what it tells you about the solutions.

(2)
(c)

Find the rate per pallet for each zone.

(3)
(d)

A fourth customer sends 3, 1 and 2 pallets to zones 1, 2 and 3. Find the cost.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 A = [[2, 1, 3], [1, 3, 2], [3, 2, 1]] and v = (x, y, z)T✓ 1
A1 c = (66, 75, 75)T✓ 1
Part (b)
A1 det A = −18✓ 1
R1 non-zero, so A is invertible and there is a unique solution✓ 1
Part (c)
M1 v = A−1c, or elimination✓ 1
A1 x = €12 and y = €15✓ 1
A1 z = €9✓ 1
Part (d)
M1 3x + y + 2z (ECF from (c))✓ 1
A1 €69✓ 1

Common error: Check your solution by substituting it back into each of the original equations, not only the first one.

423IBMI3H-1-06r37-08Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Medium9 marks

Calculator

The population of a reef fish species is modelled each year by the vector (J, A)T of juveniles and adults, with (Jn+1, An+1)T = L(Jn, An)T. Each adult produces on average 1.5 juveniles that survive to the end of the year, 40% of juveniles survive to become adults, and 70% of adults survive. So L = [[0, 1.5], [0.4, 0.7]].

(a)

Find the eigenvalues of L.

(3)
(b)

Find an eigenvector of L corresponding to the positive eigenvalue.

(2)
(c)

Interpret the positive eigenvalue and its eigenvector in this context.

(2)
(d)

A population has 500 juveniles and 400 adults. Find the total population after 5 years, to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 solves det(L − λI) = 0✓ 1
A1 λ2 − 0.7λ − 0.6 = 0✓ 1
A1 λ = 1.2 and λ = −0.5✓ 1
Part (b)
M1 (L − 1.2I)v = 0, giving −1.2J + 1.5A = 0✓ 1
A1 (5, 4)T or any multiple✓ 1
Part (c)
R1 in the long run the population grows by a factor of 1.2 per year (20% growth)✓ 1
R1 with juveniles and adults in the ratio 5 : 4✓ 1
Part (d)
M1 (500, 400)T is an eigenvector, so the population is multiplied by 1.25✓ 1
A1 2240✓ 1

Exam tip: If the starting vector is an eigenvector, the population is simply multiplied by the eigenvalue each step, which avoids repeated matrix powers.

424IBMI3H-1-06r37-10Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Medium9 marks

Calculator

Design software applies a stretch to every point (x, y) of a building's floor plan using the matrix M = [[p, q], [1, 3]]. The vector (2, 1)T is an eigenvector of M. The other eigenvalue of M is 2.

(a)

Find the eigenvalue of M that corresponds to the eigenvector (2, 1)T.

(2)
(b)

Find the values of p and q.

(3)
(c)

Find an eigenvector of M corresponding to the eigenvalue 2.

(2)
(d)

A rectangular room on the plan has area 6 m2. Find its area after the transformation, and explain your method.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 second row of M(2, 1)T = (2 + 3) = 5✓ 1
A1 eigenvalue 5 (since the second component of the eigenvector is 1)✓ 1
Part (b)
M1 uses sum of eigenvalues = trace: 5 + 2 = p + 3, or the first row equation 2p + q = 10✓ 1
A1 p = 4✓ 1
A1 q = 2✓ 1
Part (c)
M1 (M − 2I)v = 0, giving 2x + 2y = 0✓ 1
A1 (1, −1)T or any multiple✓ 1
Part (d)
R1 area scale factor = |det M| = 4 × 3 − 2 × 1✓ 1
A1 60 m2✓ 1

Common error: The eigenvalue 5 means points along (2, 1) are stretched by a factor of 5 from the origin; it is not the area scale factor, which is the determinant.

425IBMI3H-1-04r40-08Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Medium10 marks

A logistics company has a fleet of 100 vans shared between two depots, A and B. Each night, 20% of the vans at A move to B and 30% of the vans at B move to A; all other vans stay where they are. At the start of day 0, all 100 vans are at depot A. Let T be the transition matrix, written so that (an+1, bn+1)T = T(an, bn)T, where an and bn are the numbers of vans at A and B at the start of day n.

(a)

Write down T.

(1)
(b)

Find the number of vans at each depot at the start of day 3.

(2)
(c)

Find the eigenvalues of T.

(2)
(d)

Find an eigenvector for λ = 1, and hence the number of vans at each depot in the long run.

(3)
(e)

The difference between the number of vans at A and the long-run value of 60 is 40 on day 0. Using the second eigenvalue, show that this difference is 5 on day 3, and so confirm your answer to (b).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 T = [[0.8, 0.3], [0.2, 0.7]]✓ 1
Part (b)
M1 T3(100, 0)T (or three successive multiplications)✓ 1
A1 65 vans at A and 35 vans at B✓ 1
Part (c)
M1 det(T − λI) = 0, giving λ2 − 1.5λ + 0.5 = 0✓ 1
A1 λ = 1 and λ = 0.5✓ 1
Part (d)
M1 solve (T − I)v = 0, e.g. −0.2a + 0.3b = 0✓ 1
A1 eigenvector a multiple of (3, 2)T✓ 1
A1 60 vans at A and 40 vans at B✓ 1
Part (e)
M1 the difference is multiplied by 0.5 each day, so 40 × 0.53✓ 1
A1 = 5, so vans at A = 60 + 5 = 65, agreeing with (b)✓ 1

Exam tip: The eigenvalue 1 of a transition matrix (columns sum to 1) gives the long-run split; the other eigenvalue controls how fast the fleet approaches it.

426IBMI3H-1-07r44-09Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Medium9 marks

A shipping company has 2320 containers that are moved between three ports, X, Y and Z, once a week. The proportions of containers at each port that are moved to each port are given by the transition matrix
  T = rows (0.7, 0.2, 0.1), (0.2, 0.6, 0.3), (0.1, 0.2, 0.6),
where the columns give the starting port (X, Y, Z) and the rows give the destination port (X, Y, Z). The numbers of containers at X, Y and Z at the start of a week form a column vector c.

(a)

At the start of one week, there are 1000 containers at X, 900 at Y and 420 at Z. Find the numbers at each port at the start of the next week.

(4)
(b)

The matrix T has an eigenvalue of 1. Find the corresponding eigenvector whose components sum to 1.

(3)
(c)

Hence find the number of containers at each port in the long term.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to calculate Tc✓ 1
A1 X: 0.7 × 1000 + 0.2 × 900 + 0.1 × 420 = 922✓ 1
A1 Y: 0.2 × 1000 + 0.6 × 900 + 0.3 × 420 = 866✓ 1
A1 Z: 0.1 × 1000 + 0.2 × 900 + 0.6 × 420 = 532✓ 1
Part (b)
M1 attempt to solve (T − I)v = 0 (or Tv = v) with technology or by hand✓ 1
A1 eigenvector proportional to (10, 11, 8)✓ 1
A1 (10/29, 11/29, 8/29) = (0.345, 0.379, 0.276)✓ 1
Part (c)
M1 multiplying each component by 2320✓ 1
A1 X: 800, Y: 880, Z: 640 (all three correct)✓ 1

Exam tip: A steady state is an eigenvector for eigenvalue 1. Scale it so the components sum to 1 before multiplying by the total number of items.

427IBMI3H-1-05r46-05Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Medium9 marks

Calculator

A data centre uses three types of server, A, B and C, which draw a, b and c watts respectively. Rack 1 holds 2 of type A, 1 of type B and 3 of type C and draws 745 W. Rack 2 holds 1 of type A, 3 of type B and 2 of type C and draws 1040 W. Rack 3 holds 4 of type A and 1 of type C and draws 565 W.

(a)

Write the information as a matrix equation.

(2)
(b)

Find the determinant of the coefficient matrix and state what it tells you about the solution.

(2)
(c)

Find \(a\), \(b\) and \(c\).

(3)
(d)

A new rack holds 6 of type A, 4 of type B and 8 of type C. Find the power it draws.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 coefficient matrix \(\begin{pmatrix} 2 & 1 & 3 \\ 1 & 3 & 2 \\ 4 & 0 & 1 \end{pmatrix}\) times \(\begin{pmatrix} a \\ b \\ c \end{pmatrix}\)✓ 1
A1 equal to \(\begin{pmatrix} 745 \\ 1040 \\ 565 \end{pmatrix}\)✓ 1
Part (b)
M1 attempt at the determinant✓ 1
A1 \(-23\), which is non-zero so the matrix is invertible and there is exactly one solution✓ 1
Part (c)
M1 solve the system, e.g. \(\begin{pmatrix} a \\ b \\ c \end{pmatrix} = \mathbf{M}^{-1}\begin{pmatrix} 745 \\ 1040 \\ 565 \end{pmatrix}\)✓ 1
A1 \(a = 120\) and \(b = 250\)✓ 1
A1 \(c = 85\)✓ 1
Part (d)
M1 \(6a + 4b + 8c\) using the values from (c)✓ 1
A1 2400 W✓ 1

Exam tip: For a 3 × 3 system, enter the matrices on your calculator rather than inverting by hand; write down the matrices you entered so method marks are available.

428IBMI3H-1-05r46-11Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Medium9 marks

Calculator

A data centre adds 40 petabytes (PB) of storage in year 1. In each later year the amount added is 12% more than in the previous year.

(a)

Find the amount of storage added in year 6, correct to 3 significant figures.

(2)
(b)

Find the total storage added in the first 10 years, correct to 3 significant figures.

(2)
(c)

Find the least whole number of years \(N\) for the total storage added to exceed 1500 PB.

(3)
(d)

Find the total storage added in the first \(N\) years, correct to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(40 \times 1.12^{5}\)✓ 1
A1 70.5 PB✓ 1
Part (b)
M1 \(S_{10} = \frac{40(1.12^{10}-1)}{1.12-1}\)✓ 1
A1 702 PB✓ 1
Part (c)
M1 \(\frac{40(1.12^N-1)}{0.12} > 1500\)✓ 1
M1 solve inequality or trial values, e.g. total after 15 years = 1491 PB✓ 1
A1 \(N = 16\)✓ 1
Part (d)
M1 use \(N = 16\) in the sum formula (ECF from (c))✓ 1
A1 1710 PB✓ 1

Exam tip: Identify the first term and common ratio before using the geometric series formula; a ratio of 1.12 comes from a 12% increase.

429IBMI3H-1-05r47-06Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Medium9 marks

The vertical motion of a ship's stabiliser is modelled by a system whose characteristic equation is \(\lambda^2+4\lambda+13=0\), where time is measured in seconds.

(a)

Solve the equation, giving the roots in the form a + bi.

(2)
(b)

Write the root with positive imaginary part in the form \(re^{i\theta}\), giving r and θ to 3 significant figures, with \(0<\theta<\pi\).

(3)
(c)

The oscillation has angular frequency equal to the imaginary part of the roots. Find the period of the oscillation, in seconds, to 3 significant figures.

(2)
(d)

The amplitude is multiplied by \(e^{-2t}\) after time t seconds. Find the factor by which the amplitude is multiplied after one complete period. Give your answer to 3 significant figures, using the unrounded period.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 use of the quadratic formula or completing the square✓ 1
A1 \(\lambda=-2\pm3i\)✓ 1
Part (b)
M1 attempt to find the modulus \(\sqrt{(-2)^2+3^2}\)✓ 1
A1 \(r=3.61\)✓ 1
A1 \(\theta=2.16\)✓ 1
Part (c)
M1 use \(T=\frac{2\pi}{3}\)✓ 1
A1 \(2.09\) s✓ 1
Part (d)
M1 substitute \(t=\frac{2\pi}{3}\) into \(e^{-2t}\)✓ 1
A1 \(0.0152\)✓ 1

Common error: Use the exact period \(\frac{2\pi}{3}\) in part (d) rather than the rounded 2.09, otherwise the third significant figure can change.

430IBMI-030-00-12Approximation, percentage error and standard form Paper 2Hard12 marks

The pH of a solution is defined by \(\text{pH}=-\log_{10}c\), where \(c\) is the hydrogen ion concentration in mol L−1.

(a)

Lemon juice has \(c=6.3\times10^{-3}\) mol L−1. Calculate its pH.

(2)
(b)

Seawater has a pH of 8.1. Find its hydrogen ion concentration, giving your answer in the form \(a\times10^k\) where \(1\le a<10\) and \(k\in\mathbb{Z}\).

(2)
(c)

Find how many times greater the hydrogen ion concentration of lemon juice is than that of seawater. Give your answer in the form \(a\times10^k\) where \(1\le a<10\) and \(k\in\mathbb{Z}\).

(2)
(d)

The hydrogen ion concentration of a solution is multiplied by a factor \(k\), where \(k>0\). Show that the pH decreases by \(\log_{10}k\).

(2)
(e)

A student measures the pH of a sample as 4.6 when its true value is 4.5. Calculate the percentage error in the measured pH and the percentage error in the hydrogen ion concentration deduced from it. Comment on your answers.

(4)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 for \(-\log_{10}(6.3\times10^{-3})\)✓ 1
A1 2.20✓ 1
Part (b)
M1 for \(c=10^{-8.1}\)✓ 1
A1 7.94×10−9 mol L−1✓ 1
Part (c)
M1 for \(\frac{6.3\times10^{-3}}{7.943...\times10^{-9}}\) or \(10^{8.1-2.20}\)✓ 1
A1 7.93×105; accept 7.94×105, which comes from using the rounded pH✓ 1
Part (d)
M1 for \(-\log_{10}(kc)=-(\log_{10}k+\log_{10}c)\), using \(\log(ab)=\log a+\log b\)✓ 1
A1 \(=\text{pH}-\log_{10}k\), so the pH decreases by \(\log_{10}k\) AG. If \(0✓ 1
Part (e)
A1 percentage error in pH: \(\frac{4.6-4.5}{4.5}\times100=2.22\%\)✓ 1
M1 for the concentrations \(10^{-4.6}=2.51\times10^{-5}\) and \(10^{-4.5}=3.16\times10^{-5}\)✓ 1
A1 percentage error in concentration: \(\left|\frac{10^{-4.6}-10^{-4.5}}{10^{-4.5}}\right|\times100=20.6\%\)✓ 1
R1 a small percentage error in pH produces a much larger percentage error in concentration because the scale is logarithmic✓ 1

Common error: A pH change of 0.1 is not a 0.1% or 10% change in concentration. It multiplies the concentration by 10⁰.¹, which is about 1.26.

431IBMI-030-04-r2-03Complex numbers and matrices Paper 2Hard15 marks

A car-sharing company has 200 cars, which are parked overnight either at the Airport (A) or in the City (C). Each day:
• 70% of cars that start the day at A end the day at A; the rest end at C.
• 80% of cars that start the day at C end the day at C; the rest end at A.
Initially there are 120 cars at A and 80 cars at C. Let \(s_n = \begin{pmatrix} A_n \\ C_n \end{pmatrix}\) be the numbers of cars at each location after \(n\) days, so that \(s_{n} = T^n s_0\).

(a)

Write down the transition matrix \(T\).

(2)
(b)

Find the number of cars at each location after 3 days.

(2)
(c)

Find the eigenvalues of \(T\) and a corresponding eigenvector for each.

(4)
(d)

Hence show that \(A_n = 80 + 40(0.5)^n\).

(3)
(e)

Find the least number of days after which there are fewer than 81 cars at A.

(2)
(f)

State the long-term number of cars in the City, justifying your answer.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
\(T = \begin{pmatrix} 0.7 & 0.2 \\ 0.3 & 0.8 \end{pmatrix}\)✓ 2
(one mark for each correct column)—
Part (b)
Uses \(T^3\begin{pmatrix} 120 \\ 80 \end{pmatrix}\)✓ 1
85 cars at A and 115 cars at C✓ 1
Part (c)
Characteristic equation \(\lambda^2 - 1.5\lambda + 0.5 = 0\)✓ 1
\(\lambda = 1\) and \(\lambda = 0.5\)✓ 1
For \(\lambda = 1\): eigenvector \(\begin{pmatrix} 2 \\ 3 \end{pmatrix}\) (or any multiple)✓ 1
For \(\lambda = 0.5\): eigenvector \(\begin{pmatrix} 1 \\ -1 \end{pmatrix}\) (or any multiple)✓ 1
Part (d)
Writes \(s_0 = a\begin{pmatrix} 2 \\ 3 \end{pmatrix} + b\begin{pmatrix} 1 \\ -1 \end{pmatrix}\), giving \(2a + b = 120\), \(3a - b = 80\)✓ 1
\(a = 40\), \(b = 40\)✓ 1
\(s_n = 40\begin{pmatrix} 2 \\ 3 \end{pmatrix} + 40(0.5)^n\begin{pmatrix} 1 \\ -1 \end{pmatrix}\), so \(A_n = 80 + 40(0.5)^n\)✓ 1
Part (e)
Solves \(80 + 40(0.5)^n < 81\), giving \(n > 5.32\)✓ 1
\(n = 6\) days✓ 1
Part (f)
As \(n \to \infty\), \((0.5)^n \to 0\), so \(s_n \to \begin{pmatrix} 80 \\ 120 \end{pmatrix}\)✓ 1
In the long term there are 120 cars in the City (and 80 at the Airport)✓ 1

Exam tip: The eigenvector for \(\lambda = 1\) gives the steady-state proportions. Scaling \(\begin{pmatrix} 2 \\ 3 \end{pmatrix}\) to total 200 is a quick check on your long-term answer.

432IBMI-030-03-r2-06Exponents and logarithms Paper 2Hard13 marks

The populations of two towns, \(t\) years after the start of 2025, are modelled by
Town A: \(P_A = 24\,000\times1.032^{t}\)
Town B: \(P_B = 41\,000\times0.987^{t}\)

(a)

Using logarithms, find the value of \(t\) when the two towns have equal populations.

(4)
(b)

The model for town A can be written as \(P_A = 24\,000\,e^{kt}\). Find the value of \(k\).

(2)
(c)

Find the time taken for the population of town A to double.

(2)
(d)

A third town, C, had a population of 15 000 at the start of 2025. Its population is modelled as doubling every 18 years. Find the value of \(t\) when the population of town C equals the population of town A, and comment on the reliability of your answer.

(5)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1: sets up the equation \(24\,000\times1.032^{t} = 41\,000\times0.987^{t}\)✓ 1
M1: rearranges to \(\left(\dfrac{1.032}{0.987}\right)^{t} = \dfrac{41}{24}\), or takes logs of both sides✓ 1
M1: \(t = \dfrac{\ln(41/24)}{\ln(1.032/0.987)}\)✓ 1
A1: \(t = 12.0\) years✓ 1
Part (b)
M1: \(e^{k} = 1.032\)✓ 1
A1: \(k = \ln1.032 = 0.0315\)✓ 1
Part (c)
M1: \(1.032^{t} = 2 \Rightarrow t = \dfrac{\ln2}{\ln1.032}\)✓ 1
A1: 22.0 years✓ 1
Part (d)
M1: writes the model \(P_C = 15\,000\times2^{t/18}\)✓ 1
M1: sets up the equation \(15\,000\times2^{t/18} = 24\,000\times1.032^{t}\) and takes logarithms✓ 1
M1: \(t\left(\dfrac{\ln2}{18} - \ln1.032\right) = \ln1.6\)✓ 1
A1: \(t = 67.1\) years✓ 1
R1: unreliable, because this is a long-term extrapolation; constant growth rates are unlikely to continue for about 67 years✓ 1

Exam tip: When the unknown appears in an exponent on both sides, take logarithms of both sides and collect the terms in t before dividing. A GDC intersection is a good way to check your answer.

433IBMI-030-04-r1-06Complex numbers and matrices Paper 2Hard14 marks

A wildlife team models a deer population in two stages: juveniles \(J_n\) and adults \(A_n\), where \(n\) is the number of years after the start of the study.
• Each year, every adult produces on average 1.8 juveniles that survive to the next count. No juveniles remain juveniles for more than one year.
• 20% of juveniles survive and become adults the next year.
• 90% of adults survive to the next year.
Initially \(J_0 = 200\) and \(A_0 = 300\).

(a)

Find the matrix \(M\) such that \(\begin{pmatrix}J_{n+1}\\A_{n+1}\end{pmatrix} = M\begin{pmatrix}J_n\\A_n\end{pmatrix}\).

(2)
(b)

Find the eigenvalues of \(M\) and a corresponding eigenvector for each.

(4)
(c)

Hence show that \(A_n = \frac{800}{3}(1.2)^n + \frac{100}{3}(-0.3)^n\).

(4)
(d)

Describe the long-term behaviour of the population, with reference to the eigenvalues.

(2)
(e)

Find the number of adults predicted after 10 years.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 top row \((0 \;\; 1.8)\)✓ 1
A1 bottom row \((0.2 \;\; 0.9)\), so \(M = \begin{pmatrix}0 & 1.8\\0.2 & 0.9\end{pmatrix}\)✓ 1
Part (b)
M1 characteristic equation \(\lambda^2 - 0.9\lambda - 0.36 = 0\)✓ 1
A1 \(\lambda = 1.2\) and \(\lambda = -0.3\)✓ 1
A1 for \(\lambda = 1.2\): eigenvector \(\begin{pmatrix}3\\2\end{pmatrix}\) (or any non-zero multiple)✓ 1
A1 for \(\lambda = -0.3\): eigenvector \(\begin{pmatrix}-6\\1\end{pmatrix}\) (or any non-zero multiple)✓ 1
Part (c)
M1 writes \(\begin{pmatrix}200\\300\end{pmatrix} = p\begin{pmatrix}3\\2\end{pmatrix} + q\begin{pmatrix}-6\\1\end{pmatrix}\), or uses \(M^n = PD^nP^{-1}\)✓ 1
A1 \(p = \frac{400}{3}\)✓ 1
A1 \(q = \frac{100}{3}\)✓ 1
A1 \(A_n = 2p(1.2)^n + q(-0.3)^n = \frac{800}{3}(1.2)^n + \frac{100}{3}(-0.3)^n\) (AG)✓ 1
Part (d)
R1 \((-0.3)^n \to 0\), so the term with \(\lambda = 1.2\) dominates for large \(n\)✓ 1
A1 the population grows by about 20% per year, with juveniles and adults tending to the ratio \(3:2\)✓ 1
Part (e)
M1 substitutes \(n = 10\) into the expression for \(A_n\), or calculates \(M^{10}\begin{pmatrix}200\\300\end{pmatrix}\)✓ 1
A1 \(A_{10} = 1651\) adults (accept 1650)✓ 1

Exam tip: Check your general formula against a value worked out directly. Here \(A_1 = 0.2(200) + 0.9(300) = 310\), and the formula gives \(320 - 10 = 310\), which confirms the coefficients.

434IBMI-030-08-r2-06Exponents and logarithms Paper 2Hard14 marks

A biologist models the resting metabolic rate, R watts, of a mammal of mass M kg by
\(R = aM^{b}\)
where \(a\) and \(b\) are positive constants.

(a)

Show that this model gives a linear relationship between \(\log_{10} R\) and \(\log_{10} M\).

(2)
(b)

For data from a range of mammals, the regression line of \(y = \log_{10} R\) on \(x = \log_{10} M\) is \(y = 0.74x + 0.53\). Find the value of b and the value of a.

(3)
(c)

Use the model to estimate the resting metabolic rate of a 65 kg mammal.

(2)
(d)

Use the model to estimate the mass of a mammal with a resting metabolic rate of 500 W.

(3)
(e)

Show that, according to the model, doubling the mass of a mammal multiplies its resting metabolic rate by a constant factor, and find this factor.

(2)
(f)

Interpret your answer to the previous part in terms of the energy used per kilogram of body mass.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1: takes logarithms of both sides, \(\log_{10} R = \log_{10}(aM^{b})\)✓ 1
A1: \(\log_{10} R = b\log_{10} M + \log_{10} a\), which has the form \(y = mx + c\) with \(y = \log_{10} R\) and \(x = \log_{10} M\), so the relationship is linear✓ 1
Part (b)
A1: \(b = 0.74\)✓ 1
M1: \(\log_{10} a = 0.53\)✓ 1
A1: \(a = 10^{0.53} = 3.39\)✓ 1
Part (c)
M1: substitutes, \(R = 10^{0.53} \times 65^{0.74}\)✓ 1
A1: \(R = 74.4\) W✓ 1
Part (d)
M1: forms the equation \(500 = 10^{0.53}M^{0.74}\), or \(\log_{10} 500 = 0.74\log_{10} M + 0.53\)✓ 1
M1: rearranges correctly, \(\log_{10} M = \frac{\log_{10} 500 - 0.53}{0.74}\), or solves with GDC✓ 1
A1: \(M = 853\) kg✓ 1
Part (e)
M1: forms the ratio \(\frac{a(2M)^{b}}{aM^{b}} = 2^{b}\), which is independent of M✓ 1
A1: factor \(= 2^{0.74} = 1.67\)✓ 1
Part (f)
R1: the rate is multiplied by less than 2 when the mass doubles, so the rate per kilogram, \(\frac{R}{M} = aM^{-0.26}\), decreases as M increases✓ 1
R1: larger mammals therefore use less energy per kilogram of body mass than smaller mammals✓ 1

Exam tip: In the line \(y = 0.74x + 0.53\), the intercept is \(\log_{10} a\), not a itself. Undo the logarithm, \(a = 10^{0.53}\), before you substitute into the power model.

435IBMI3H-1-02-12Complex numbers in Cartesian, polar and exponential form; matrices, inverses and solving systems; eigenvalues and eigenvectors in context Paper 2Hard14 marks

A workshop makes three products P, Q and R. The weekly resources used per batch are:
Machine hours: 2, 1, 3
Labour hours: 1, 2, 1
Wood (kg): 3, 1, 2
This week 85 machine hours, 65 labour hours and 80 kg of wood are available, and all are used. Let \(x\), \(y\) and \(z\) be the numbers of batches of P, Q and R. The coefficient matrix is \(A=\begin{pmatrix}2&1&3\\1&2&1\\3&1&2\end{pmatrix}\).

(a)

Write the information as a matrix equation \(A\mathbf{v}=\mathbf{b}\).

(2)
(b)

Find \(\det A\) and explain why a unique solution exists.

(2)
(c)

Write down \(A^{-1}\).

(2)
(d)

Hence find \(x\), \(y\) and \(z\).

(3)
(e)

Next week the available resources are 91, 68 and 85. Find the new numbers of batches.

(2)
(f)

Find the percentage increase in the total number of batches made, to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(\mathbf{v}=\begin{pmatrix}x\\y\\z\end{pmatrix}\)✓ 1
A1 \(\mathbf{b}=\begin{pmatrix}85\\65\\80\end{pmatrix}\)✓ 1
Part (b)
A1 \(\det A=-8\)✓ 1
R1 \(\det A\ne0\) so \(A^{-1}\) exists✓ 1
Part (c)
M1 use of technology or adjugate method✓ 1
A1 \(A^{-1}=\frac18\begin{pmatrix}-3&-1&5\\-1&5&-1\\5&-1&-3\end{pmatrix}\)✓ 1
Part (d)
M1 \(\mathbf{v}=A^{-1}\mathbf{b}\)✓ 1
A1 \(x=10\) and \(y=20\)✓ 1
A1 \(z=15\)✓ 1
Part (e)
M1 \(A^{-1}\begin{pmatrix}91\\68\\85\end{pmatrix}\)✓ 1
A1 \(x=10.5\), \(y=20.5\), \(z=16.5\)✓ 1
Part (f)
A1 totals 45 and 47.5✓ 1
M1 \(\frac{47.5-45}{45}\times100\)✓ 1
A1 5.56 %✓ 1

Exam tip: Check an inverse by confirming \(AA^{-1}=I\) for one row and column before using it.

436IBMI3H-1-02-14Complex numbers in Cartesian, polar and exponential form; matrices, inverses and solving systems; eigenvalues and eigenvectors in context Paper 2Hard14 marks

The populations (in thousands) of two fish species in a reservoir in year \(n\) are \(A_n\) and \(B_n\), and \(\begin{pmatrix}A_{n+1}\\B_{n+1}\end{pmatrix}=M\begin{pmatrix}A_n\\B_n\end{pmatrix}\) where \(M=\begin{pmatrix}1&0.3\\0.1&0.8\end{pmatrix}\). Initially \(A_0=4\) and \(B_0=0\).

(a)

Find the eigenvalues of \(M\).

(3)
(b)

Find an eigenvector for each eigenvalue.

(3)
(c)

Express \(\begin{pmatrix}4\\0\end{pmatrix}\) as a combination of the two eigenvectors.

(2)
(d)

Hence write down \(A_n\) and \(B_n\) in terms of \(n\).

(2)
(e)

Find the limit of \(\frac{A_n}{B_n}\) as \(n\to\infty\).

(2)
(f)

Find \(A_{10}\), to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(\det(M-\lambda I)=0\)✓ 1
A1 \(\lambda^{2}-1.8\lambda+0.77=0\)✓ 1
A1 \(\lambda=1.1\) and \(\lambda=0.7\)✓ 1
Part (b)
M1 solving \((M-\lambda I)\mathbf{v}=\mathbf{0}\)✓ 1
A1 \(\lambda=1.1\): \(\begin{pmatrix}3\\1\end{pmatrix}\)✓ 1
A1 \(\lambda=0.7\): \(\begin{pmatrix}1\\-1\end{pmatrix}\)✓ 1
Part (c)
M1 \(3p+q=4\) and \(p-q=0\)✓ 1
A1 \(\begin{pmatrix}4\\0\end{pmatrix}=\begin{pmatrix}3\\1\end{pmatrix}+\begin{pmatrix}1\\-1\end{pmatrix}\)✓ 1
Part (d)
M1 \(M^{n}\) acting on each eigenvector scales it by \(\lambda^{n}\)✓ 1
A1 \(A_n=3(1.1)^{n}+0.7^{n}\) and \(B_n=1.1^{n}-0.7^{n}\)✓ 1
Part (e)
M1 \(0.7^{n}\to0\) relative to \(1.1^{n}\)✓ 1
A1 3✓ 1
Part (f)
M1 \(3(1.1)^{10}+0.7^{10}\)✓ 1
A1 7.81 thousand✓ 1

Exam tip: Writing the starting vector in terms of eigenvectors turns repeated matrix multiplication into powers of the eigenvalues.

437IBMI3H-1-02b-09Complex numbers in Cartesian, polar and exponential form; matrices, inverses and solving systems; eigenvalues and eigenvectors in context Paper 2Hard14 marks

The populations of a pest, in thousands, in two fields satisfy \(\begin{pmatrix} a_{n+1} \\ b_{n+1} \end{pmatrix}\) = A\(\begin{pmatrix} a_n \\ b_n \end{pmatrix}\) each week, where an is the population in field A, bn is the population in field B and A = \(\begin{pmatrix} 0.96 & 0.36 \\ 0.24 & 0.84 \end{pmatrix}\). Initially a0 = 50 and b0 = 30.

(a)

Show that the eigenvalues of A are 1.2 and 0.6.

(3)
(b)

Find an eigenvector for each eigenvalue.

(3)
(c)

Write down matrices P and D such that A = PDP−1.

(2)
(d)

Use An = PDnP−1 to find the populations in the two fields after 10 weeks, to the nearest thousand.

(3)
(e)

Explain what happens to the proportion of the pest population in field A as n becomes very large, and find this proportion.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 det(A − λI) = 0✓ 1
A1 (0.96 − λ)(0.84 − λ) − 0.36 × 0.24 = λ2 − 1.8λ + 0.72 = 0✓ 1
A1 (λ − 1.2)(λ − 0.6) = 0, so λ = 1.2 or 0.6✓ 1
Part (b)
M1 solves (A − λI)v = 0 for each λ✓ 1
A1 λ = 1.2: \(\begin{pmatrix} 3 \\ 2 \end{pmatrix}\)✓ 1
A1 λ = 0.6: \(\begin{pmatrix} 1 \\ -1 \end{pmatrix}\)✓ 1
Part (c)
A1 P = \(\begin{pmatrix} 3 & 1 \\ 2 & -1 \end{pmatrix}\)✓ 1
A1 D = \(\begin{pmatrix} 1.2 & 0 \\ 0 & 0.6 \end{pmatrix}\) (columns in the same order as in P)✓ 1
Part (d)
M1 uses A10 = PD10P−1✓ 1
M1 multiplies by \(\begin{pmatrix} 50 \\ 30 \end{pmatrix}\)✓ 1
A1 field A: 297 thousand, field B: 198 thousand (297.2 and 198.1 before rounding)✓ 1
Part (e)
R1 as n increases, 0.6n → 0 while 1.2n grows, so the populations approach a multiple of \(\begin{pmatrix} 3 \\ 2 \end{pmatrix}\)✓ 1
M1 proportion in field A = 3 ÷ (3 + 2)✓ 1
A1 3/5 = 0.6 (60 %)✓ 1

Exam tip: The columns of P and the diagonal entries of D must be in the same order, otherwise PDP⁻¹ is not equal to A.

438IBMI3H-1-02b-10Complex numbers in Cartesian, polar and exponential form; matrices, inverses and solving systems; eigenvalues and eigenvectors in context Paper 2Hard15 marks

A drone flies from a start point 6 km due east of a control tower and moves in a series of legs. The position of the drone after n legs is the complex number zn, where the tower is at the origin, east is the positive real axis and north is the positive imaginary axis. The drone starts at z0 = 6. After each leg, zn+1 = w zn where w = 1.04eiπ/12.

(a)

State the effect of multiplying by w on the distance of the drone from the tower and on its direction from the tower.

(2)
(b)

Show that zn = 6 × 1.04neinπ/12.

(2)
(c)

Find z8 in the form a + bi, with a and b correct to 3 significant figures.

(3)
(d)

Find the number of legs after which the drone is first more than 15 km from the tower.

(3)
(e)

The drone is observed to be due north of the tower more than 15 km away. Find the distance from the tower at the first leg for which this happens.

(5)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 the distance increases by 4 % (is multiplied by 1.04)✓ 1
A1 the direction rotates anticlockwise by π/12 (15°)✓ 1
Part (b)
M1 zn = wnz0 = 6wn✓ 1
A1 de Moivre: wn = 1.04neinπ/12✓ 1
Part (c)
M1 modulus 6 × 1.048 = 8.21 and argument 8π/12 = 2π/3✓ 1
M1 8.21cos(2π/3) and 8.21sin(2π/3)✓ 1
A1 −4.11 + 7.11i✓ 1
Part (d)
M1 6 × 1.04n > 15✓ 1
A1 n > 23.4 (6 × 1.0423 < 15)✓ 1
A1 n = 24✓ 1
Part (e)
M1 due north means nπ/12 = π/2 + 2πk✓ 1
A1 n = 6 + 24k, for integer k ≥ 0✓ 1
M1 compares with n ≥ 24✓ 1
A1 n = 30✓ 1
A1 6 × 1.0430 = 19.5 km✓ 1

Exam tip: For repeated multiplication by a complex number, keep the modulus and argument separate: the modulus is multiplied each time and the argument is added each time.

439IBMI3H-1-02b-13Complex numbers in Cartesian, polar and exponential form; matrices, inverses and solving systems; eigenvalues and eigenvectors in context Paper 2Hard14 marks

A farmer plants an orchard at a start-up cost of €60 000. In year n the revenue from the orchard is €18 000 × 1.12n−1, and the running costs are €(14 000 + 900(n − 1)).
In the parts below, work in thousands of euros.

(a)

Find the revenue in year 10.

(2)
(b)

Find the total revenue in the first 10 years.

(2)
(c)

Find the total running costs in the first 10 years.

(2)
(d)

The cumulative profit after n years, after the start-up cost, is Pn = 150(1.12n − 1) − n/2 (28 + 0.9(n − 1)) − 60. Find the least number of years for the orchard to have made a cumulative profit.

(3)
(e)

Suggest one reason why the revenue model may not be valid for a very large number of years.

(1)
(f)

The annual profit in year n is the revenue minus the running costs for that year. Find the first year in which the annual profit exceeds €30 000, and the annual profit in that year.

(4)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 18 × 1.129✓ 1
A1 49.9 (thousand euros)✓ 1
Part (b)
M1 S10 = 18(1.1210 − 1) ÷ (1.12 − 1)✓ 1
A1 316 (thousand euros)✓ 1
Part (c)
M1 arithmetic series with first term 14 and common difference 0.9: S10 = 10/2 (2 × 14 + 9 × 0.9)✓ 1
A1 180.5 (thousand euros)✓ 1
Part (d)
M1 evaluates Pn for successive n, or uses a graph✓ 1
A1 P6 = −11.4 (< 0) and P7 = 4.7 (> 0)✓ 1
A1 7 years✓ 1
Part (e)
R1 revenue cannot keep growing by 12 % each year, as it is limited by the number of trees or the market (or trees age and yield falls)✓ 1
Part (f)
M1 18 × 1.12n−1 − (14 + 0.9(n − 1)) > 30 solved by table or graph✓ 1
A1 year 10 gives 27.8 (< 30)✓ 1
A1 year 11✓ 1
A1 annual profit = 32.9 thousand euros, so €32900✓ 1

Exam tip: Check units when a question changes from euros to thousands of euros: write the unit alongside each answer to avoid a factor of 1000 error.

440IBMI3H-1-03r7-09Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Hard13 marks

A livestock feed is a mixture of three types of grain, A, B and C. The amounts of three nutrients in one bag of each grain are given by the columns of the matrix
\(\mathbf{N}=\begin{pmatrix}2&1&k\\1&3&2\\1&0&1\end{pmatrix}\),
where the rows are protein, energy and fibre (in units per bag) and \(k>0\) is the protein content of a bag of grain C.
A cow needs a daily diet of 20 units of protein, 17 units of energy and 7 units of fibre. The cow is given \(x\) bags of A, \(y\) bags of B and \(z\) bags of C each day.

(a)

Show that \(\det\mathbf{N}=7-3k\).

(3)
(b)

Find the value of \(k\) for which the diet cannot be found as a unique solution, and explain what this means for the three grains.

(3)
(c)

For \(k=3\), find \(\mathbf{N}^{-1}\).

(3)
(d)

Hence find \(x\), \(y\) and \(z\) when \(k=3\).

(2)
(e)

A bag of A costs £1.20, a bag of B costs £0.95 and a bag of C costs £1.50. Find the daily cost of the diet.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to expand the determinant of \(\mathbf{N}\)✓ 1
A1 \(2(3-0)-1(1-2)+k(0-3)\)✓ 1
A1 \(=7-3k\) (AG)✓ 1
Part (b)
M1 set \(\det\mathbf{N}=0\)✓ 1
A1 \(k=\frac73\)✓ 1
R1 the nutrient content of C is then a combination of those of A and B (columns dependent), so the grains cannot be combined to give every required diet uniquely✓ 1
Part (c)
M1 attempt to find the inverse (GDC or adjugate)✓ 1
A1 two correct rows✓ 1
A1 \(\mathbf{N}^{-1}=\begin{pmatrix}-1.5&0.5&3.5\\-0.5&0.5&0.5\\1.5&-0.5&-2.5\end{pmatrix}\)✓ 1
Part (d)
M1 attempt to find \(\mathbf{N}^{-1}\begin{pmatrix}20\\17\\7\end{pmatrix}\)✓ 1
A1 \(x=3\), \(y=2\) and \(z=4\) (all three correct)✓ 1
Part (e)
M1 \(3\times1.20+2\times0.95+4\times1.50\) (ECF from (d))✓ 1
A1 £11.50✓ 1

Common error: A zero determinant means no inverse exists; it does not by itself tell you whether the system has no solutions or infinitely many.

441IBMI3H-1-03r7-10Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Hard13 marks

The population of a beetle species in a field is divided into juveniles \(J_n\) and adults \(A_n\) in year \(n\). Each adult produces on average 2.4 juveniles per year, half of the juveniles survive to become adults the next year, and 60% of the adults survive from one year to the next. The model is
\(\begin{pmatrix}J_{n+1}\\A_{n+1}\end{pmatrix}=\mathbf{L}\begin{pmatrix}J_n\\A_n\end{pmatrix}\).

(a)

Write down the matrix \(\mathbf{L}\).

(2)
(b)

Initially there are 200 juveniles and 100 adults. Find the number of adults after 3 years, to the nearest whole number.

(3)
(c)

Find the dominant eigenvalue of \(\mathbf{L}\), and hence the long-term percentage increase in population per year, correct to 3 significant figures.

(3)
(d)

Show that, in the long term, the ratio of juveniles to adults is approximately \(1.67:1\).

(2)
(e)

A treatment lowers the juvenile survival rate to \(s\) (replacing 0.5), with all else unchanged. Find the value of \(s\) for which the dominant eigenvalue is exactly 1, so that the population stops growing. Give your answer correct to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 correct structure with \(J_{n+1}=2.4A_n\)✓ 1
A1 \(\mathbf{L}=\begin{pmatrix}0&2.4\\0.5&0.6\end{pmatrix}\)✓ 1
Part (b)
M1 attempt to find \(\mathbf{L}^3\begin{pmatrix}200\\100\end{pmatrix}\) or to iterate three times✓ 1
A1 after three years \(J=518.4\) and \(A=321.6\)✓ 1
A1 322 adults✓ 1
Part (c)
M1 \(\lambda^2-0.6\lambda-1.2=0\)✓ 1
A1 \(\lambda=1.4358\ldots\)✓ 1
A1 43.6%✓ 1
Part (d)
M1 attempt to use an eigenvector, \(-\lambda J+2.4A=0\) (ECF from (c))✓ 1
A1 \(J:A=2.4:1.4358\ldots=1.67:1\) (AG)✓ 1
Part (e)
M1 set \(\det(\mathbf{L}-\mathbf{I})=0\) with \(\mathbf{L}=\begin{pmatrix}0&2.4\\s&0.6\end{pmatrix}\)✓ 1
A1 \((-1)(0.6-1)-2.4s=0\)✓ 1
A1 \(s=0.167\)✓ 1

Exam tip: If \(\lambda=1\) is an eigenvalue then \(\det(\mathbf{L}-\mathbf{I})=0\), which avoids having to solve the full characteristic equation.

442IBMI3H-1-03r7-14Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Hard12 marks

A tractor costs £84 000 when new. After 3 years its value is £61 500. A dealer models the value \(V\) (in £) after \(t\) years by \(V=84000e^{-kt}\).

(a)

Find the value of \(k\), correct to 3 significant figures.

(3)
(b)

Use the model to find the value of the tractor after 8 years, correct to the nearest £100.

(2)
(c)

The model can be written as \(V=84000(1-\frac{r}{100})^t\). Find \(r\), correct to 3 significant figures.

(2)
(d)

The tractor is replaced when its value first falls to 30% of its new price. Using the exponential model with unrounded \(k\), find the time taken, in years, correct to 2 decimal places.

(2)
(e)

A second dealer uses a linear model passing through the same two data points, \((0, 84000)\) and \((3, 61500)\). Find how many years earlier than in (d) this model reaches 30% of the new price, correct to 2 decimal places. Use the unrounded answer to (d).

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(61500=84000e^{-3k}\)✓ 1
M1 attempt to take natural logarithms, \(k=\frac13\ln\frac{84000}{61500}\)✓ 1
A1 \(k=0.104\) (0.103927…)✓ 1
Part (b)
M1 \(84000e^{-8k}\) with unrounded \(k\)✓ 1
A1 £36 600✓ 1
Part (c)
M1 \(1-\frac{r}{100}=e^{-k}\)✓ 1
A1 \(r=9.87\)✓ 1
Part (d)
M1 \(84000e^{-kt}=0.3\times84000\)✓ 1
A1 \(t=11.58\) years✓ 1
Part (e)
M1 gradient \(-7500\) per year, so \(V=84000-7500t\)✓ 1
M1 \(84000-7500t=25200\), \(t=7.84\)✓ 1
A1 \(3.74\) years✓ 1

Exam tip: Keep unrounded values in your calculator memory between parts; rounding \(k\) to 3 significant figures first would change the later answers.

443IBMI3H-1-04r10-06Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Hard11 marks

A streaming company tracks weekly active users, in thousands, in two regions. If xn is the column vector (users in region 1, users in region 2) in week n, then xn+1 = Mxn, where M = rows (1.0, 0.1), (0.2, 0.9). In week 0, x0 = (30, 60).

(a)

Show that the eigenvalues of M are 1.1 and 0.8.

(3)
(b)

Find an eigenvector for each eigenvalue.

(2)
(c)

Write x0 as a linear combination of these eigenvectors.

(2)
(d)

Hence find the number of users in region 1 in week 10, in thousands, to 3 significant figures.

(3)
(e)

Explain why the ratio of users in the two regions approaches 1 : 1 in the long term.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
characteristic equation (1 − λ)(0.9 − λ) − 0.02 = 0✓ 1
λ2 − 1.9λ + 0.88 = 0✓ 1
(λ − 1.1)(λ − 0.8) = 0 (AG)✓ 1
Part (b)
λ = 1.1: eigenvector (1, 1)✓ 1
λ = 0.8: eigenvector (1, −2)✓ 1
Part (c)
(30, 60) = p(1, 1) + q(1, −2) leading to p + q = 30 and p − 2q = 60✓ 1
p = 40, q = −10✓ 1
Part (d)
xn = 40(1.1)n(1, 1) − 10(0.8)n(1, −2)✓ 1
region 1 users = 40(1.1)10 − 10(0.8)10✓ 1
102.68 so 103 thousand✓ 1
Part (e)
(0.8)n → 0 faster than (1.1)n grows, so the (1, 1) term dominates✓ 1

Exam tip: Writing the starting vector in terms of eigenvectors turns Mⁿ into powers of the eigenvalues, which avoids repeated multiplication.

444IBMI3H-1-04r10-09Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Hard9 marks

A server in a data centre is checked each hour and found to be in one of three states: Up (U), Degraded (D) or Down (W). The state probabilities form a column vector (u, d, w) and the next hour's vector is T times the current one, where T = rows (0.90, 0.40, 0.50), (0.08, 0.45, 0.30), (0.02, 0.15, 0.20). The server is Up at the first check.

(a)

Find the probability that the server is Down at the third check (two hours later).

(3)
(b)

Explain why the long-term probabilities form an eigenvector of T and state the corresponding eigenvalue.

(2)
(c)

Find the long-term probability that the server is Up. Give your answer to 3 significant figures.

(4)
Show mark scheme
Marking pointMarkNotes
Part (a)
first step vector (0.90, 0.08, 0.02) (T applied to (1, 0, 0))✓ 1
second step: w = 0.02(0.90) + 0.15(0.08) + 0.20(0.02)✓ 1
0.034✓ 1
Part (b)
long-term vector s satisfies Ts = s✓ 1
eigenvalue 1✓ 1
Part (c)
equations from the second and third rows: 0.55d = 0.08u + 0.30w and 0.80w = 0.02u + 0.15d✓ 1
d = 0.177u, w = 0.0582u (ratios from solving)✓ 1
u + d + w = 1 used✓ 1
u = 0.809✓ 1

Exam tip: The entries of the steady-state vector must sum to 1, so scale the eigenvector at the end by dividing by its sum.

445IBMI3H-1-03r11-06Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Hard9 marks

Three teleport pads in a game are placed at the solutions of \(z^3 = 64i\) in the complex plane, where one unit represents one metre.

(a)

Write \(64i\) in the form \(re^{i\theta}\), with \(r > 0\) and \(0 < \theta < \pi\).

(2)
(b)

Find the three solutions of \(z^3 = 64i\) in the form \(re^{i\theta}\) with \(-\pi < \theta \le \pi\).

(4)
(c)

Find the distance between two neighbouring pads, in metres, to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(r = 64\)✓ 1
A1 \(\theta = \frac{\pi}{2}\), so \(64i = 64e^{i\pi/2}\)✓ 1
Part (b)
M1 modulus \(\sqrt[3]{64} = 4\)✓ 1
M1 arguments \(\frac{\pi/2 + 2\pi k}{3}\) for \(k = 0, 1, 2\)✓ 1
A1 \(4e^{i\pi/6}\) and \(4e^{5i\pi/6}\)✓ 1
A1 \(4e^{-i\pi/2}\)✓ 1
Part (c)
M1 adjacent roots are \(\frac{2\pi}{3}\) apart on a circle of radius 4✓ 1
M1 cosine rule \(\sqrt{4^2 + 4^2 - 2(4)(4)\cos\frac{2\pi}{3}}\)✓ 1
A1 \(4\sqrt{3} = 6.93\) m✓ 1

Exam tip: Roots of \(z^n = w\) are equally spaced around a circle, which gives a quick check: neighbouring arguments differ by \(\frac{2\pi}{n}\).

446IBMI3H-1-04r14-10Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Hard10 marks

Two pastures, lowland (\(x\)) and upland (\(y\)), hold grass biomass in tonnes. Year by year, \(\begin{pmatrix}x_{n+1}\\y_{n+1}\end{pmatrix}=P\begin{pmatrix}x_n\\y_n\end{pmatrix}\), where \(P=\begin{pmatrix}1.1&0.2\\0.3&0.6\end{pmatrix}\). In year 0, \(x_0=35\) and \(y_0=35\).

(a)

Show that the eigenvalues of \(P\) are 1.2 and 0.5.

(3)
(b)

Find an eigenvector for each eigenvalue.

(2)
(c)

Show that \(\begin{pmatrix}35\\35\end{pmatrix}=20\begin{pmatrix}2\\1\end{pmatrix}-5\begin{pmatrix}1\\-3\end{pmatrix}\) and hence write down \(x_n\) in terms of \(n\).

(2)
(d)

Find \(x_5\) to 3 significant figures, and state the long-term ratio \(y_n:x_n\).

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 sets \(\det(P-\lambda I)=0\)✓ 1
A1 \(\lambda^2-1.7\lambda+0.6=0\)✓ 1
A1 \((\lambda-1.2)(\lambda-0.5)=0\) so \(\lambda=1.2,\ 0.5\)✓ 1
Part (b)
A1 \(\lambda=1.2\): \(\begin{pmatrix}2\\1\end{pmatrix}\) (any multiple)✓ 1
A1 \(\lambda=0.5\): \(\begin{pmatrix}1\\-3\end{pmatrix}\) (any multiple)✓ 1
Part (c)
A1 \(20\begin{pmatrix}2\\1\end{pmatrix}-5\begin{pmatrix}1\\-3\end{pmatrix}=\begin{pmatrix}35\\35\end{pmatrix}\)✓ 1
A1 \(x_n=40(1.2)^n-5(0.5)^n\)✓ 1
Part (d)
M1 \(40(1.2)^5-5(0.5)^5\)✓ 1
A1 \(x_5=99.4\) tonnes✓ 1
A1 ratio tends to \(1:2\) (the eigenvector for 1.2)✓ 1

Exam tip: Write the starting vector as a combination of eigenvectors; then each term is multiplied by its own eigenvalue to the power \(n\).

447IBMI3H-1-04r15-09Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Hard9 marks

The numbers of juvenile and adult fish in a managed fishery are modelled each year by \(\begin{pmatrix}J_{n+1}\\A_{n+1}\end{pmatrix} = \mathbf{L}\begin{pmatrix}J_n\\A_n\end{pmatrix}\), where \(\mathbf{L} = \begin{pmatrix}0&2\\0.5&0.8\end{pmatrix}\).

(a)

Interpret the entries 2 and 0.8 in the matrix L.

(2)
(b)

Show that the eigenvalues λ of L satisfy \(\lambda^2 - 0.8\lambda - 1 = 0\).

(2)
(c)

Find the dominant eigenvalue to 4 significant figures, and interpret it for the fish population.

(2)
(d)

Find an eigenvector corresponding to the dominant eigenvalue and hence find the long-term percentage of the population that is juvenile, to the nearest whole number.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 each adult produces 2 juveniles per year (that survive to the next count)✓ 1
A1 80% of adults survive each year✓ 1
Part (b)
M1 \(\det(\mathbf{L} - \lambda\mathbf{I}) = (-\lambda)(0.8-\lambda) - 2(0.5)\)✓ 1
A1 expansion to the given equation (AG)✓ 1
Part (c)
A1 \(\lambda = 1.477\)✓ 1
R1 in the long run the population grows by about 47.7% each year✓ 1
Part (d)
M1 solve \(-\lambda J + 2A = 0\) or \((\mathbf{L}-\lambda\mathbf{I})\mathbf{v}=\mathbf{0}\) with \(\lambda = 1.477\) (ECF from (c))✓ 1
A1 \(J : A = 1.354 : 1\) (e.g. eigenvector \(\begin{pmatrix}1.354\\1\end{pmatrix}\))✓ 1
A1 \(\dfrac{1.354}{2.354} \approx 57.5\%\), so 58% (accept 57.5%)✓ 1

Exam tip: An eigenvector can be scaled by any non-zero constant, so choose a convenient scale and convert to a proportion at the end.

448IBMI3H-1-04r17-12Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Hard9 marks

A cycling club keeps a record of its junior members \(J_n\) and senior members \(S_n\) at the start of year \(n\). The membership is modelled by $\(\begin{pmatrix}J_{n+1}\\S_{n+1}\end{pmatrix}=\mathbf{M}\begin{pmatrix}J_n\\S_n\end{pmatrix},\qquad \mathbf{M}=\begin{pmatrix}0.5&1.5\\0.4&0.2\end{pmatrix}.\)$

(a)

Show that the eigenvalues of \(\mathbf{M}\) satisfy \(\lambda^2-0.7\lambda-0.5=0\) and find them, to 3 significant figures.

(4)
(b)

Explain which eigenvalue determines the long-term behaviour, and state the long-term percentage yearly growth in membership, to 3 significant figures.

(2)
(c)

Find the ratio \(J:S\) in the long term, giving it in the form \(r:1\) with \(r\) to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(\det(\mathbf{M}-\lambda\mathbf{I})=(0.5-\lambda)(0.2-\lambda)-0.6\)✓ 1
A1 \(=\lambda^2-0.7\lambda-0.5=0\) (AG)✓ 1
M1 solves the quadratic (or GDC)✓ 1
A1 \(\lambda=1.14\) or \(\lambda=-0.439\)✓ 1
Part (b)
R1 the eigenvalue of greatest modulus, 1.139, dominates as \(n\) increases✓ 1
A1 membership grows by about 13.9% each year✓ 1
Part (c)
M1 uses the eigenvector for \(\lambda=1.1389\ldots\) from \(0.4x+(0.2-\lambda)y=0\) (unrounded \(\lambda\))✓ 1
M1 \(\frac{x}{y}=\frac{\lambda-0.2}{0.4}\)✓ 1
A1 \(r=2.35\)✓ 1

Exam tip: Use the unrounded eigenvalue when finding the eigenvector; a rounded eigenvalue gives a vector that only approximately satisfies the equations.

449IBMI3H-1-04r17-13Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Hard10 marks

On a grid plan of an excavation, an archaeologist reconstructs the original positions of artefacts that have been rotated and enlarged by a landslip. The movement is represented by the matrix $\(\mathbf{A}=\begin{pmatrix}\sqrt3&-1\\1&\sqrt3\end{pmatrix},\)$ which maps a position $(x,y)$ to $\mathbf{A}\begin{pmatrix}x\\y\end{pmatrix}$.

(a)

An artefact is originally at \((4,2)\). Find its new position, giving your answers to 3 significant figures.

(2)
(b)

By writing the position \((x,y)\) as the complex number \(z=x+iy\), show that the transformation is equivalent to multiplication by \(w=\sqrt3+i\), and write \(w\) in the form \(re^{i\theta}\).

(2)
(c)

Show that the eigenvalues of \(\mathbf{A}\) are complex and find them.

(3)
(d)

The landslip is repeated six times in total. Show that the combined effect is \(\mathbf{A}^6=k\mathbf{I}\) and find \(k\).

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(\mathbf{A}\begin{pmatrix}4\\2\end{pmatrix}=\begin{pmatrix}4\sqrt3-2\\4+2\sqrt3\end{pmatrix}\)✓ 1
A1 \((4.93, 7.46)\)✓ 1
Part (b)
M1 \((\sqrt3+i)(x+iy)=(\sqrt3x-y)+i(x+\sqrt3y)\), matching the matrix product✓ 1
A1 \(w=2e^{i\pi/6}\) (modulus 2, argument \(\frac{\pi}{6}\))✓ 1
Part (c)
M1 \(\det(\mathbf{A}-\lambda\mathbf{I})=(\sqrt3-\lambda)^2+1=0\)✓ 1
M1 \(\lambda-\sqrt3=\pm i\)✓ 1
A1 \(\lambda=\sqrt3\pm i\) (complex, so no real eigenvector: no direction is unchanged by a rotation of \(30^\circ\))✓ 1
Part (d)
M1 uses \(w^6=2^6e^{i\pi}\) (six rotations of \(30^\circ\), six enlargements of factor 2)✓ 1
A1 \(w^6=-64\)✓ 1
A1 \(\mathbf{A}^6=-64\mathbf{I}\), so \(k=-64\)✓ 1

Exam tip: A matrix of the form \(\begin{pmatrix}a&-b\\b&a\end{pmatrix}\) behaves like multiplication by \(a+bi\), so powers can be found from the polar form.

450IBMI3H-1-04r17-14Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Hard10 marks

The loads on two server clusters, in thousands of requests per second, are \(\mathbf{x}_n=\begin{pmatrix}a_n\\b_n\end{pmatrix}\) in week \(n\), and evolve by \(\mathbf{x}_{n+1}=\mathbf{M}\mathbf{x}_n\), where $\(\mathbf{M}=\begin{pmatrix}1.2&0.3\\0.2&1.1\end{pmatrix}.\)$ In week 0 the loads are $\mathbf{x}_0=\begin{pmatrix}50\\40\end{pmatrix}$.

(a)

Find the eigenvalues of \(\mathbf{M}\).

(3)
(b)

Find an eigenvector for each eigenvalue.

(2)
(c)

Express \(\mathbf{x}_0\) as a linear combination of these two eigenvectors.

(2)
(d)

Hence find the load on the first cluster in week 10, to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(\det(\mathbf{M}-\lambda\mathbf{I})=(1.2-\lambda)(1.1-\lambda)-0.06=0\)✓ 1
M1 \(\lambda^2-2.3\lambda+1.26=0\)✓ 1
A1 \(\lambda=1.4\) and \(\lambda=0.9\)✓ 1
Part (b)
A1 for \(\lambda=1.4\): \(\begin{pmatrix}3\\2\end{pmatrix}\) (or any multiple)✓ 1
A1 for \(\lambda=0.9\): \(\begin{pmatrix}1\\-1\end{pmatrix}\) (or any multiple)✓ 1
Part (c)
M1 \(3p+q=50\) and \(2p-q=40\) (with \(\mathbf{x}_0=p\begin{pmatrix}3\\2\end{pmatrix}+q\begin{pmatrix}1\\-1\end{pmatrix}\))✓ 1
A1 \(p=18,\ q=-4\)✓ 1
Part (d)
M1 \(\mathbf{x}_{10}=18(1.4)^{10}\begin{pmatrix}3\\2\end{pmatrix}-4(0.9)^{10}\begin{pmatrix}1\\-1\end{pmatrix}\)✓ 1
M1 first component \(54(1.4)^{10}-4(0.9)^{10}\)✓ 1
A1 \(1560\) thousand requests per second✓ 1

Common error: Powers of \(\mathbf{M}\) act on each eigenvector component separately by \(\lambda^n\); do not raise the entries of \(\mathbf{M}\) to the power \(n\).

451IBMI3H-1-04r18-04Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Hard9 marks

A hospital drone delivers supplies between buildings. The drone starts at the origin of a complex plane (units: metres) and flies outwards along the straight line \(\arg z=\frac{\pi}{6}\). A wireless relay at \(z=60\) gives coverage in the region \(|z-60|\le50\).

(a)

Write the drone's position as \(z=re^{i\pi/6}\) with \(r>0\). Show that the drone is on the boundary of the coverage region when \(r^2-60\sqrt3\,r+1100=0\).

(3)
(b)

Solve the equation in part (a), giving the roots in exact form.

(3)
(c)

Find the distance the drone travels while inside the coverage region.

(2)
(d)

Find the greatest distance from the origin at which the drone still has coverage, to 3 significant figures.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(|re^{i\pi/6}-60|^2=2500\)✓ 1
M1 expands using \(\left|re^{i\pi/6}-60\right|^2=r^2-2(60)r\cos\frac{\pi}{6}+3600\)✓ 1
A1 \(r^2-60\sqrt3\,r+1100=0\) (AG)✓ 1
Part (b)
M1 uses the quadratic formula (or completes the square)✓ 1
A1 discriminant \(=10800-4400=6400\)✓ 1
A1 \(r=30\sqrt3\pm40\)✓ 1
Part (c)
M1 difference between the two roots✓ 1
A1 \(80\) m✓ 1
Part (d)
A1 \(30\sqrt3+40=92.0\) m✓ 1

Exam tip: For a locus question, substitute the polar form into \(|z-a|=k\) and expand using \(|u|^2=u\bar u\) or the cosine rule. The result is a quadratic in \(r\).

452IBMI3H-1-04r18-08Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Hard10 marks

During a race simulation, an engineer tracks the lap-by-lap deviations of a car's front tyre temperature and brake temperature from their target values, written as a vector \(\mathbf{d}_n\) in °C on lap n. The model is \(\mathbf{d}_{n+1}=M\mathbf{d}_n\), where \(M=\begin{pmatrix}0.8&-0.3\\0.3&0.8\end{pmatrix}\).

(a)

Find the eigenvalues of \(M\).

(3)
(b)

Find the modulus and argument of the eigenvalue \(0.8+0.3i\), to 3 significant figures.

(2)
(c)

The engineer wants the size of the deviation \(|\mathbf{d}_n|\) to fall below 5% of \(|\mathbf{d}_0|\). Given that \(|M^n\mathbf{d}_0|=(0.73)^{n/2}|\mathbf{d}_0|\), find the smallest whole number of laps needed.

(3)
(d)

State how many laps one full cycle of the oscillation takes, to 3 significant figures, and describe the long-term behaviour of the deviations.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(\det(M-\lambda I)=(0.8-\lambda)^2+0.09=0\)✓ 1
A1 \((0.8-\lambda)^2=-0.09\)✓ 1
A1 \(\lambda=0.8\pm0.3i\)✓ 1
Part (b)
A1 modulus \(=\sqrt{0.73}=0.854\)✓ 1
A1 argument \(=\arctan\frac{0.3}{0.8}=0.359\) rad✓ 1
Part (c)
M1 \((0.73)^{n/2}<0.05\)✓ 1
M1 takes logarithms, \(n>\frac{2\ln0.05}{\ln0.73}=19.04\) (reverses inequality as \(\ln0.73<0\))✓ 1
A1 \(n=20\)✓ 1
Part (d)
A1 \(\frac{2\pi}{0.359}=17.5\) laps (using unrounded argument)✓ 1
R1 the deviations spiral (oscillate) with decreasing size, tending to zero✓ 1

Common error: Concluding that complex eigenvalues mean no solution. Complex eigenvalues describe rotation, and the modulus decides whether the motion grows (\(>1\)) or decays (\(<1\)).

453IBMI3H-1-04r18-12Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Hard12 marks

A Formula 1 team plans its tyre allocation. The numbers of sets of soft, medium and hard tyres, \(x\), \(y\) and \(z\), must satisfy
\(x+y+z=13\)
\(x+2y+3z=26\)
\(x+4y+kz=40\)
where \(k\) is a constant that depends on the wear rating of the hard compound.

(a)

Write the system as \(A\mathbf{v}=\mathbf{b}\) and show that \(\det A=k-7\).

(3)
(b)

Find the value of \(k\) for which the system does not have a unique solution.

(1)
(c)

For \(k=5\), use the inverse of \(A\) to find \(x\), \(y\) and \(z\).

(3)
(d)

For \(k=7\), show that the system has no solution.

(3)
(e)

With \(k=7\), the number 40 in the third equation is replaced by \(p\). Find the value of \(p\) for which there are infinitely many solutions.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 expands the determinant of \(A=\begin{pmatrix}1&1&1\\1&2&3\\1&4&k\end{pmatrix}\)✓ 1
M1 \(1(2k-12)-1(k-3)+1(4-2)\)✓ 1
A1 \(\det A=k-7\) (AG)✓ 1
Part (b)
A1 \(k=7\)✓ 1
Part (c)
M1 \(\mathbf{v}=A^{-1}\mathbf{b}\)✓ 1
A1 \(x=6\)✓ 1
A1 \(y=1\) and \(z=6\)✓ 1
Part (d)
M1 subtracts equations: the second minus the first gives \(y+2z=13\)✓ 1
M1 the third minus the first gives \(3y+6z=27\), i.e. \(y+2z=9\)✓ 1
R1 these contradict each other, so the system is inconsistent✓ 1
Part (e)
M1 consistency needs \(p-13=3\times(26-13)\)✓ 1
A1 \(p=52\)✓ 1

Exam tip: A zero determinant means no unique solution, but the system may have either no solutions or infinitely many. Eliminate variables to decide which.

454IBMI3H-1-05r19-07Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Hard12 marks

A rover on a seabed survey makes a sequence of moves. Each move is represented by a complex number, in metres. The first move is \(d_0=5\) (due east) and each later move is obtained from the one before by \(d_{n+1}=r\,d_n\), where \(r=0.9\,e^{0.4i}\). So \(d_n=5r^n\) for \(n=0,1,2,\ldots\)

(a)

Find \(d_3\) in the form \(a+bi\), with \(a\) and \(b\) correct to 3 significant figures.

(3)
(b)

The rover stops as soon as the length of a move is less than 1 m. Find the value of \(n\) for the first move that is shorter than 1 m.

(3)
(c)

If the rover could continue for ever, its total displacement from its starting point would be \(S=d_0+d_1+d_2+\cdots\). Find \(S\) in the form \(a+bi\), with \(a\) and \(b\) to 3 significant figures.

(4)
(d)

Find \(|S|\), the distance of this limiting point from the starting point, in metres to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(d_3=5\times0.9^3\,e^{1.2i}=3.645\,e^{1.2i}\)✓ 1
A1 \(a=1.32\) (1.3208…)✓ 1
A1 \(b=3.40\) (3.3973…)✓ 1
Part (b)
M1 \(|d_n|=5\times0.9^n\)✓ 1
M1 solve \(5\times0.9^n<1\) using logarithms or a table, \(n>15.28\) (or \(0.9^{15}\times5=1.03\), \(0.9^{16}\times5=0.927\))✓ 1
A1 \(n=16\)✓ 1
Part (c)
M1 use of \(S_\infty=\frac{d_0}{1-r}\)✓ 1
M1 substitute \(r=0.9e^{0.4i}=0.829\ldots+0.350\ldots i\) and simplify✓ 1
A1 \(a=5.62\) (5.6231…)✓ 1
A1 \(b=11.5\) (11.522…)✓ 1
Part (d)
M1 \(\sqrt{5.623^2+11.52^2}\) or \(\frac{5}{|1-r|}\)✓ 1
A1 12.8 m (12.82…)✓ 1

Exam tip: A geometric series works with a complex common ratio exactly as with a real one; it converges when \(|r|<1\), which is why the factor 0.9 matters, not the angle 0.4.

455IBMI3H-1-05r19-09Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Hard10 marks

A lake is split into a northern basin (P) and a southern basin (Q). The number of fish in each basin in year \(n\) is \(P_n\) and \(Q_n\). Allowing for growth and movement, the model is
\(\begin{pmatrix}P_{n+1}\\Q_{n+1}\end{pmatrix}=\mathbf{T}\begin{pmatrix}P_{n}\\Q_{n}\end{pmatrix},\quad \mathbf{T}=\begin{pmatrix}1.2&0.3\\0.2&1.1\end{pmatrix}.\)
Initially \(P_0=500\) and \(Q_0=300\).

(a)

Find the eigenvalues of \(\mathbf{T}\).

(3)
(b)

Find an eigenvector for each eigenvalue.

(3)
(c)

Write \(\begin{pmatrix}500\\300\end{pmatrix}\) as a combination of the two eigenvectors in (b).

(2)
(d)

Hence find \(P_8\), the number of fish in the northern basin after 8 years, to the nearest whole number.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(\det(\mathbf{T}-\lambda\mathbf{I})=0\)✓ 1
A1 \(\lambda^2-2.3\lambda+1.26=0\)✓ 1
A1 \(\lambda=1.4\) and \(\lambda=0.9\)✓ 1
Part (b)
M1 solve \((\mathbf{T}-\lambda\mathbf{I})\mathbf{v}=\mathbf{0}\) for one eigenvalue✓ 1
A1 for \(\lambda=1.4\): any multiple of \(\begin{pmatrix}3\\2\end{pmatrix}\)✓ 1
A1 for \(\lambda=0.9\): any multiple of \(\begin{pmatrix}1\\-1\end{pmatrix}\)✓ 1
Part (c)
M1 \(3a+b=500\) and \(2a-b=300\)✓ 1
A1 \(a=160\), \(b=20\), so \(\begin{pmatrix}500\\300\end{pmatrix}=160\begin{pmatrix}3\\2\end{pmatrix}+20\begin{pmatrix}1\\-1\end{pmatrix}\)✓ 1
Part (d)
M1 \(P_8=160\times3\times1.4^8+20\times1\times0.9^8\) ECF from (c)✓ 1
A1 7092 (7092.4…)✓ 1

Exam tip: Each eigenvector component scales by its own eigenvalue: if \(\mathbf{x}_0=a\mathbf{v}_1+b\mathbf{v}_2\) then \(\mathbf{T}^n\mathbf{x}_0=a\lambda_1^n\mathbf{v}_1+b\lambda_2^n\mathbf{v}_2\), which avoids repeated matrix multiplication.

456IBMI3H-1-05r20-12Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Hard9 marks

Calculator

Two pressure oscillations in a water main are represented by the complex numbers (in kPa)
\(z_1=5e^{0.4i}\) and \(z_2=3e^{1.9i}\),
where the argument is the phase in radians.

(a)

Write \(z_1+z_2\) in the form \(a+bi\), giving \(a\) and \(b\) to 3 significant figures.

(3)
(b)

Hence find the amplitude and the phase of the combined oscillation, to 3 significant figures. Use the unrounded sum from part (a).

(3)
(c)

A damper adds a third oscillation \(z_3\) so that \(z_1+z_2+z_3=0\). Write \(z_3\) in the form \(re^{i\theta}\) with \(-\pi<\theta\le\pi\), giving \(r\) and \(\theta\) to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 convert \(z_1\) and \(z_2\) to Cartesian form, e.g. \(5\cos0.4+5i\sin0.4\)✓ 1
A1 \(z_1=4.61+1.95i\), \(z_2=-0.970+2.84i\)✓ 1
A1 \(z_1+z_2=3.64+4.79i\)✓ 1
Part (b)
M1 modulus \(\sqrt{a^2+b^2}\)✓ 1
A1 amplitude \(=6.01\) kPa✓ 1
A1 phase \(=\arctan\frac{b}{a}=0.921\) rad✓ 1
Part (c)
M1 \(z_3=-(z_1+z_2)\), so the modulus is unchanged and the argument differs by \(\pi\)✓ 1
A1 \(r=6.01\)✓ 1
A1 \(\theta=0.9212-\pi=-2.22\)✓ 1

Common error: When you negate a complex number, its modulus stays the same and its argument changes by \(\pi\). Choose the version that lies in the stated interval.

457IBMI3H-1-05r20-14Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Hard11 marks

Calculator

Two storage tanks A and B in a water supply network are connected. Each hour, 30% of the water in tank A is pumped to tank B and 40% of the water in tank B is pumped to tank A. The volumes (in m3) at the start of hour \(n\) are \(\mathbf{v}_n=\begin{pmatrix}A_n\\B_n\end{pmatrix}\), so that \(\mathbf{v}_{n+1}=\mathbf{M}\mathbf{v}_n\) with
\(\mathbf{M}=\begin{pmatrix}0.7&0.4\\0.3&0.6\end{pmatrix}.\)
Initially tank A holds 700 m3 and tank B is empty.

(a)

Show that the eigenvalues of \(\mathbf{M}\) are 1 and 0.3.

(3)
(b)

Find an eigenvector for each eigenvalue.

(2)
(c)

Write \(\begin{pmatrix}700\\0\end{pmatrix}\) as a linear combination of your two eigenvectors.

(2)
(d)

Hence show that \(A_n=400+300(0.3)^n\), and find the smallest value of \(n\) for which \(A_n\) is within 1 m3 of its long-term value.

(3)
(e)

State the long-term volume in tank B, in m3.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(\det(\mathbf{M}-\lambda\mathbf{I})=0\)✓ 1
A1 \(\lambda^2-1.3\lambda+0.3=0\)✓ 1
A1 \((\lambda-1)(\lambda-0.3)=0\), so \(\lambda=1\) or \(0.3\)✓ 1
Part (b)
A1 for \(\lambda=1\): e.g. \(\begin{pmatrix}4\\3\end{pmatrix}\)✓ 1
A1 for \(\lambda=0.3\): e.g. \(\begin{pmatrix}1\\-1\end{pmatrix}\)✓ 1
Part (c)
M1 \(a\begin{pmatrix}4\\3\end{pmatrix}+b\begin{pmatrix}1\\-1\end{pmatrix}=\begin{pmatrix}700\\0\end{pmatrix}\), so \(4a+b=700\) and \(3a-b=0\)✓ 1
A1 \(a=100\), \(b=300\)✓ 1
Part (d)
M1 \(\mathbf{v}_n=100\cdot1^n\begin{pmatrix}4\\3\end{pmatrix}+300(0.3)^n\begin{pmatrix}1\\-1\end{pmatrix}\) so \(A_n=400+300(0.3)^n\)✓ 1
M1 \(300(0.3)^n<1\), i.e. \(n>\frac{\ln300}{\ln(1/0.3)}=4.74\)✓ 1
A1 \(n=5\)✓ 1
Part (e)
A1 \(B_n=300-300(0.3)^n\to300\) m3✓ 1

Exam tip: Once an initial vector is split into eigenvectors, each part just gets multiplied by its own eigenvalue to the power \(n\). The term with eigenvalue less than 1 dies away, leaving the steady state.

458IBMI3H-1-04r21-06Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Hard14 marks

A drug is taken in a single dose. Each hour, the amounts bn (in the blood) and tn (in the tissue), both in mg, change according to \(\begin{pmatrix}b_{n+1}\\t_{n+1}\end{pmatrix} = \mathbf{M}\begin{pmatrix}b_n\\t_n\end{pmatrix}\) where \(\mathbf{M} = \begin{pmatrix}0.8&0.1\\0.2&0.7\end{pmatrix}\). Initially 30 mg is in the blood and none in the tissue.

(a)

Find the eigenvalues of M.

(3)
(b)

Find an eigenvector corresponding to each eigenvalue.

(3)
(c)

Write down a matrix P and a diagonal matrix D such that M = PDP−1.

(2)
(d)

Express the initial vector \(\begin{pmatrix}30\\0\end{pmatrix}\) as a linear combination of the two eigenvectors.

(2)
(e)

Find the total amount of drug in the body after 10 hours. Give your answer to 3 significant figures.

(3)
(f)

Explain why, after many hours, the amounts of drug in the blood and in the tissue become almost equal.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 det(M − λI) = 0 attempted✓ 1
A1 λ2 − 1.5λ + 0.54 = 0✓ 1
A1 λ = 0.9 and λ = 0.6✓ 1
Part (b)
M1 solve (M − λI)v = 0 for one eigenvalue✓ 1
A1 for 0.9: any non-zero multiple of \(\begin{pmatrix}1\\1\end{pmatrix}\)✓ 1
A1 for 0.6: any non-zero multiple of \(\begin{pmatrix}1\\-2\end{pmatrix}\)✓ 1
Part (c)
A1 P = \(\begin{pmatrix}1&1\\1&-2\end{pmatrix}\) (eigenvectors as columns)✓ 1
A1 D = \(\begin{pmatrix}0.9&0\\0&0.6\end{pmatrix}\) with order matching P✓ 1
Part (d)
M1 set up a + b = 30 and a − 2b = 0✓ 1
A1 a = 20 and b = 10✓ 1
Part (e)
M1 use M10 applied to the initial vector, or 20(0.9)10\(\begin{pmatrix}1\\1\end{pmatrix}\) + 10(0.6)10\(\begin{pmatrix}1\\-2\end{pmatrix}\)✓ 1
A1 blood 7.03 mg and tissue 6.85 mg✓ 1
A1 total 13.9 mg✓ 1
Part (f)
R1 the 0.6n term decays faster than the 0.9n term, so the vector tends to a multiple of (1, 1)✓ 1

Exam tip: When a repeated matrix process is written using eigenvectors, the eigenvalue with the largest size controls the long-term behaviour.

459IBMI3H-1-04r21-09Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Hard12 marks

Two pump motors in a water-supply station have complex impedances Z1 = 3 + 4i Ω and Z2 = 4 − 3i Ω.

(a)

The motors are connected in series, so the total impedance is Z1 + Z2. Write down the total impedance, its modulus to 3 significant figures, and its argument to 3 significant figures.

(3)
(b)

The motors are instead connected in parallel, so that 1/Z = 1/Z1 + 1/Z2. Show that the total impedance is Z = 3.5 + 0.5i Ω.

(4)
(c)

A voltage of 120 V (taken as a real number) is applied across the parallel combination. The current is I = 120/Z. Find I in the form a + bi and find its modulus to 3 significant figures.

(3)
(d)

Find the angle, in degrees to 3 significant figures, by which the current lags behind the voltage.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 7 + i✓ 1
A1 modulus √50 = 7.07 Ω✓ 1
A1 argument tan−1(1/7) = 0.142 rad✓ 1
Part (b)
M1 Z = Z1Z2/(Z1 + Z2)✓ 1
A1 Z1Z2 = 24 + 7i✓ 1
M1 multiply numerator and denominator by the conjugate 7 − i✓ 1
A1 (175 + 25i)/50 = 3.5 + 0.5i✓ 1
Part (c)
M1 multiply by the conjugate 3.5 − 0.5i of Z✓ 1
A1 I = 33.6 − 4.8i✓ 1
A1 modulus 33.9 A✓ 1
Part (d)
M1 angle = tan−1(0.5/3.5) or tan−1(4.8/33.6)✓ 1
A1 8.13°✓ 1

Common error: Dividing by a complex number without multiplying top and bottom by the conjugate. Real and imaginary parts cannot be separated until the denominator is real.

460IBMI3H-1-04r22-06Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Hard11 marks

A coach plans the weekly hours of strength (x), endurance (y) and skills (z) training for a squad. The plan must satisfy
\(x + 2y + z = 20\)
\(2x + ky + 3z = b\)
\(x + y + 2z = 15\)
where k and b are constants.

(a)

Find, in terms of k, the determinant of the matrix of coefficients \(\begin{pmatrix} 1 & 2 & 1 \\ 2 & k & 3 \\ 1 & 1 & 2 \end{pmatrix}\).

(2)
(b)

Hence write down the value of k for which the system does not have a unique solution.

(1)
(c)

Let \(k = 3\) and \(b = 35\). Show that the system has infinitely many solutions and find the solutions in terms of z.

(3)
(d)

The coach needs x, y and z to be non-negative whole numbers of hours. Find all such solutions.

(3)
(e)

Let \(k = 3\) and \(b = 40\). Explain why the system has no solution.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 expands the determinant, e.g. \(1(2k - 3) - 2(4 - 3) + 1(2 - k)\)✓ 1
A1 \(k - 3\)✓ 1
Part (b)
A1 \(k = 3\)✓ 1
Part (c)
M1 subtracts the third equation from the first: \(y - z = 5\) (or notes that row 2 = row 1 + row 3 and \(35 = 20 + 15\), so one equation is redundant)✓ 1
A1 \(y = z + 5\)✓ 1
A1 \(x = 20 - 2y - z = 10 - 3z\), so infinitely many solutions (one for each z)✓ 1
Part (d)
M1 \(x \ge 0\) gives \(10 - 3z \ge 0\), so \(z \le 3.33\ldots\) (and \(y = z + 5 > 0\))✓ 1
A1 \(z \in \{0, 1, 2, 3\}\)✓ 1
A1 \((x, y, z) = (10, 5, 0), (7, 6, 1), (4, 7, 2), (1, 8, 3)\)✓ 1
Part (e)
M1 eliminates in the same way: row 2 − (row 1 + row 3) gives \(0 = 40 - 35\)✓ 1
R1 \(0 = 5\) is a contradiction, so the equations are inconsistent✓ 1

Common error: A zero determinant does not tell you which case you are in: always substitute and eliminate to decide between no solution and infinitely many solutions.

461IBMI3H-1-04r22-08Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Hard12 marks

The population of a bird species on a nature reserve is modelled using juveniles (J) and adults (A). Each year, every adult produces on average 1.8 juveniles. Half of the juveniles survive to become adults the next year, and 70% of adults survive to the next year. Juveniles do not breed. The populations in year n and year n + 1 are related by \(\begin{pmatrix} J_{n+1} \\ A_{n+1} \end{pmatrix} = L\begin{pmatrix} J_{n} \\ A_{n} \end{pmatrix}\).

(a)

Write down the matrix L. In year 0 there are 200 juveniles and 300 adults. Find the populations in year 1.

(2)
(b)

Show that the eigenvalues of L satisfy \(\lambda^2 - 0.7\lambda - 0.9 = 0\).

(2)
(c)

Find the eigenvalues of L, to 3 significant figures.

(3)
(d)

Find an eigenvector for the larger eigenvalue, and hence find the long-term percentage of the population that are juveniles. Use the unrounded eigenvalue and give your answer to 3 significant figures.

(3)
(e)

Interpret the larger eigenvalue in this context. Give the long-term percentage growth in the total population each year, to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(L = \begin{pmatrix} 0 & 1.8 \\ 0.5 & 0.7 \end{pmatrix}\)✓ 1
A1 \(J_1 = 540\) and \(A_1 = 310\)✓ 1
Part (b)
M1 \(\det(L - \lambda I) = (-\lambda)(0.7 - \lambda) - 1.8 \times 0.5\)✓ 1
A1 \(= \lambda^2 - 0.7\lambda - 0.9 = 0\) (AG)✓ 1
Part (c)
M1 solves the quadratic (calculator or formula)✓ 1
A1 \(\lambda = 1.36\)✓ 1
A1 \(\lambda = -0.661\)✓ 1
Part (d)
M1 solves \(-\lambda J + 1.8A = 0\) so \(J : A = 1.8 : \lambda\)✓ 1
A1 eigenvector proportional to \(\begin{pmatrix} 1.8 \\ 1.361 \end{pmatrix}\) (ratio \(J : A = 1.322 : 1\))✓ 1
A1 juveniles are \(\dfrac{1.322}{1.322 + 1}\) = 56.9%✓ 1
Part (e)
R1 in the long term the population is multiplied by about 1.36 each year (population grows)✓ 1
A1 growth of 36.1% per year✓ 1

Exam tip: In a population model the dominant eigenvalue gives the long-term growth factor each step, and its eigenvector gives the long-term proportions.

462IBMI3H-1-04r22-12Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Hard12 marks

Planets orbiting the Sun are observed to satisfy \(T = kr^{b}\), where T is the orbital period in years, r is the mean distance from the Sun in astronomical units (AU), and k and b are constants. Venus has \(r = 0.723\) and \(T = 0.615\). Neptune has \(r = 30.1\) and \(T = 165\).

(a)

Show that \(\ln T = b\ln r + \ln k\).

(2)
(b)

Use the data for Venus and Neptune to find b, to 3 significant figures.

(3)
(c)

Use b = 1.50 and the data for Venus to find k, to 3 significant figures.

(2)
(d)

Use b = 1.50 and k = 1.00 to find the mean distance of an asteroid whose period is 6.5 years. Give your answer in AU to 3 significant figures.

(3)
(e)

Use the same values to estimate the period of a comet whose mean distance is 17.8 AU. Give your answer in years to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 takes natural logarithms of both sides: \(\ln T = \ln(kr^{b})\)✓ 1
A1 uses the laws of logarithms: \(\ln k + b\ln r\) (AG)✓ 1
Part (b)
M1 forms two equations \(\ln 0.615 = b\ln 0.723 + \ln k\) and \(\ln 165 = b\ln 30.1 + \ln k\)✓ 1
M1 subtracts: \(b = \dfrac{\ln 165 - \ln 0.615}{\ln 30.1 - \ln 0.723}\)✓ 1
A1 \(b = 1.50\)✓ 1
Part (c)
M1 \(k = \dfrac{0.615}{0.723^{1.5}}\)✓ 1
A1 \(k = 1.00\)✓ 1
Part (d)
M1 \(6.5 = r^{1.5}\)✓ 1
M1 \(r = 6.5^{1/1.5} = 6.5^{2/3}\)✓ 1
A1 3.48 AU✓ 1
Part (e)
M1 \(T = 1.00 \times 17.8^{1.5}\)✓ 1
A1 75.1 years✓ 1

Exam tip: Taking logarithms of both sides turns a power model \(T = kr^{b}\) into a linear relationship between \(\ln T\) and \(\ln r\), with gradient b and intercept \(\ln k\).

463IBMI3H-1-05r22-03Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Hard12 marks

In a sample of dough, the yeast population is modelled in hourly stages. At the start of hour n there are Yn thousand young cells and Mn thousand mature cells. Then
Yn+1 = 0.4Yn + 1.5Mn
Mn+1 = 0.6Yn + 0.5Mn

(a)

Write the model as \(\mathbf{x}_{n+1}=L\mathbf{x}_n\), stating the matrix \(L\).

(1)
(b)

Show that the eigenvalues of \(L\) are 1.4 and −0.5.

(4)
(c)

Find an eigenvector of L corresponding to the eigenvalue 1.4.

(2)
(d)

Hence describe the long-term behaviour of the population: state the hourly growth factor and the long-term percentage of cells that are young. Explain why the other eigenvalue does not affect this.

(3)
(e)

Initially Y0 = 10 and M0 = 5. Find the total number of cells, in thousands, after two hours.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(L=\begin{pmatrix}0.4&1.5\\0.6&0.5\end{pmatrix}\)✓ 1
Part (b)
M1 \(\det(L-\lambda I)=0\)✓ 1
A1 \((0.4-\lambda)(0.5-\lambda)-0.9=0\)✓ 1
A1 \(\lambda^2-0.9\lambda-0.7=0\)✓ 1
M1 solving the quadratic, leading to 1.4 and −0.5✓ 1
Part (c)
M1 \((L-1.4I)\mathbf{v}=\mathbf{0}\), e.g. \(-v_1+1.5v_2=0\)✓ 1
A1 any multiple of (3, 2)✓ 1
Part (d)
A1 growth factor 1.4 per hour✓ 1
A1 young : mature = 3 : 2, so 60 % are young✓ 1
R1 |−0.5| < 1, so that component decays and the dominant eigenvalue takes over✓ 1
Part (e)
M1 applying the model twice (or using \(L^2\)): (11.5, 8.5) then (17.35, 11.15)✓ 1
A1 28.5 thousand cells✓ 1

Common error: Eigenvectors are only defined up to a scalar multiple, so give a ratio (or one convenient vector) and then interpret it as proportions.

464IBMI3H-1-05r22-10Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Hard11 marks

A food scientist blends three powders: x g of X, y g of Y and z g of Z. The blend must meet three nutrient targets, which give the system
  x + 2y + z = 30
  2x + 3y + kz = 55
  x + y + 2z = 25
where k is a constant that depends on the powder Z.

(a)

Write the system as a matrix equation and show that the determinant of the coefficient matrix is k − 3.

(3)
(b)

State the value of k for which the system does not have a unique solution.

(1)
(c)

For k = 3, show that the system is consistent and find x and y in terms of z.

(4)
(d)

Masses cannot be negative. For k = 3, find the largest possible mass of Z in the blend, to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 coefficient matrix \(\begin{pmatrix}1&2&1\\2&3&k\\1&1&2\end{pmatrix}\) with \(\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}30\\55\\25\end{pmatrix}\)✓ 1
M1 expansion of the determinant, e.g. \(1(6-k)-2(4-k)+1(2-3)\)✓ 1
A1 k − 3 (AG)✓ 1
Part (b)
A1 k = 3✓ 1
Part (c)
M1 eliminate, e.g. subtract equation 3 from equation 1✓ 1
A1 y = z + 5✓ 1
A1 x = 20 − 3z✓ 1
R1 equation 2 is the sum of equations 1 and 3 in the left-hand side and also in the right-hand side (30 + 25 = 55), so the equations are consistent with infinitely many solutions✓ 1
Part (d)
M1 require x ≥ 0✓ 1
A1 \(20-3z\ge 0\), so \(z\le\frac{20}{3}\)✓ 1
A1 6.67 g (with y = z + 5 positive for all such z)✓ 1

Exam tip: A zero determinant does not by itself tell you whether there are no solutions or infinitely many: check whether the right-hand sides are consistent with the dependence between the rows.

465IBMI3H-1-05r22-14Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Hard13 marks

Two amplifier channels share a cable, and at each stage the signal strengths (x, y), in volts, mix according to
  x′ = 2x + y
  y′ = x + 2y.
The mixing at one stage is represented by the matrix A.

(a)

Write down A.

(1)
(b)

Show that the eigenvalues of A are 1 and 3.

(3)
(c)

Find an eigenvector for each eigenvalue.

(2)
(d)

Write A in the form \(PDP^{-1}\), where D is a diagonal matrix.

(2)
(e)

Hence find \(A^5\).

(3)
(f)

The input is (4, 2). Find the output after five stages, and the value of x + y for that output.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(A=\begin{pmatrix}2&1\\1&2\end{pmatrix}\)✓ 1
Part (b)
M1 \(\det(A-\lambda I)=0\)✓ 1
A1 \((2-\lambda)^2-1=0\)✓ 1
A1 \(\lambda=1\) or \(\lambda=3\)✓ 1
Part (c)
A1 for \(\lambda=3\): any multiple of (1, 1)✓ 1
A1 for \(\lambda=1\): any multiple of (1, −1)✓ 1
Part (d)
A1 \(P=\begin{pmatrix}1&1\\1&-1\end{pmatrix}\) (columns may be in either order if D matches)✓ 1
A1 \(D=\begin{pmatrix}3&0\\0&1\end{pmatrix}\) and \(P^{-1}=\frac{1}{2}\begin{pmatrix}1&1\\1&-1\end{pmatrix}\)✓ 1
Part (e)
M1 \(A^5=PD^5P^{-1}\)✓ 1
A1 \(3^5=243\)✓ 1
A1 \(\begin{pmatrix}122&121\\121&122\end{pmatrix}\)✓ 1
Part (f)
A1 output (730, 728)✓ 1
A1 x + y = 1458 V✓ 1

Exam tip: Powers of a diagonalisable matrix are found by raising only the diagonal entries to the power: \(A^n=PD^nP^{-1}\).

466IBMI3H-1-04r25-08Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Hard11 marks

A population of beetles found at an excavation site is split into juveniles (J) and adults (A). Each year, each adult produces on average a juveniles that survive to the next year; half of the juveniles become adults; and 40% of adults survive. The populations satisfy
Jn+1 = aAn
An+1 = 0.5Jn + 0.4An
with matrix L = [[0, a], [0.5, 0.4]] acting on (J, A)T.

(a)

Show that the eigenvalues λ of L satisfy λ2 − 0.4λ − 0.5a = 0.

(2)
(b)

In the long run the total population grows by 10% each year, so the dominant eigenvalue is 1.1. Find a.

(3)
(c)

Using a = 1.54 and λ = 1.1, find an eigenvector of L and hence the long-term ratio J : A.

(4)
(d)

Find the number of juveniles when there are 500 adults, once the population has settled to this long-term ratio.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 det(L − λI) = (−λ)(0.4 − λ) − 0.5a✓ 1
A1 expansion to λ2 − 0.4λ − 0.5a = 0 (AG)✓ 1
Part (b)
M1 substitute λ = 1.1✓ 1
A1 1.21 − 0.44 = 0.5a✓ 1
A1 a = 1.54✓ 1
Part (c)
M1 solve (L − 1.1I)v = 0, e.g. −1.1J + 1.54A = 0✓ 1
A1 J = 1.4A, eigenvector (1.4, 1) (or any multiple)✓ 1
A1 check with second equation: 0.5 × 1.4 + 0.4 = 1.1✓ 1
A1 J : A = 1.4 : 1 (that is, 7 : 5)✓ 1
Part (d)
M1 J = 1.4 × 500 (ECF from (c))✓ 1
A1 700✓ 1

Exam tip: Substituting the known eigenvalue into the characteristic equation is far quicker than solving for the unknown parameter by any other route.

467IBMI3H-1-04r25-14Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Hard12 marks

The flow rates (in litres per second) x, y and z in three coolant loops of a data centre satisfy the system
x + y + z = 6
2x + ky + z = 11
x + y + (k + 3)z = 10
where k is a constant. The system is written as Av = b, where A = [[1, 1, 1], [2, k, 1], [1, 1, k + 3]].

(a)

Show that det A = k2 − 4.

(3)
(b)

Hence write down the values of k for which the system does not have a unique solution.

(2)
(c)

For k = 2, show that the system has infinitely many solutions and find them, writing x and z in terms of y.

(3)
(d)

For k = −2, show that the system has no solution.

(2)
(e)

For k = 1, find the value of z to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 expansion along a row or column, e.g. 1(k(k + 3) − 1) − 1(2(k + 3) − 1) + 1(2 − k)✓ 1
A1 k2 + 3k − 1 − 2k − 5 + 2 − k✓ 1
A1 = k2 − 4 (AG)✓ 1
Part (b)
M1 det A = 0✓ 1
A1 k = 2 or k = −2✓ 1
Part (c)
M1 eliminate: second equation minus twice the first gives −z = −1, so z = 1✓ 1
A1 third equation minus the first gives 4z = 4, consistent with z = 1✓ 1
A1 x = 5 − y, z = 1 (with y free)✓ 1
Part (d)
M1 the first and third equations become x + y + z = 6 and x + y + z = 10✓ 1
R1 these contradict each other, so there is no solution✓ 1
Part (e)
M1 v = A−1b (or elimination), with det A = −3✓ 1
A1 z = 4/3 = 1.33 (with x = 5, y = −1/3)✓ 1

Exam tip: When the determinant is zero, the system has either no solutions or infinitely many; row-reduce to decide which.

468IBMI3H-1-08r27-08Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Hard10 marks

A survey robot moves on a plane with the origin as a beacon. Its positions are the complex numbers z0 = 10, zn+1 = wzn, where w = 0.9eiπ/6 and distances are in metres. Between positions it travels in a straight line.

(a)

Find z3 in exponential form and in the form a + bi.

(3)
(b)

Find the smallest value of n for which the robot is less than 1 m from the beacon.

(3)
(c)

Show that the length of the straight path from zn to zn+1 is |zn||w − 1|. Hence find the total length of the path travelled by the robot if it continues indefinitely, to 3 significant figures.

(4)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 z3 = 10w3✓ 1
A1 7.29eiπ/2 (0.93 × 10 = 7.29)✓ 1
A1 7.29i✓ 1
Part (b)
M1 |zn| = 10 × 0.9n < 1✓ 1
M1 n > ln 0.1 ÷ ln 0.9 = 21.86 (reverse inequality when dividing by ln 0.9 < 0)✓ 1
A1 n = 22✓ 1
Part (c)
M1 |zn+1 − zn| = |zn(w − 1)| = |zn||w − 1|✓ 1
A1 |w − 1| = √(0.92 + 1 − 1.8cos(π/6)) = 0.5012✓ 1
M1 geometric series with first term 10|w − 1| and common ratio 0.9, sum = first term ÷ (1 − 0.9)✓ 1
A1 50.1 m✓ 1

Exam tip: Multiplying by w = reiθ scales distances from the origin by r and turns by θ; the lengths of the successive steps then form a geometric sequence.

469IBMI3H-1-08r27-10Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Hard9 marks

A fish stock has three age classes (ages 0–1, 1–2 and 2–3 years). Let xn = (x1, x2, x3)T be the numbers in each class in year n, with xn+1 = Lxn where
L = [[0, 0.8, 4.0], [s, 0, 0], [0, 0.6, 0]] (rows listed in turn).
Here s is the proportion of the youngest class that survives to the next year. Managers want the population to be stable in the long term, which requires 1 to be an eigenvalue of L.

(a)

Use det(L − I) = 0 to find the value of s.

(4)
(b)

Using this value of s, find an eigenvector of L corresponding to the eigenvalue 1.

(3)
(c)

Hence find the percentage of the stable population that is in the oldest age class.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 det(L − I) = 0 with L − I = [[−1, 0.8, 4], [s, −1, 0], [0, 0.6, −1]]✓ 1
M1 expansion of the 3 × 3 determinant✓ 1
A1 3.2s − 1 = 0✓ 1
A1 s = 0.3125✓ 1
Part (b)
M1 (L − I)v = 0, giving x2 = 0.3125x1✓ 1
A1 x3 = 0.6x2 = 0.1875x1✓ 1
A1 v = (1, 0.3125, 0.1875)T (or any multiple); check 0.8 × 0.3125 + 4 × 0.1875 = 1✓ 1
Part (c)
M1 0.1875 ÷ (1 + 0.3125 + 0.1875)✓ 1
A1 12.5%✓ 1

Exam tip: A stable population corresponds to the eigenvalue 1; its eigenvector gives the long-term proportions, which you find by dividing each entry by the sum of all entries.

470IBMI3H-1-08r27-12Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Hard11 marks

A fish cooperative sorts a catch of three grades of fish. Let x, y and z be the masses in kg of grades 1, 2 and 3. The conditions are
x + y + z = 120 (total mass),
2x + 3y + 5z = 380 (processing time in minutes),
4x + 5y + kz = 680 (refrigeration units), where k depends on the machine setting.

(a)

Show that the system does not have a unique solution only when k = 7.

(3)
(b)

When k = 7, show that the system has no solution.

(3)
(c)

When k = 9, find x, y and z.

(3)
(d)

Grade 1 sells for €4.20 per kg, grade 2 for €3.50 per kg and grade 3 for €6.10 per kg. Find the value of the catch.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 coefficient determinant set equal to zero✓ 1
A1 det = 1(3k − 25) − 1(2k − 20) + 1(10 − 12) = k − 7✓ 1
A1 no unique solution only when k = 7 (AG)✓ 1
Part (b)
M1 combine equations: 2 × (first) + (second)✓ 1
A1 4x + 5y + 7z = 620✓ 1
R1 this contradicts the third equation (680), so there is no solution✓ 1
Part (c)
M1 matrix inverse or GDC solver (or elimination)✓ 1
A1 x = 40 and y = 50✓ 1
A1 z = 30✓ 1
Part (d)
M1 4.20 × 40 + 3.50 × 50 + 6.10 × 30 (ECF from (c))✓ 1
A1 €526.00✓ 1

Exam tip: For "no unique solution", set the determinant to zero first; then decide between no solution and infinitely many solutions by checking whether the equations are consistent.

471IBMI3H-1-04r30-12Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Hard13 marks

A comet's position relative to a star, in a plane, is modelled by a sequence of complex numbers \(z_n\) (in astronomical units, AU). The comet starts at \(z_0=10\) and each step multiplies the position by \(w=0.9e^{i\pi/8}\), so \(z_{n+1}=wz_n\).

(a)

Find \(z_1\) in the form \(a+bi\), with \(a\) and \(b\) to 3 significant figures.

(2)
(b)

Write down expressions for \(|z_n|\) and \(\arg z_n\) in terms of \(n\).

(2)
(c)

Find the smallest value of \(n\) for which \(|z_n|<2\).

(3)
(d)

Show that \(z_{16}\) is a real number, and find its value to 3 significant figures.

(3)
(e)

The comet's positions are summed: \(z_0+z_1+z_2+\dots\). State why the sum converges, and find the modulus of the sum to infinity, to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(z_1=9e^{i\pi/8}=9\cos\frac{\pi}{8}+9i\sin\frac{\pi}{8}\)✓ 1
A1 \(8.31+3.44i\)✓ 1
Part (b)
A1 \(|z_n|=10\times0.9^{n}\)✓ 1
A1 \(\arg z_n=\frac{n\pi}{8}\)✓ 1
Part (c)
M1 \(10\times0.9^{n}<2\)✓ 1
M1 \(n>\frac{\ln0.2}{\ln0.9}=15.3\ldots\) (or table of values)✓ 1
A1 \(n=16\)✓ 1
Part (d)
M1 \(\arg z_{16}=\frac{16\pi}{8}=2\pi\)✓ 1
R1 so \(z_{16}\) lies on the positive real axis (real)✓ 1
A1 1.85✓ 1
Part (e)
R1 geometric series with common ratio \(w\) and \(|w|=0.9<1\)✓ 1
M1 \(S_\infty=\frac{10}{1-0.9e^{i\pi/8}}\) and modulus of the denominator \(|0.1685-0.3444i|\) (or \(\sqrt{(1-0.9\cos\frac{\pi}{8})^2+(0.9\sin\frac{\pi}{8})^2}\))✓ 1
A1 26.1 (26.08...)✓ 1

Exam tip: A geometric sequence with a complex ratio \(w\) behaves like a spiral: \(|w|\) controls the shrinking and \(\arg w\) controls the turning. The sum to infinity needs \(|w|<1\).

472IBMI3H-1-04r30-13Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Hard12 marks

A coach plans a week of training with \(x\) sprint sessions, \(y\) strength sessions and \(z\) endurance sessions. Three load measures (in suitable units) must have the following totals:
\(x+2y+z=T_1\)
\(2x+ky+3z=T_2\)
\(x+y+2z=T_3\)
where \(k\) is a constant that depends on the strength programme chosen. The system is written \(A\mathbf{v}=\mathbf{t}\), where \(A=\begin{pmatrix}1&2&1\\2&k&3\\1&1&2\end{pmatrix}\).

(a)

Show that \(\det A=k-3\).

(3)
(b)

State the value of \(k\) for which \(A^{-1}\) does not exist.

(1)
(c)

Let \(k=3\), \(T_1=8\), \(T_2=15\) and \(T_3=7\). Show that the system has infinitely many solutions, and find \(x\) and \(y\) in terms of \(z\).

(3)
(d)

For the system in part (c), \(x\), \(y\) and \(z\) must be non-negative integers. List all possible values of \((x,y,z)\).

(2)
(e)

Now let \(k=5\), \(T_1=8\), \(T_2=19\) and \(T_3=7\). Find the number of endurance sessions, \(z\).

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 expansion of the \(3\times3\) determinant along a row or column✓ 1
A1 \(1(2k-3)-2(4-3)+1(2-k)\)✓ 1
A1 \(=k-3\) (AG)✓ 1
Part (b)
A1 \(k=3\)✓ 1
Part (c)
M1 eliminate \(x\) from equations 1 and 3 to obtain \(y-z=1\)✓ 1
R1 the sum of equations 1 and 3 equals equation 2 (\(8+7=15\)), so equation 2 gives no new information and there are infinitely many solutions✓ 1
A1 \(y=1+z\), \(x=6-3z\)✓ 1
Part (d)
M1 \(x\ge0\) gives \(z\le2\), with \(z\) a non-negative integer✓ 1
A1 \((6,1,0)\), \((3,2,1)\), \((0,3,2)\)✓ 1
Part (e)
A1 \(\det A=2\neq0\) so a unique solution exists✓ 1
M1 \(\mathbf{v}=A^{-1}\mathbf{t}\) or use of a solver✓ 1
A1 \(z=1\) (with \(x=3\), \(y=2\))✓ 1

Common error: When a determinant is zero, do not stop: check whether the equations are consistent, because a singular system has either no solution or infinitely many.

473IBMI3H-1-04r30-14Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Hard12 marks

Rail wagons are moved weekly between three depots, A, B and C. Each week, of the wagons at A, 70% stay, 20% go to B and 10% go to C. Of the wagons at B, 10% go to A, 80% stay and 10% go to C. Of the wagons at C, 20% go to A, 20% go to B and 60% stay. The numbers at the depots, \(\mathbf{x}_n=\begin{pmatrix}A_n\\B_n\\C_n\end{pmatrix}\) after \(n\) weeks, satisfy \(\mathbf{x}_{n+1}=T\mathbf{x}_n\) with \(T=\begin{pmatrix}0.7&0.1&0.2\\0.2&0.8&0.2\\0.1&0.1&0.6\end{pmatrix}\). There are 1200 wagons in total. At week 0, \(\mathbf{x}_0=\begin{pmatrix}600\\400\\200\end{pmatrix}\).

(a)

Find \(\mathbf{x}_1\).

(2)
(b)

Use technology to find the eigenvalues of \(T\).

(2)
(c)

Find an eigenvector for \(\lambda=1\), and hence find the long-term number of wagons at each depot.

(3)
(d)

Find the smallest number of weeks after which the number of wagons at depot A is within 1 wagon of its long-term value.

(3)
(e)

Depot B can hold 580 wagons. Find the number of extra wagons of capacity that B needs to cope with the long-term demand.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(T\mathbf{x}_0\)✓ 1
A1 \(\begin{pmatrix}500\\480\\220\end{pmatrix}\)✓ 1
Part (b)
M1 use of eigenvalue function or \(\det(T-\lambda I)=0\)✓ 1
A1 \(\lambda=1,\ 0.6,\ 0.5\)✓ 1
Part (c)
M1 eigenvector \(\begin{pmatrix}3\\5\\2\end{pmatrix}\) (or any multiple)✓ 1
M1 scale so entries sum to 1200 (factor \(\frac{1200}{10}\))✓ 1
A1 360 at A, 600 at B, 240 at C✓ 1
Part (d)
M1 evaluate \(T^{n}\mathbf{x}_0\) for successive \(n\) (or use the eigen-decomposition)✓ 1
A1 \(A_{10}=361.25\) (1.25 from 360)✓ 1
A1 \(A_{11}=360.75\) so \(n=11\)✓ 1
Part (e)
M1 \(600-580\) (ECF from (c))✓ 1
A1 20 wagons✓ 1

Exam tip: The eigenvalue 1 of a transition matrix gives the long-term split; the other eigenvalues (here 0.6 and 0.5) have modulus less than 1 and control how fast the system settles.

474IBMI3H-1-06r30-11Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Hard12 marks

A strain-gauge circuit on a bridge has a resistor of 25 Ω, an inductor of inductance 0.08 H and a capacitor of capacitance 2.0 × 10−5 F in series. At angular frequency ω rad s−1 its impedance is the complex number Z = 25 + i(0.08ω − 50 000/ω) ohms.

(a)

Write Z in the form a + bi when ω = 500.

(2)
(b)

Find |Z| and arg Z when ω = 500. Give the argument in radians to 3 significant figures.

(2)
(c)

The circuit is driven by a voltage V = 130 (a real number, in volts) and the current is I = V/Z. Find the magnitude of I in amperes when ω = 500, and state the argument of I to 3 significant figures.

(3)
(d)

Find the value of ω, to 3 significant figures, at which Z is a real number.

(2)
(e)

Find the positive value of ω other than 500 for which |Z| = 65.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 0.08 × 500 = 40 and 50 000/500 = 100✓ 1
A1 Z = 25 − 60i✓ 1
Part (b)
A1 |Z| = √(252 + 602) = 65✓ 1
A1 arg Z = −arctan(60/25) = −1.18✓ 1
Part (c)
M1 |I| = 130/|Z|✓ 1
A1 |I| = 2 A✓ 1
A1 arg I = +1.18 radians, since V has argument 0 and arg(1/Z) = −arg Z✓ 1
Part (d)
M1 sets 0.08ω = 50 000/ω so ω2 = 625 000✓ 1
A1 ω = 791 rad s−1✓ 1
Part (e)
M1 sets 0.08ω − 50 000/ω = ±60✓ 1
A1 0.08ω2 − 60ω − 50 000 = 0 for the +60 case✓ 1
A1 ω = 1250 rad s−1 (the case −60 only gives ω = 500 and a negative value)✓ 1

Exam tip: The impedance of a series circuit is real exactly when the imaginary part, the net reactance, is zero.

475IBMI3H-1-06r30-12Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Hard12 marks

A space probe is either in Nominal mode or in Safe mode at midnight each day. If Nominal, the probability that it is Nominal the next midnight is 0.9. If in Safe mode, the probability that it is Nominal the next midnight is 0.2. The transition matrix on the column vector (Nominal, Safe) is T = \(\begin{pmatrix}0.9&0.2\\0.1&0.8\end{pmatrix}\). The probe is Nominal at midnight on day 0.

(a)

Find the eigenvalues of T.

(3)
(b)

Find an eigenvector for each eigenvalue.

(2)
(c)

Write T = PDP−1, giving P, D and P−1.

(2)
(d)

Use T5 = PD5P−1 to find the probability that the probe is Nominal at midnight on day 5. Give your answer to 3 significant figures.

(3)
(e)

Find the long-term probability that the probe is Nominal, to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 det(T − λI) = 0✓ 1
A1 λ2 − 1.7λ + 0.7 = 0✓ 1
A1 λ = 1 and λ = 0.7✓ 1
Part (b)
A1 for λ = 1: \(\begin{pmatrix}2\\1\end{pmatrix}\) or a multiple✓ 1
A1 for λ = 0.7: \(\begin{pmatrix}1\\-1\end{pmatrix}\) or a multiple✓ 1
Part (c)
A1 P = \(\begin{pmatrix}2&1\\1&-1\end{pmatrix}\) and D = \(\begin{pmatrix}1&0\\0&0.7\end{pmatrix}\)✓ 1
A1 P−1 = (1/3)\(\begin{pmatrix}1&1\\1&-2\end{pmatrix}\)✓ 1
Part (d)
M1 uses D5 = diag(1, 0.75) = diag(1, 0.16807)✓ 1
A1 top-left entry of T5 = (2 + 0.16807)/3✓ 1
A1 0.723✓ 1
Part (e)
M1 0.7n tends to 0, so the top-left entry of Tn tends to 2/3✓ 1
A1 0.667✓ 1

Exam tip: As n grows, the powers of an eigenvalue smaller than 1 in size tend to zero, so only the eigenvalue 1 contributes to the long-term behaviour.

476IBMI3H-1-06r30-13Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Hard11 marks

The characteristic equation of a model of vibrations of a bridge is p(z) = z3 + az2 + bz + c = 0, where a, b and c are real constants. Two of the roots are z = −1 + 2i and z = 3.

(a)

Write down the third root.

(1)
(b)

Find the values of a, b and c.

(4)
(c)

Find the modulus and the argument of −1 + 2i. Give the argument in radians to 3 significant figures.

(3)
(d)

Write −1 + 2i in the form reiθ, using 3 significant figures.

(1)
(e)

Find |(−1 + 2i)4|.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 −1 − 2i, the complex conjugate✓ 1
Part (b)
M1 (z − 3)(z + 1 − 2i)(z + 1 + 2i) or the quadratic factor z2 + 2z + 5✓ 1
M1 expands (z − 3)(z2 + 2z + 5)✓ 1
A1 z3 − z2 − z − 15✓ 1
A1 a = −1, b = −1, c = −15✓ 1
Part (c)
A1 modulus √5 = 2.24✓ 1
M1 argument = π − arctan 2 (second quadrant)✓ 1
A1 2.03 radians✓ 1
Part (d)
A1 2.24e2.03i✓ 1
Part (e)
M1 |z4| = |z|4 = (√5)4✓ 1
A1 25✓ 1

Common error: Complex roots of a polynomial with real coefficients come in conjugate pairs, so the second root of the pair is the conjugate, not the negative.

477IBMI3H-1-05r32-14Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Hard13 marks

A sourdough starter contains yeast and bacteria. Let \(y_n\) and \(b_n\) be the numbers of yeast and bacteria cells, in millions, at the start of day \(n\). The populations are modelled by \(\begin{pmatrix}y_{n+1}\\b_{n+1}\end{pmatrix}=M\begin{pmatrix}y_n\\b_n\end{pmatrix}\) with \(M=\begin{pmatrix}1.2&0.3\\0.2&1.1\end{pmatrix}\). Initially \(y_0=50\) and \(b_0=30\).

(a)

Show that the eigenvalues of \(M\) are \(1.4\) and \(0.9\).

(3)
(b)

Find an eigenvector for each eigenvalue.

(3)
(c)

Express \(\begin{pmatrix}50\\30\end{pmatrix}\) as \(a\begin{pmatrix}3\\2\end{pmatrix}+c\begin{pmatrix}1\\-1\end{pmatrix}\).

(2)
(d)

Find the number of yeast cells, in millions to the nearest million, at the start of day 10.

(3)
(e)

In the long run, state the daily percentage growth of each population and the percentage of the cells that are yeast.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(\det(M-\lambda I)=0\)✓ 1
A1 \(\lambda^2-2.3\lambda+1.26=0\)✓ 1
A1 \(\lambda=1.4\), \(\lambda=0.9\)✓ 1
Part (b)
M1 solves \((M-1.4I)\mathbf{v}=\mathbf{0}\)✓ 1
A1 \(\begin{pmatrix}3\\2\end{pmatrix}\) for \(\lambda=1.4\)✓ 1
A1 \(\begin{pmatrix}1\\-1\end{pmatrix}\) for \(\lambda=0.9\) (any non-zero multiples)✓ 1
Part (c)
M1 \(3a+c=50\) and \(2a-c=30\)✓ 1
A1 \(a=16\), \(c=2\)✓ 1
Part (d)
M1 \(M^{10}\mathbf{x}_0=16(1.4)^{10}\begin{pmatrix}3\\2\end{pmatrix}+2(0.9)^{10}\begin{pmatrix}1\\-1\end{pmatrix}\) (ECF from (c))✓ 1
M1 yeast component \(48(1.4)^{10}+2(0.9)^{10}\)✓ 1
A1 \(1389\) million✓ 1
Part (e)
A1 growth \(40\%\) per day (dominant eigenvalue \(1.4\))✓ 1
A1 yeast fraction \(\frac{3}{3+2}=60\%\)✓ 1

Exam tip: The dominant eigenvalue gives the long-run growth factor and its eigenvector gives the long-run ratio of the populations.

478IBMI3H-1-08r34-06Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Hard9 marks

A theatre lighting circuit has two components in series with impedances \(Z_1 = 8e^{i\pi/6}\) ohms and \(Z_2 = 5e^{-i\pi/3}\) ohms. The supply voltage is \(V = 230\) volts (a real number), and the current is given by \(I = \frac{V}{Z}\) where \(Z = Z_1 + Z_2\).

(a)

Write \(Z_1\) and \(Z_2\) in the form \(a + bi\), giving a and b to four significant figures.

(2)
(b)

Find Z in the form \(a + bi\), and hence find its modulus and argument. Give the modulus to three significant figures and the argument to three significant figures in radians.

(4)
(c)

Write the current in the form \(Ie^{i\varphi}\), giving I and \(\varphi\) to three significant figures. Use unrounded values of \(|Z|\) and \(\arg Z\).

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(Z_1 = 6.928 + 4.000i\)✓ 1
A1 \(Z_2 = 2.500 - 4.330i\)✓ 1
Part (b)
M1 add the real parts and add the imaginary parts✓ 1
A1 \(Z = 9.428 - 0.3301i\)✓ 1
A1 \(|Z| = 9.43\) ohms✓ 1
A1 \(\arg Z = -0.0350\) radians✓ 1
Part (c)
M1 \(I = \frac{230}{|Z|}e^{-i\arg Z}\) (modulus divided, argument negated)✓ 1
A1 \(I = 24.4\) amperes (24.380…)✓ 1
A1 \(\varphi = 0.0350\) radians✓ 1

Common error: Dividing by a complex number in polar form means dividing the moduli and subtracting the arguments; do not divide the arguments.

479IBMI3H-1-08r34-08Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Hard10 marks

A shipping company moves containers between two ports, P and Q, every day. Each day, 30% of the containers at P are moved to Q, and 20% of the containers at Q are moved to P; all other containers stay where they are. Let \(\mathbf{x}_n = \begin{pmatrix}P_n\\Q_n\end{pmatrix}\) be the numbers of containers at the two ports after n days, so that \(\mathbf{x}_{n+1} = M\mathbf{x}_n\) with \(M = \begin{pmatrix}0.7 & 0.2\\0.3 & 0.8\end{pmatrix}\). Initially there are 1000 containers at P and none at Q.

(a)

Find the eigenvalues of M.

(3)
(b)

Find an eigenvector for each eigenvalue.

(2)
(c)

Write \(\mathbf{x}_0\) as a combination of the two eigenvectors, and hence show that \(P_n = 400 + 600(0.5)^n\).

(3)
(d)

Find the number of containers at port P after 8 days, to the nearest whole container.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(\det(M - \lambda I) = 0\)✓ 1
A1 \(\lambda^2 - 1.5\lambda + 0.5 = 0\)✓ 1
A1 \(\lambda = 1\) and \(\lambda = 0.5\)✓ 1
Part (b)
A1 \(\lambda = 1\): eigenvector proportional to \(\begin{pmatrix}2\\3\end{pmatrix}\)✓ 1
A1 \(\lambda = 0.5\): eigenvector proportional to \(\begin{pmatrix}1\\-1\end{pmatrix}\)✓ 1
Part (c)
M1 \(\begin{pmatrix}1000\\0\end{pmatrix} = a\begin{pmatrix}2\\3\end{pmatrix} + b\begin{pmatrix}1\\-1\end{pmatrix}\), so \(a=200\), \(b=600\)✓ 1
M1 \(\mathbf{x}_n = 200\times 1^n\begin{pmatrix}2\\3\end{pmatrix} + 600(0.5)^n\begin{pmatrix}1\\-1\end{pmatrix}\)✓ 1
A1 first component \(P_n = 400 + 600(0.5)^n\) (AG)✓ 1
Part (d)
M1 \(400 + 600(0.5)^8\)✓ 1
A1 402 containers (402.34 before rounding)✓ 1

Exam tip: Express the starting vector in the eigenvector basis first; then each eigenvalue simply gets raised to the power n.

480IBMI3H-1-07r36-09Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Hard13 marks

Two cycling clubs, A and B, recruit members and also exchange members each year. Let \(x_n\) and \(y_n\) be the numbers of members of clubs A and B at the start of year n. A model is \(\begin{pmatrix}x_{n+1}\\y_{n+1}\end{pmatrix}=\mathbf{M}\begin{pmatrix}x_n\\y_n\end{pmatrix}\) where \(\mathbf{M}=\begin{pmatrix}1.3&0.4\\0.2&1.1\end{pmatrix}\). At the start of year 0, club A has 260 members and club B has 100 members.

(a)

Find the eigenvalues of \(\mathbf{M}\).

(3)
(b)

Find an eigenvector for each eigenvalue.

(3)
(c)

Express \(\begin{pmatrix}260\\100\end{pmatrix}\) in the form \(p\begin{pmatrix}2\\1\end{pmatrix}+q\begin{pmatrix}1\\-1\end{pmatrix}\).

(2)
(d)

Hence find the number of members of club A at the start of year 5, correct to the nearest whole number.

(3)
(e)

For large n, the proportion of all members who belong to club A approaches a constant value. Find this value as a percentage, correct to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 sets \(\det(\mathbf{M}-\lambda\mathbf{I})=0\)✓ 1
A1 \(\lambda^2-2.4\lambda+1.35=0\), or equivalent✓ 1
A1 \(\lambda=1.5\) and \(\lambda=0.9\)✓ 1
Part (b)
M1 solves \((\mathbf{M}-\lambda\mathbf{I})\mathbf{v}=\mathbf{0}\) for at least one eigenvalue✓ 1
A1 \(\lambda=1.5\): \(\begin{pmatrix}2\\1\end{pmatrix}\) or a multiple✓ 1
A1 \(\lambda=0.9\): \(\begin{pmatrix}1\\-1\end{pmatrix}\) or a multiple✓ 1
Part (c)
M1 \(2p+q=260\) and \(p-q=100\)✓ 1
A1 \(p=120\), \(q=20\)✓ 1
Part (d)
M1 uses \(\mathbf{M}^5\mathbf{v}=p(1.5)^5\mathbf{v}_1+q(0.9)^5\mathbf{v}_2\)✓ 1
A1 \(x_5=2(120)(1.5)^5+20(0.9)^5\)✓ 1
A1 \(1834\)✓ 1
Part (e)
M1 the term with the larger eigenvalue dominates, so the ratio is \(2:1\)✓ 1
A1 \(\frac{2}{3}=66.7\%\)✓ 1

Exam tip: For large powers of a matrix, the eigenvalue with the greatest absolute value controls the long-term behaviour; the other term shrinks if its eigenvalue lies between -1 and 1.

481IBMI3H-1-07r36-11Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Hard12 marks

An archaeological team marks points on a spiral. The position of marker n, in metres from a central peg, is the complex number \(z_n\), where \(z_0=20\) and \(z_{n+1}=wz_n\) with \(w=0.8e^{i\pi/3}\).

(a)

Find \(z_1\) and \(z_2\) in the form \(a+bi\), with \(a\) and \(b\) correct to 3 significant figures.

(3)
(b)

Find the modulus and argument of \(z_6\). Give the modulus correct to 3 significant figures.

(3)
(c)

Explain why the infinite sum \(z_0+z_1+z_2+\dots\) converges.

(1)
(d)

Find the value of the sum in the form \(a+bi\), with \(a\) and \(b\) correct to 3 significant figures.

(3)
(e)

Find the distance of the sum from the central peg, correct to 3 significant figures. Use the unrounded value of the sum.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(z_1=wz_0=16e^{i\pi/3}\)✓ 1
A1 \(z_1=8.00+13.9i\)✓ 1
A1 \(z_2=-6.40+11.1i\)✓ 1
Part (b)
M1 \(z_6=20w^6=20(0.8)^6e^{6i\pi/3}\)✓ 1
A1 modulus \(5.24\) m✓ 1
A1 argument \(2\pi\) (or \(0\)), so \(z_6\) lies on the positive real axis✓ 1
Part (c)
R1 geometric series with common ratio \(w\) and \(|w|=0.8<1\)✓ 1
Part (d)
M1 \(S=\frac{z_0}{1-w}\)✓ 1
M1 \(1-w=0.6-0.693i\) and divides \(\frac{20}{0.6-0.693i}\)✓ 1
A1 \(14.3+16.5i\)✓ 1
Part (e)
M1 \(\left|\frac{20}{1-w}\right|=\frac{20}{\sqrt{0.84}}\), or \(\sqrt{14.29^2+16.50^2}\)✓ 1
A1 \(21.8\) m✓ 1

Exam tip: A geometric series of complex numbers converges when the modulus of the common ratio is less than 1, and the sum to infinity is the first term divided by (1 minus the ratio).

482IBMI3H-1-06r37-07Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Hard11 marks

Calculator

A logistics planner allocates numbers x, y and z of three types of pallet to vehicles. The vehicle limits give the system
x + 2y + z = p
2x + ky + 3z = q
x + y + 2z = r
where k is a constant depending on pallet size and p, q, r are the limits.

(a)

Find the determinant of the coefficient matrix in terms of k, and hence the value of k for which the system does not have a unique solution.

(3)
(b)

For k = 3, (p, q, r) = (9, 16, 7). Show that the system is consistent and describe the set of solutions.

(3)
(c)

For k = 3, (p, q, r) = (9, 17, 7). Explain why the system has no solution.

(2)
(d)

For k = 4 and (p, q, r) = (9, 19, 7), find x, y and z.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to expand the 3 × 3 determinant✓ 1
A1 det = k − 3✓ 1
A1 k = 3✓ 1
Part (b)
M1 shows row 2 = row 1 + row 3 for the coefficients, so equation 2 is redundant if 9 + 7 = 16✓ 1
R1 9 + 7 = 16 so the system is consistent✓ 1
A1 infinitely many solutions, e.g. z = t, y = t + 2, x = 5 − 3t✓ 1
Part (c)
M1 row 2 of the coefficients = row 1 + row 3, so the left sides require q = p + r = 16✓ 1
R1 17 ≠ 16, so the equations contradict and there is no solution✓ 1
Part (d)
M1 uses the inverse matrix or elimination (equation 2 − 2 × equation 1 gives z)✓ 1
A1 z = 1✓ 1
A1 x = 2 and y = 3✓ 1

Common error: A zero determinant does not always mean no solutions: the right-hand side decides between infinitely many solutions and none.

483IBMI3H-1-06r37-09Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Hard15 marks

Calculator

Containers move weekly between two hubs, P and Q. Each week, 30% of the containers at P move to Q and 20% of the containers at Q move to P; the rest stay. The numbers (pn, qn)T after n weeks satisfy vn+1 = Avn, where A = [[0.7, 0.2], [0.3, 0.8]]. Initially there are 500 containers at P and 100 at Q.

(a)

Show that the eigenvalues of A are 1 and 0.5.

(3)
(b)

Find an eigenvector for each eigenvalue.

(2)
(c)

Express v0 = (500, 100)T as a combination c1(2, 3)T + c2(1, −1)T.

(3)
(d)

Hence show that pn = 240 + 260 × 0.5n.

(2)
(e)

Find the smallest number of weeks after which the number of containers at P is within 1 of its long-term value.

(3)
(f)

Find the number of containers at P after 4 weeks, to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 det(A − λI) = 0✓ 1
A1 λ2 − 1.5λ + 0.5 = 0✓ 1
A1 (λ − 1)(λ − 0.5) = 0, so λ = 1 or 0.5✓ 1
Part (b)
A1 λ = 1: (2, 3)T✓ 1
A1 λ = 0.5: (1, −1)T (or any multiples)✓ 1
Part (c)
M1 forms 2c1 + c2 = 500 and 3c1 − c2 = 100✓ 1
A1 c1 = 120✓ 1
A1 c2 = 260✓ 1
Part (d)
M1 vn = 120 × 1n(2, 3)T + 260(0.5)n(1, −1)T✓ 1
A1 first component 240 + 260 × 0.5n✓ 1
Part (e)
M1 260 × 0.5n < 1✓ 1
A1 n = 8 gives 1.016 (not less than 1)✓ 1
A1 n = 9 (gives 0.508)✓ 1
Part (f)
M1 240 + 260 × 0.54✓ 1
A1 256✓ 1

Exam tip: Writing the starting vector as a combination of eigenvectors turns a matrix power into a sum of simple powers of the eigenvalues.

484IBMI3H-1-06r39-05Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Hard9 marks

A swimming school runs three kinds of weekly session: \(x\) beginner, \(y\) intermediate and \(z\) advanced sessions. The use of pool resources gives the system
\(x+2y+z=10\)
\(2x+ky+3z=25\)
\(x+y+2z=15\)
where \(k\) is a constant.

(a)

Show that the system does not have a unique solution when \(k=3\).

(3)
(b)

For \(k=3\), show that the system is consistent and find the general solution in terms of \(z\).

(3)
(c)

The numbers of sessions must be non-negative integers. Find the number of possible timetables, and list them as \((x,y,z)\).

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 sets the determinant of \(\begin{pmatrix}1&2&1\\2&k&3\\1&1&2\end{pmatrix}\) equal to zero✓ 1
A1 determinant \(=k-3\)✓ 1
R1 so for \(k=3\) the determinant is zero and there is no unique solution✓ 1
Part (b)
M1 eliminates \(x\) using row operations, for example \(R_2-2R_1\) and \(R_3-R_1\)✓ 1
A1 both give \(-y+z=5\), so the equations are consistent✓ 1
A1 \(y=z-5\) and \(x=20-3z\)✓ 1
Part (c)
M1 uses \(y\geq0\) and \(x\geq0\) to get \(5\leq z\leq6\frac23\)✓ 1
A1 \(z=5\) or \(z=6\)✓ 1
A1 \((5,0,5)\) and \((2,1,6)\), so 2 timetables✓ 1

Common error: Finding that the determinant is zero does not decide whether there are infinitely many solutions or none; you must test whether the equations are consistent.

485IBMI3H-1-06r39-07Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Hard9 marks

Ecologists model a population of birds near a wind farm using juveniles \(J_n\) and adults \(A_n\) in year \(n\):
\(\begin{pmatrix}J_{n+1}\\A_{n+1}\end{pmatrix}=\mathbf{L}\begin{pmatrix}J_n\\A_n\end{pmatrix}\), where \(\mathbf{L}=\begin{pmatrix}0&0.9\\0.6&0.7\end{pmatrix}\).
Each adult produces 0.9 surviving juveniles per year, 60% of juveniles survive to become adults, and 70% of adults survive each year.

(a)

Show that the eigenvalues of \(\mathbf{L}\) satisfy \(\lambda^2-0.7\lambda-0.54=0\).

(2)
(b)

Find the dominant eigenvalue to 4 significant figures, and state what it tells you about the long-term annual growth of the population, as a percentage to 3 significant figures.

(3)
(c)

Find an eigenvector for the dominant eigenvalue and hence find the long-term proportion of the population that is juvenile, to 3 significant figures.

(4)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(\det(\mathbf{L}-\lambda\mathbf{I})=(-\lambda)(0.7-\lambda)-0.9\times0.6\)✓ 1
A1 \(\lambda^2-0.7\lambda-0.54=0\) (AG)✓ 1
Part (b)
A1 \(\lambda=\dfrac{0.7+\sqrt{2.65}}{2}=1.164\)✓ 1
M1 interprets as the long-term yearly multiplier of the population✓ 1
A1 growth of \(16.4\%\) per year✓ 1
Part (c)
M1 uses \(0.9A=\lambda J\) (or \((-\lambda)J+0.9A=0\)) to relate \(J\) and \(A\)✓ 1
A1 eigenvector \(\begin{pmatrix}0.773\\1\end{pmatrix}\) (or any multiple)✓ 1
M1 proportion \(=\dfrac{0.7732\ldots}{1+0.7732\ldots}\)✓ 1
A1 \(0.436\)✓ 1

Common error: The long-term proportions come from the eigenvector direction, so divide each component by the sum of the components rather than by the eigenvalue.

486IBMI3H-1-06r39-09Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Hard10 marks

The stability of a wind turbine controller depends on the roots of the polynomial \(p(z)=z^3-9z^2+33z-65\), where \(z\) is a complex number.

(a)

Show that \(2+3i\) is a root of \(p(z)=0\).

(3)
(b)

Write down a second root of \(p(z)=0\) and justify your answer.

(2)
(c)

Find the third root.

(2)
(d)

The controller is stable only if every root has modulus less than 6. Determine whether the controller is stable.

(2)
(e)

Find the argument of the root with positive imaginary part, in radians to 3 significant figures.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 calculates \((2+3i)^2=-5+12i\)✓ 1
A1 \((2+3i)^3=-46+9i\)✓ 1
A1 substitutes to get \((-46+9i)-9(-5+12i)+33(2+3i)-65=0\)✓ 1
Part (b)
A1 \(2-3i\)✓ 1
R1 the coefficients of \(p(z)\) are real, so complex roots occur in conjugate pairs✓ 1
Part (c)
M1 uses sum of roots \(=9\) or divides by \(z^2-4z+13\)✓ 1
A1 \(z=5\)✓ 1
Part (d)
A1 \(|2\pm3i|=\sqrt{13}=3.61\) and \(|5|=5\)✓ 1
R1 all moduli are less than 6, so the controller is stable✓ 1
Part (e)
A1 \(\arctan\dfrac32=0.983\) rad✓ 1

Exam tip: When one non-real root of a real polynomial is known, its conjugate is also a root, and the quadratic factor \(z^2-(\text{sum})z+(\text{product})\) gives the remaining factor quickly.

487IBMI3H-1-04r40-07Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Hard15 marks

A population of a fish species is modelled in years. Each year the number of juveniles is three times the number of adults in the previous year. 45% of juveniles in a year become adults in the next year, and 60% of adults survive to the next year. Let Jn and An be the numbers of juveniles and adults in year n. Then
(Jn+1, An+1)T = L(Jn, An)T, where L = [[0, 3], [0.45, 0.6]].
In year 0 there are 400 juveniles and 300 adults.

(a)

Find the numbers of juveniles and adults in year 1.

(2)
(b)

Show that the eigenvalues of L satisfy λ2 − 0.6λ − 1.35 = 0, and find the eigenvalues.

(3)
(c)

Find an eigenvector of L corresponding to the eigenvalue 1.5.

(2)
(d)

An eigenvector for λ = −0.9 is (10, −3)T. Explain what the dominant eigenvalue and its eigenvector tell you about the population in the long term, including the percentage of the population that are juveniles.

(2)
(e)

Find constants p and q such that (400, 300)T = p(2, 1)T + q(10, −3)T.

(3)
(f)

Use your answers to (c) and (e) to show that the number of adults in year n is 262.5(1.5)n + 37.5(−0.9)n, and hence find the number of adults in year 10, to the nearest whole number.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 L(400, 300)T✓ 1
A1 900 juveniles and 360 adults✓ 1
Part (b)
M1 det(L − λI) = 0, i.e. (−λ)(0.6 − λ) − 3 × 0.45 = 0✓ 1
A1 λ2 − 0.6λ − 1.35 = 0 (AG)✓ 1
A1 λ = 1.5 and λ = −0.9✓ 1
Part (c)
M1 solve (L − 1.5I)v = 0, e.g. −1.5J + 3A = 0✓ 1
A1 any multiple of (2, 1)T✓ 1
Part (d)
A1 the population grows by a factor 1.5 each year, i.e. 50% per year in the long term✓ 1
A1 the ratio of juveniles to adults approaches 2 : 1, so about 66.7% (2/3) are juveniles✓ 1
Part (e)
M1 2p + 10q = 400 and p − 3q = 300✓ 1
A1 eliminating one unknown✓ 1
A1 p = 262.5 and q = −12.5✓ 1
Part (f)
M1 (Jn, An)T = p(1.5)n(2, 1)T + q(−0.9)n(10, −3)T✓ 1
A1 adults: 262.5(1.5)n + (−12.5)(−3)(−0.9)n = 262.5(1.5)n + 37.5(−0.9)n (AG)✓ 1
A1 15150✓ 1

Exam tip: Write the initial vector as a combination of eigenvectors: then each eigenvalue is applied n times as λⁿ, with no repeated matrix multiplication.

488IBMI3H-1-04r40-10Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Hard12 marks

Two sonar echoes arriving at a receiver are modelled by the complex numbers z1 = 6e0.4i and z2 = 4e1.9i, where the modulus is the amplitude and the argument is the phase in radians. The combined echo is represented by z1 + z2.

(a)

Write z1 and z2 in the form a + bi, with a and b correct to 3 significant figures.

(2)
(b)

Find the amplitude and the phase of the combined echo. Use unrounded values of the real and imaginary parts and give each answer to 3 significant figures.

(4)
(c)

The phase of z2 can be changed while its amplitude stays 4. State the greatest and least possible amplitudes of the combined echo, and give the condition on the phases for each.

(2)
(d)

The phase of z2 is now changed to 0.4 + φ, where 0 ≤ φ ≤ π. Find the value of φ, correct to 3 significant figures, for which the amplitude of the combined echo is exactly 7.

(4)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 z1 = 5.53 + 2.34i✓ 1
A1 z2 = -1.29 + 3.79i✓ 1
Part (b)
A1 z1 + z2 = 4.23 + 6.12i✓ 1
A1 amplitude 7.44✓ 1
M1 argument = arctan(imaginary ÷ real), real part positive✓ 1
A1 phase 0.966✓ 1
Part (c)
A1 greatest amplitude 10, when the two phases are equal✓ 1
A1 least amplitude 2, when the phases differ by π✓ 1
Part (d)
M1 |z1 + z2|2 = 62 + 42 + 2 × 6 × 4 cos φ (or the cosine rule)✓ 1
A1 52 + 48 cos φ = 49✓ 1
A1 cos φ = −0.0625✓ 1
A1 φ = 1.63✓ 1

Exam tip: To add complex numbers given in modulus-argument form, convert both to Cartesian form, add, then convert the sum back.

489IBMI3H-1-06r40-07Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Hard10 marks

The characteristic equation for the vertical motion of a floating marine platform is \(z^3 - 4z^2 + 14z - 20 = 0\). Each root \(\lambda\) gives a motion component proportional to \(e^{\lambda t}\), where t is in seconds.

(a)

Show that \(1 + 3i\) is a root of the equation.

(3)
(b)

Write down a second root, and find the third root.

(3)
(c)

Explain whether the platform's motion is stable.

(2)
(d)

Find the period of the oscillating component, to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \((1+3i)^2 = -8 + 6i\)✓ 1
A1 \((1+3i)^3 = -26 - 18i\)✓ 1
A1 \((-26-18i) - 4(-8+6i) + 14(1+3i) - 20 = 0\)✓ 1
Part (b)
A1 \(1 - 3i\), since the coefficients are real✓ 1
M1 sum of roots equals 4: \((1+3i) + (1-3i) + \lambda_3 = 4\)✓ 1
A1 third root \(2\)✓ 1
Part (c)
R1 the real part of \(1\pm3i\) is \(+1\), and the root 2 is also positive, so the components grow like \(e^{t}\) and \(e^{2t}\)✓ 1
R1 the amplitude increases without bound, so the platform is unstable✓ 1
Part (d)
M1 angular frequency 3 rad s−1, so period \(\frac{2\pi}{3}\)✓ 1
A1 2.09 s✓ 1

Exam tip: Complex roots of a polynomial with real coefficients always come in conjugate pairs.

490IBMI3H-1-07r40-13Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Hard13 marks

A colony of seabirds nests on a volcanic island. Let \(J_n\) and \(A_n\) be the numbers of juveniles and adults in year \(n\). The model is \(\begin{pmatrix}J_{n+1}\\A_{n+1}\end{pmatrix}=\mathbf{M}\begin{pmatrix}J_n\\A_n\end{pmatrix}\), where \(\mathbf{M}=\begin{pmatrix}0&1.8\\0.4&0.6\end{pmatrix}\). In year 0 there are 30 juveniles and 50 adults.

(a)

Show that the eigenvalues of \(\mathbf{M}\) are 1.2 and \(-0.6\).

(3)
(b)

Find an eigenvector for each eigenvalue.

(3)
(c)

Write \(\begin{pmatrix}30\\50\end{pmatrix}=a\begin{pmatrix}3\\2\end{pmatrix}+b\begin{pmatrix}3\\-1\end{pmatrix}\) and find \(a\) and \(b\).

(3)
(d)

Hence write down an expression for \(\begin{pmatrix}J_n\\A_n\end{pmatrix}\) and find the number of juveniles and adults in year 10, each to the nearest whole number.

(3)
(e)

For large \(n\), find the percentage of the colony that are juveniles.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(\det(\mathbf{M}-\lambda\mathbf{I})=(-\lambda)(0.6-\lambda)-0.72\)✓ 1
A1 \(\lambda^2-0.6\lambda-0.72=0\)✓ 1
A1 \((\lambda-1.2)(\lambda+0.6)=0\) so \(\lambda=1.2\) or \(-0.6\) (AG)✓ 1
Part (b)
M1 solve \((\mathbf{M}-\lambda\mathbf{I})\mathbf{v}=\mathbf{0}\) for one eigenvalue✓ 1
A1 \(\lambda=1.2\): \(\begin{pmatrix}3\\2\end{pmatrix}\) (or multiple)✓ 1
A1 \(\lambda=-0.6\): \(\begin{pmatrix}3\\-1\end{pmatrix}\) (or multiple)✓ 1
Part (c)
M1 form simultaneous equations \(3a+3b=30\), \(2a-b=50\)✓ 1
A1 \(a=20\)✓ 1
A1 \(b=-10\)✓ 1
Part (d)
A1 \(\begin{pmatrix}J_n\\A_n\end{pmatrix}=20(1.2)^n\begin{pmatrix}3\\2\end{pmatrix}-10(-0.6)^n\begin{pmatrix}3\\-1\end{pmatrix}\) (ECF from (c))✓ 1
A1 \(J_{10}=371\)✓ 1
A1 \(A_{10}=248\)✓ 1
Part (e)
A1 the \(1.2^n\) term dominates, so the ratio \(J:A\to3:2\), giving \(\frac{3}{5}=60\%\)✓ 1

Exam tip: For large \(n\) the term with the largest-magnitude eigenvalue dominates, so the population ratio approaches that eigenvector's ratio.

491IBMI3H-1-07r40-14Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Hard12 marks

The roll angle of a ship in rough water is modelled using the equation \(z^3+5z^2+17z+13=0\), where \(z\in\mathbb{C}\).

(a)

Show that \(z=-1\) is a root of the equation.

(2)
(b)

Given that \(-2+3\mathrm{i}\) is also a root, write down a third root and give a reason for your answer.

(2)
(c)

Show that the sum and the product of the three roots are consistent with the coefficients of the equation.

(3)
(d)

Write \(-2+3\mathrm{i}\) in the form \(re^{\mathrm{i}\theta}\), giving \(r\) and \(\theta\) (in radians) to 3 significant figures.

(3)
(e)

The oscillating part of the ship's motion is \(e^{-2t}\left(B\cos3t+C\sin3t\right)\), with \(t\) in seconds. Find the period of the oscillation, to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 substitute \(z=-1\): \((-1)^3+5(-1)^2+17(-1)+13\)✓ 1
A1 \(-1+5-17+13=0\) (AG)✓ 1
Part (b)
A1 \(-2-3\mathrm{i}\)✓ 1
R1 the coefficients are real so complex roots occur in conjugate pairs✓ 1
Part (c)
A1 sum: \(-1+(-2+3\mathrm{i})+(-2-3\mathrm{i})=-5\), which equals \(-\frac{5}{1}\)✓ 1
M1 product of conjugates \((-2+3\mathrm{i})(-2-3\mathrm{i})=4+9=13\)✓ 1
A1 \(-1\times13=-13\), which equals \(-\frac{13}{1}\)✓ 1
Part (d)
A1 \(r=\sqrt{13}=3.61\)✓ 1
M1 argument in the second quadrant, e.g. \(\pi-\arctan\frac{3}{2}\)✓ 1
A1 \(\theta=2.16\), so \(3.61e^{2.16\mathrm{i}}\)✓ 1
Part (e)
M1 period \(=\frac{2\pi}{3}\)✓ 1
A1 2.09 s✓ 1

Common error: Giving the argument as \(\arctan\frac{3}{-2}=-0.983\) puts the number in the fourth quadrant; sketch the point first.

492IBMI3H-1-04r41-10Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Hard12 marks

Calculator

After a drug is injected, the amounts (in mg) in the blood and in body tissue after n hours are given by the vector an = (bn, tn)T, where an+1 = Man and M = \(\begin{pmatrix}0.7&0.2\\0.1&0.6\end{pmatrix}\). Initially, 30 mg of the drug is in the blood and none is in the tissue, so a0 = (30, 0)T.

(a)

Find the eigenvalues of M.

(3)
(b)

Find an eigenvector for each eigenvalue.

(2)
(c)

Write a0 as a linear combination of these two eigenvectors.

(2)
(d)

Explain why, in the long term, the ratio of the amount in the blood to the amount in the tissue approaches 2 : 1.

(2)
(e)

Find the amount of the drug in the blood after 6 hours. Give your answer correct to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 det(M − λI) = 0✓ 1
M1 λ2 − 1.3λ + 0.4 = 0✓ 1
A1 λ = 0.8 and λ = 0.5✓ 1
Part (b)
A1 for λ = 0.8: any multiple of (2, 1)T✓ 1
A1 for λ = 0.5: any multiple of (1, −1)T✓ 1
Part (c)
M1 (30, 0)T = c1(2, 1)T + c2(1, −1)T, giving 2c1 + c2 = 30 and c1 − c2 = 0 (ECF from (b))✓ 1
A1 c1 = c2 = 10✓ 1
Part (d)
M1 an = 10(0.8)n(2, 1)T + 10(0.5)n(1, −1)T (ECF)✓ 1
R1 the 0.5n term decays faster than the 0.8n term, so the vector approaches a multiple of (2, 1)T✓ 1
Part (e)
M1 blood component of a6 = 10(0.8)6 × 2 + 10(0.5)6 × 1 (ECF)✓ 1
A1 5.24288 + 0.15625 (or 5.399…)✓ 1
A1 5.40 mg✓ 1

Exam tip: Write the starting vector in the eigenvector basis: then each eigen-component is simply multiplied by its eigenvalue at every step.

493IBMI3H-1-04r41-11Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Hard15 marks

Calculator

A controller for a water tank reacts to the water level in steps. Its stability is analysed using the polynomial P(z) = z3 − 1.7z2 + 1.45z − 0.425, where z is complex. The controller is stable if every root of P(z) = 0 has modulus less than 1.

(a)

Show that z = 0.5 is a root of P(z) = 0.

(2)
(b)

Hence express P(z) as a product of a linear factor and a quadratic factor.

(2)
(c)

Find the other two roots of P(z) = 0, in the form a + bi.

(2)
(d)

For the root 0.6 + 0.7i, find the modulus and the argument, each correct to 3 significant figures.

(3)
(e)

Hence state, with a reason, whether the controller is stable.

(2)
(f)

The oscillating part of the controller's response is multiplied by the modulus of the complex roots at each step. Using the exact modulus √0.85, find the least whole number of steps n after which this part is less than 0.01 of its initial size.

(3)
(g)

Each step lasts 0.25 seconds. Find the time taken for the number of steps found in part (f).

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 substitutes: 0.125 − 0.425 + 0.725 − 0.425✓ 1
A1 = 0, so z = 0.5 is a root✓ 1
Part (b)
M1 divides by (z − 0.5) or equates coefficients✓ 1
A1 (z − 0.5)(z2 − 1.2z + 0.85)✓ 1
Part (c)
M1 quadratic formula with discriminant 1.44 − 3.4 = −1.96✓ 1
A1 z = 0.6 + 0.7i and z = 0.6 − 0.7i✓ 1
Part (d)
M1 modulus = √(0.62 + 0.72) = √0.85✓ 1
A1 0.922✓ 1
A1 argument = arctan(0.7/0.6) = 0.862 radians✓ 1
Part (e)
R1 the three roots have moduli 0.5, 0.922 and 0.922, all less than 1 (ECF)✓ 1
A1 the controller is stable✓ 1
Part (f)
M1 (√0.85)n < 0.01✓ 1
M1 n > ln 0.01 / (0.5 ln 0.85) = 56.67… (or table/graph)✓ 1
A1 n = 57✓ 1
Part (g)
A1 57 × 0.25 = 14.25 seconds (ECF from (f))✓ 1

Exam tip: Check that a root really satisfies the polynomial before dividing: a quick substitution confirms the factor and prevents long division errors.

494IBMI3H-1-04r41-12Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Hard12 marks

No calculator

In a game, the position vector of a sprite is multiplied each frame by the matrix A = \(\begin{pmatrix}1&-\sqrt{3}\\\sqrt{3}&1\end{pmatrix}\).

(a)

Find the eigenvalues of A, giving them in the form a + bi.

(3)
(b)

Write the eigenvalue 1 + √3 i in the form reiθ.

(2)
(c)

Given that \(\mathbf{A}^n = 2^n\begin{pmatrix}\cos\frac{n\pi}{3}&-\sin\frac{n\pi}{3}\\\sin\frac{n\pi}{3}&\cos\frac{n\pi}{3}\end{pmatrix}\), find \(\mathbf{A}^6\).

(3)
(d)

Find the factor by which areas are multiplied after 6 frames.

(2)
(e)

Find the smallest positive integer n for which \(\mathbf{A}^n\) is a negative multiple of the identity matrix.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 det(A − λI) = (1 − λ)2 + 3 = 0✓ 1
M1 λ2 − 2λ + 4 = 0✓ 1
A1 λ = 1 ± √3 i✓ 1
Part (b)
A1 r = 2✓ 1
A1 θ = π/3✓ 1
Part (c)
M1 substitutes n = 6 so that the angle is 2π✓ 1
M1 cos 2π = 1 and sin 2π = 0 and 26 = 64✓ 1
A1 A6 = 64I (the matrix with 64 on the leading diagonal and 0 elsewhere)✓ 1
Part (d)
M1 det A = 4 so det A6 = 46 (or det(64I) = 642)✓ 1
A1 4096✓ 1
Part (e)
M1 nπ/3 = π (cos = −1, sin = 0)✓ 1
A1 n = 3 (so A3 = −8I)✓ 1

Exam tip: Matrices of the form [[a, −b],[b, a]] behave like the complex number a + bi: modulus gives the scale factor and argument gives the rotation.

495IBMI3H-1-04r43-05Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Hard12 marks

In a game, a point starts at z0 = 1 on the complex plane. At each move the point is multiplied by w = 1.2eiπ⁄8, so that after n moves it is at zn = wn. The distance travelled in a move is the distance between consecutive points.

(a)

Write zn in the form reiθ, giving r and θ in terms of n.

(2)
(b)

The point first lies on the negative real axis after n moves. Find n and the distance of the point from the origin at that moment, to 3 significant figures.

(3)
(c)

Find the smallest n for which the point is more than 10 units from the origin.

(2)
(d)

Show that the distance travelled in move k + 1, from zk to zk+1, equals 1.2k|w − 1|, and hence find the total distance travelled in the first 12 moves. Give your answer to 3 significant figures.

(5)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 r = 1.2n✓ 1
A1 θ = nπ⁄8✓ 1
Part (b)
M1 sets nπ⁄8 = π✓ 1
A1 n = 8✓ 1
A1 |z8| = 1.28 = 4.30✓ 1
Part (c)
M1 solves 1.2n > 10 (n > 12.6…)✓ 1
A1 n = 13✓ 1
Part (d)
M1 zk+1 − zk = zk(w − 1), so the distance is |zk||w − 1|✓ 1
A1 |zk| = 1.2k, giving 1.2k|w − 1|✓ 1
A1 |w − 1| = 0.472 (0.4719…) using 1.2 cos(π⁄8) − 1 and 1.2 sin(π⁄8)✓ 1
M1 total = |w − 1| × (1 + 1.2 + … + 1.211), a geometric series with 12 terms✓ 1
A1 total = 18.7 units✓ 1

Common error: |w − 1| is not |w| − 1: the distance between two points is the modulus of their difference, so you must use the real and imaginary parts of w.

496IBMI3H-1-04r43-08Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Hard12 marks

A drug is given by injection. Each hour, the amounts of the drug in the blood, b mg, and in body tissue, s mg, change according to \(\begin{pmatrix}b_{n+1}\\s_{n+1}\end{pmatrix}=\mathbf{M}\begin{pmatrix}b_n\\s_n\end{pmatrix}\), where \(\mathbf{M}=\begin{pmatrix}0.7&0.2\\0.1&0.6\end{pmatrix}\) and n is the number of hours since the injection. An initial dose of 30 mg is injected directly into the blood, so \((b_0, s_0) = (30, 0)\).

(a)

Show that the eigenvalues of M are 0.8 and 0.5.

(3)
(b)

Find an eigenvector corresponding to each eigenvalue.

(3)
(c)

Write (30, 0) as a linear combination of the two eigenvectors found in part (b), using (2, 1) and (1, −1).

(2)
(d)

Find the amount of the drug in the blood 4 hours after the injection. Give your answer to 3 significant figures.

(2)
(e)

Describe what happens to the proportions of the drug in the blood and in the tissue as n becomes large, and by what percentage the total amount of the drug falls each hour in the long term.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(\det(\mathbf{M}-\lambda\mathbf{I})=0\): \((0.7-\lambda)(0.6-\lambda)-0.02=0\)✓ 1
A1 λ2 − 1.3λ + 0.4 = 0✓ 1
A1 (λ − 0.8)(λ − 0.5) = 0, so λ = 0.8 and λ = 0.5✓ 1
Part (b)
M1 solves \((\mathbf{M}-0.8\mathbf{I})\mathbf{v}=\mathbf{0}\): \(-0.1b+0.2s=0\)✓ 1
A1 eigenvector for 0.8: any multiple of (2, 1)✓ 1
A1 eigenvector for 0.5: any multiple of (1, −1)✓ 1
Part (c)
M1 forms 2a + c = 30 and a − c = 0 (30 = 2a + c, 0 = a − c)✓ 1
A1 (30, 0) = 10(2, 1) + 10(1, −1)✓ 1
Part (d)
M1 amount in the blood = 10(0.8)4 × 2 + 10(0.5)4 × 1✓ 1
A1 8.82 mg✓ 1
Part (e)
R1 the 0.5n term becomes negligible, so the amounts approach the ratio 2 : 1 (two thirds of the drug in the blood, one third in the tissue)✓ 1
A1 the amount falls by 20% each hour (multiplied by 0.8)✓ 1

Exam tip: The eigenvalue with the larger absolute value controls the long-term behaviour: in the long run the state vector lines up with its eigenvector.

497IBMI3H-1-05r43-06Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Hard10 marks

Two connected lakes at a fish farm hold xn and yn thousand fish in year n (lake A and lake B). Growth and movement between the lakes give \(\binom{x_{n+1}}{y_{n+1}}=\mathbf{M}\binom{x_n}{y_n}\) where \(\mathbf{M}=\begin{pmatrix}0.95&0.15\\0.10&0.90\end{pmatrix}\).

(a)

Show that the eigenvalues of M are 1.05 and 0.8.

(3)
(b)

Find an eigenvector for each eigenvalue.

(3)
(c)

Initially lake A holds 40 thousand fish and lake B holds 10 thousand. Write (40, 10) as a combination of the two eigenvectors and hence find the number of fish in lake A after 10 years, to 3 significant figures.

(4)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to set det(M − λI) = 0✓ 1
A1 (0.95 − λ)(0.90 − λ) − 0.015 = 0, giving λ2 − 1.85λ + 0.84 = 0✓ 1
A1 (λ − 1.05)(λ − 0.8) = 0 so λ = 1.05 or 0.8 (AG)✓ 1
Part (b)
M1 attempt to solve (M − λI)v = 0 for one eigenvalue✓ 1
A1 λ = 1.05: (3, 2) or any multiple✓ 1
A1 λ = 0.8: (1, −1) or any multiple✓ 1
Part (c)
M1 solve 3a + b = 40 and 2a − b = 10✓ 1
A1 a = 10 and b = 10✓ 1
M1 x10 = 30(1.05)10 + 10(0.8)10✓ 1
A1 49.9 thousand fish✓ 1

Exam tip: Once the starting vector is written in eigenvector form, each eigenvector component is just multiplied by its eigenvalue raised to the power n.

498IBMI3H-1-07r44-07Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Hard9 marks

Three cables support a canopy. The tensions in the cables, in kN, are x, y and z, and the equilibrium conditions give the system
  x + 2y + z = 9
  2x + ky + 3z = 14
  x + y + 2z = 5
where k is a constant that depends on the design of the canopy.

(a)

Show that the determinant of the coefficient matrix is k − 3.

(2)
(b)

Write down the value of k for which the system does not have a unique solution.

(1)
(c)

For this value of k, show that the system is consistent and find its general solution in terms of a parameter t, where z = t.

(4)
(d)

Tensions cannot be negative. Find the range of values of t for which all three tensions are non-negative.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to expand the 3 × 3 determinant✓ 1
A1 1(2k − 3) − 2(4 − 3) + 1(2 − k) = k − 3 (or equivalent)✓ 1
Part (b)
A1 k = 3✓ 1
Part (c)
M1 attempt to eliminate (e.g. subtract the first equation from the second to obtain x + y + 2z = 5)✓ 1
R1 this is identical to the third equation, so the system is consistent with infinitely many solutions✓ 1
A1 y = 4 + t✓ 1
A1 x = 1 − 3t✓ 1
Part (d)
M1 attempt to use x ≥ 0, y ≥ 0, z ≥ 0 (ECF from (c))✓ 1
A1 0 ≤ t ≤ 1/3 (ECF from (c))✓ 1

Common error: When the determinant is zero, the system has either no solution or infinitely many; you must test consistency before stating which one it is.

499IBMI3H-1-07r44-08Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Hard14 marks

A population of short-lived squid has two stages: young and mature. Each year, every young squid produces on average 0.5 new young squid and every mature squid produces on average 1.4 new young squid. A fraction 0.45 of the young squid survive to become mature the following year. All mature squid die after spawning. The numbers of young and mature squid in year n are Jn and Mn, and the model is
  (Jn+1, Mn+1)T = L(Jn, Mn)T.

(a)

Write down the matrix L.

(2)
(b)

In one year there are 800 young and 500 mature squid. Find the numbers of young and mature squid in the following year.

(2)
(c)

Show that the eigenvalues of L satisfy λ² − 0.5λ − 0.63 = 0, and find both eigenvalues to 3 significant figures.

(4)
(d)

Explain what the larger eigenvalue tells you about the squid population in the long term, including the approximate annual percentage change.

(2)
(e)

In the long term, the ratio of young to mature squid settles to a constant. Find the proportion of the squid population that is young, as a percentage to 3 significant figures. Use the unrounded eigenvalue.

(4)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 first row 0.5, 1.4✓ 1
A1 second row 0.45, 0✓ 1
Part (b)
A1 young: 0.5 × 800 + 1.4 × 500 = 1100✓ 1
A1 mature: 0.45 × 800 = 360✓ 1
Part (c)
M1 attempt to set det(L − λI) = 0✓ 1
A1 (0.5 − λ)(−λ) − 1.4 × 0.45 = 0, leading to λ² − 0.5λ − 0.63 = 0✓ 1
A1 λ = 1.08✓ 1
A1 λ = −0.582✓ 1
Part (d)
R1 the long-term population is multiplied by about 1.082 each year (dominant eigenvalue > 1)✓ 1
A1 an increase of about 8.2% per year (accept 8%)✓ 1
Part (e)
M1 attempt to find an eigenvector for the dominant eigenvalue, e.g. (0.5 − λ)J + 1.4M = 0✓ 1
A1 J : M = 1.4 : (λ − 0.5) = 1.4 : 0.582…✓ 1
M1 proportion = 1.4/(1.4 + 0.582…)✓ 1
A1 70.6%✓ 1

Exam tip: An eigenvector gives a ratio, not an actual population; convert it to a proportion by dividing one component by the sum of the components.

500IBMI3H-1-07r44-10Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Hard11 marks

Two species of algae, P and Q, grow together in a marine sediment sample. The numbers of cells (in thousands) of each species at the start of day n are given by the vector sn = (pn, qn)T and sn+1 = Msn, where M = rows (1.2, 0.3), (0.6, 0.9). At the start of day 0, s0 = (300, 0)T.

(a)

Show that the eigenvalues of M are 1.5 and 0.6.

(3)
(b)

Find an eigenvector corresponding to each eigenvalue.

(3)
(c)

Write s0 in the form a(1, 1)T + b(1, −2)T.

(2)
(d)

Hence find the total number of cells of both species at the start of day 6, in thousands, to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to form det(M − λI) = 0✓ 1
A1 (1.2 − λ)(0.9 − λ) − 0.18 = 0, i.e. λ² − 2.1λ + 0.9 = 0✓ 1
A1 factorising or solving to give λ = 1.5 and λ = 0.6✓ 1
Part (b)
M1 attempt to solve (M − λI)v = 0 for one eigenvalue✓ 1
A1 λ = 1.5: (1, 1) (or any multiple)✓ 1
A1 λ = 0.6: (1, −2) (or any multiple)✓ 1
Part (c)
M1 attempt to solve a + b = 300 and a − 2b = 0✓ 1
A1 a = 200, b = 100✓ 1
Part (d)
M1 attempt to use s6 = 200(1.5)6(1, 1)T + 100(0.6)6(1, −2)T✓ 1
A1 p6 = 2282.79… and q6 = 2268.79… (unrounded)✓ 1
A1 total = 4551.58… ≈ 4550 thousand (4.55 × 103)✓ 1

Exam tip: Writing the starting vector as a combination of eigenvectors turns a matrix power into simple powers of the eigenvalues: Mⁿs = aλ₁ⁿv₁ + bλ₂ⁿv₂.

501IBMI3H-1-04r46-05Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Hard12 marks

A logistics company packs three types of parcel, X, Y and Z, with masses \(x\) kg, \(y\) kg and \(z\) kg respectively. Three containers hold the following parcels and have the stated total masses.
Container 1: 2 of X, 1 of Y, 1 of Z, total 20 kg.
Container 2: 1 of X, 3 of Y, 2 of Z, total 35 kg.
Container 3: 1 of X, 2 of Y, 4 of Z, total 51 kg.

(a)

Write this information as a matrix equation \(M\mathbf{v} = \mathbf{b}\).

(2)
(b)

Find \(\det M\) and state what this tells you about the system.

(2)
(c)

Hence find the mass of each type of parcel.

(3)
(d)

A fourth container holds 5 of X, 2 of Y and 3 of Z. Find its total mass.

(2)
(e)

The contents of container 3 change so that it holds 1 of X, 2 of Y and \(k\) of Z. Find the value of \(k\) for which the matrix of coefficients is singular.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(M = \begin{pmatrix}2&1&1\\1&3&2\\1&2&4\end{pmatrix}\) and \(\mathbf{v} = (x, y, z)^T\)✓ 1
A1 \(\mathbf{b} = (20, 35, 51)^T\)✓ 1
Part (b)
A1 \(\det M = 13\)✓ 1
R1 the determinant is non-zero, so \(M\) is invertible and there is a unique solution✓ 1
Part (c)
M1 \(\mathbf{v} = M^{-1}\mathbf{b}\) (or other valid method)✓ 1
A1 \(x = 3\) and \(y = 4\)✓ 1
A1 \(z = 10\)✓ 1
Part (d)
M1 \(5(3) + 2(4) + 3(10)\)✓ 1
A1 53 kg✓ 1
Part (e)
M1 expands the determinant of \(\begin{pmatrix}2&1&1\\1&3&2\\1&2&k\end{pmatrix}\)✓ 1
A1 \(\det = 5k - 7\)✓ 1
A1 \(k = \tfrac75\)✓ 1

Exam tip: A system of \(n\) equations has a unique solution exactly when the determinant of its coefficient matrix is non-zero.

502IBMI3H-1-04r46-07Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Hard14 marks

A delivery firm moves vans between two depots, P and Q, each night. Each night, 80% of the vans at P stay at P and the rest go to Q; 30% of the vans at Q go to P and the rest stay at Q. The expected numbers of vans at P and Q at the start of day \(n\) are given by \(\mathbf{v}_n\), where \(\mathbf{v}_{n+1} = T\mathbf{v}_n\) and \(T = \begin{pmatrix}0.8&0.3\\0.2&0.7\end{pmatrix}\). Initially, \(\mathbf{v}_0 = \begin{pmatrix}400\\100\end{pmatrix}\).

(a)

Find the eigenvalues of \(T\).

(3)
(b)

Find an eigenvector for each eigenvalue.

(3)
(c)

Write \(\mathbf{v}_0\) in the form \(a\begin{pmatrix}3\\2\end{pmatrix} + b\begin{pmatrix}1\\-1\end{pmatrix}\).

(3)
(d)

Hence show that \(\mathbf{v}_n = \begin{pmatrix}300 + 100(0.5)^n\\200 - 100(0.5)^n\end{pmatrix}\).

(2)
(e)

Find the expected number of vans at depot P at the start of day 6, to the nearest whole number.

(2)
(f)

The firm has 500 vans in total. Find the number of vans expected at depot P in the long run.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(\det(T - \lambda I) = 0\)✓ 1
A1 \(\lambda^2 - 1.5\lambda + 0.5 = 0\)✓ 1
A1 \(\lambda = 1\) and \(\lambda = 0.5\)✓ 1
Part (b)
M1 solves \((T - \lambda I)\mathbf{v} = \mathbf{0}\) for either eigenvalue✓ 1
A1 \(\lambda = 1\): \(\begin{pmatrix}3\\2\end{pmatrix}\) (or a multiple)✓ 1
A1 \(\lambda = 0.5\): \(\begin{pmatrix}1\\-1\end{pmatrix}\) (or a multiple)✓ 1
Part (c)
M1 forms equations \(3a + b = 400\) and \(2a - b = 100\)✓ 1
A1 \(a = 100\)✓ 1
A1 \(b = 100\)✓ 1
Part (d)
M1 applies \(T^n\) to each eigenvector term, multiplying by \(1^n\) and \(0.5^n\)✓ 1
A1 combines to give the stated result (AG)✓ 1
Part (e)
M1 substitutes \(n = 6\)✓ 1
A1 301.5625… ≈ 302 vans✓ 1
Part (f)
A1 \((0.5)^n \to 0\), so P tends to 300 vans (also 3/5 of 500)✓ 1

Exam tip: The eigenvalue 1 gives the long-run state; its eigenvector fixes the proportions, and the total fleet size fixes the actual numbers.

503IBMI3H-1-05r46-06Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Hard9 marks

Calculator

On an excavation, three marker stakes are placed at the points representing the cube roots of \(w = 27i\) in the complex plane. The origin is a fixed datum and distances are in metres.

(a)

Write \(w\) in the form \(re^{i\theta}\), where \(r > 0\) and \(-\pi < \theta \le \pi\).

(2)
(b)

Find the three cube roots of \(w\), giving each in the form \(re^{i\theta}\) with \(-\pi < \theta \le \pi\).

(4)
(c)

Find the distance between two neighbouring stakes, correct to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(r = 27\)✓ 1
A1 \(\theta = \frac{\pi}{2}\), so \(w = 27e^{i\pi/2}\)✓ 1
Part (b)
M1 \(r = 27^{1/3}\) and arguments \(\frac{\pi/2 + 2k\pi}{3}\) for \(k = 0, 1, 2\)✓ 1
A1 \(3e^{i\pi/6}\)✓ 1
A1 \(3e^{5\pi i/6}\)✓ 1
A1 \(3e^{-i\pi/2}\) (accept \(3e^{3\pi i/2}\))✓ 1
Part (c)
M1 recognise that the stakes form an equilateral triangle inscribed in a circle of radius 3, or use \(|z_1 - z_2|\) with Cartesian values✓ 1
M1 chord length e.g. \(2 \times 3\sin\frac{\pi}{3}\)✓ 1
A1 5.20 m✓ 1

Exam tip: The \(n\) roots of a complex number are equally spaced on a circle, separated by an angle of \(\frac{2\pi}{n}\).

504IBMI3H-1-05r46-13Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Hard9 marks

Calculator

A population of bats in a cave is modelled with two age groups, juveniles and adults. Each year, every adult produces on average 3 juveniles (and the adults do not themselves become juveniles), 40% of juveniles survive to become adults, and 50% of adults survive to remain adults. If \(J_n\) and \(A_n\) are the numbers of juveniles and adults in year \(n\), then \(\begin{pmatrix} J_{n+1} \\ A_{n+1} \end{pmatrix} = \mathbf{L}\begin{pmatrix} J_n \\ A_n \end{pmatrix}\) where \(\mathbf{L} = \begin{pmatrix} 0 & 3 \\ 0.4 & 0.5 \end{pmatrix}\).

(a)

Show that the eigenvalues \(\lambda\) of \(\mathbf{L}\) satisfy \(\lambda^2 - 0.5\lambda - 1.2 = 0\).

(2)
(b)

Find the larger eigenvalue, correct to 3 significant figures.

(2)
(c)

Interpret the larger eigenvalue in the context of the bat population, giving the long-term percentage change per year to the nearest percent.

(1)
(d)

Find an eigenvector corresponding to the larger eigenvalue, in the form \(\begin{pmatrix} p \\ 1 \end{pmatrix}\), correct to 3 significant figures.

(2)
(e)

Hence find the long-term percentage of the population that are juveniles, correct to 3 significant figures. Use the unrounded eigenvalue.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(\det\begin{pmatrix} -\lambda & 3 \\ 0.4 & 0.5-\lambda \end{pmatrix} = 0\)✓ 1
A1 \(-\lambda(0.5 - \lambda) - 1.2 = 0\), which rearranges to the given equation✓ 1
Part (b)
M1 solve the quadratic✓ 1
A1 1.37✓ 1
Part (c)
A1 in the long run the population grows by about 37% each year✓ 1
Part (d)
M1 solve \(-\lambda p + 3 = 0\) using the unrounded eigenvalue, so \(p = \frac{3}{\lambda}\)✓ 1
A1 \(\begin{pmatrix} 2.18 \\ 1 \end{pmatrix}\)✓ 1
Part (e)
M1 \(\frac{p}{p+1}\) with \(p = \frac{3}{\lambda}\), or equivalently \(\frac{3}{3+\lambda}\)✓ 1
A1 68.6%✓ 1

Exam tip: In a long-run population model the dominant eigenvalue gives the growth factor per step and its eigenvector gives the stable age proportions.

505IBMI3H-1-05r46-14Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Hard10 marks

Calculator

The vibration of a server rack fan mount is modelled using the equation \(z^2 - 4z + 13 = 0\), where \(z\) is a complex number.

(a)

Solve the equation, giving your answers in the form \(a + bi\).

(3)
(b)

Write the root with positive imaginary part in the form \(re^{i\theta}\), giving \(r\) and \(\theta\) to 3 significant figures.

(3)
(c)

Use de Moivre's theorem on the unrounded values of \(r\) and \(\theta\) to find \(z^6\), then state the real part of \(z^6\) as an integer.

(4)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 use of the quadratic formula or completing the square✓ 1
A1 \(z = 2 + 3i\)✓ 1
A1 \(z = 2 - 3i\)✓ 1
Part (b)
A1 \(r = \sqrt{13} = 3.61\)✓ 1
M1 \(\theta = \arctan\frac{3}{2}\) (radians)✓ 1
A1 \(\theta = 0.983\)✓ 1
Part (c)
M1 \(z^6 = r^6 e^{6i\theta}\) with \(r^6 = 13^3 = 2197\)✓ 1
M1 argument \(6\theta = 5.897\ldots\)✓ 1
A1 \(z^6 = 2197(\cos 5.897\ldots + i\sin 5.897\ldots) = 2035 - 828i\)✓ 1
A1 real part 2035✓ 1

Common error: Using the rounded argument 0.983 in de Moivre's theorem gives a slightly different real part, so keep the unrounded value until the end.

506IBMI3H-1-05r47-10Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Hard11 marks

A shipping company operates 150 vessels between three ports A, B and C. At the end of each week, the proportion of vessels moving between ports is given by the transition matrix \(\mathbf{T}=\begin{pmatrix}0.6&0.2&0.2\\0.3&0.7&0.2\\0.1&0.1&0.6\end{pmatrix}\), where the columns represent the ports A, B and C that vessels start the week at and the rows represent the ports A, B and C that they end the week at. At the start there are 60 vessels at A, 55 at B and 35 at C.

(a)

State what the entry in row 1, column 3 of \(\mathbf{T}\) represents.

(1)
(b)

Find the expected number of vessels at each port after 3 weeks. Give each answer to 1 decimal place.

(2)
(c)

Show that 1 is an eigenvalue of \(\mathbf{T}\) with eigenvector \(\begin{pmatrix}5\\7\\3\end{pmatrix}\), and hence find the long-term number of vessels at each port.

(4)
(d)

Use the unrounded values from the model to find the smallest number of weeks after which the number of vessels at port A is within 0.05 of its long-term value.

(2)
(e)

The other two eigenvalues of \(\mathbf{T}\) are 0.5 and 0.4. Explain why the numbers of vessels approach the long-term values whatever the starting distribution.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 the proportion (20%) of vessels at port C that move to port A in a week✓ 1
Part (b)
M1 attempt to evaluate \(\mathbf{T}^3\begin{pmatrix}60\\55\\35\end{pmatrix}\)✓ 1
A1 A: 50.6, B: 68.7, C: 30.6✓ 1
Part (c)
M1 \(\mathbf{T}\begin{pmatrix}5\\7\\3\end{pmatrix}=\begin{pmatrix}3+1.4+0.6\\1.5+4.9+0.6\\0.5+0.7+1.8\end{pmatrix}\)✓ 1
A1 \(=\begin{pmatrix}5\\7\\3\end{pmatrix}\) so the eigenvalue is 1✓ 1
M1 scale to the total, \(\frac{150}{5+7+3}=10\) per unit✓ 1
A1 A: 50, B: 70, C: 30✓ 1
Part (d)
M1 evaluate \(\mathbf{T}^n\begin{pmatrix}60\\55\\35\end{pmatrix}\) for successive n✓ 1
A1 \(n=6\) (port A has 50.041 vessels after 6 weeks and 50.102 after 5)✓ 1
Part (e)
R1 the other eigenvalues have modulus less than 1, so those components shrink towards zero as n increases✓ 1
R1 only the component along the eigenvector for eigenvalue 1 remains, giving the long-term distribution✓ 1

Common error: Do not round the weekly numbers between weeks: the matrix power must be applied to the exact starting values, otherwise the answer to the last part can be affected.

507IBMI3H-1-05r47-13Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Hard11 marks

The concentrations of two gases, S and C, in a crater vent are measured in ppm after n days and written as the vector \(\mathbf{v}_n=\begin{pmatrix}S_n\\C_n\end{pmatrix}\). A short-term model is \(\mathbf{v}_{n+1}=\mathbf{A}\mathbf{v}_n\), where \(\mathbf{A}=\begin{pmatrix}1.1&0.3\\0.2&0.6\end{pmatrix}\), and initially \(\mathbf{v}_0=\begin{pmatrix}10\\1\end{pmatrix}\).

(a)

Show that the eigenvalues of \(\mathbf{A}\) are 1.2 and 0.5.

(3)
(b)

Find an eigenvector for each eigenvalue.

(2)
(c)

Write \(\mathbf{v}_0\) as a linear combination of your eigenvectors.

(2)
(d)

Hence show that \(S_n=9(1.2)^n+(0.5)^n\).

(2)
(e)

Find the smallest number of days for \(S_n\) to exceed 100 ppm.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(\det(\mathbf{A}-\lambda\mathbf{I})=0\)✓ 1
A1 \(\lambda^2-1.7\lambda+0.6=0\)✓ 1
A1 \((\lambda-1.2)(\lambda-0.5)=0\) so \(\lambda=1.2\) or \(0.5\)✓ 1
Part (b)
A1 \(\begin{pmatrix}3\\1\end{pmatrix}\) for \(\lambda=1.2\)✓ 1
A1 \(\begin{pmatrix}1\\-2\end{pmatrix}\) for \(\lambda=0.5\) (or non-zero multiples)✓ 1
Part (c)
M1 solve \(a\begin{pmatrix}3\\1\end{pmatrix}+b\begin{pmatrix}1\\-2\end{pmatrix}=\begin{pmatrix}10\\1\end{pmatrix}\)✓ 1
A1 \(\mathbf{v}_0=3\begin{pmatrix}3\\1\end{pmatrix}+\begin{pmatrix}1\\-2\end{pmatrix}\)✓ 1
Part (d)
M1 use \(\mathbf{v}_n=3(1.2)^n\begin{pmatrix}3\\1\end{pmatrix}+(0.5)^n\begin{pmatrix}1\\-2\end{pmatrix}\)✓ 1
A1 top row gives \(S_n=9(1.2)^n+(0.5)^n\)✓ 1
Part (e)
M1 attempt to solve \(9(1.2)^n+(0.5)^n>100\) using a table or graph✓ 1
A1 \(n=14\) (since \(S_{13}\approx96.29\) and \(S_{14}\approx115.55\))✓ 1

Exam tip: When \(\mathbf{v}_0\) is written in terms of eigenvectors, applying \(\mathbf{A}^n\) simply multiplies each component by its eigenvalue raised to the power n.

508IBMI3H-1-05r48-07Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Hard11 marks

Calculator

In a study of a seasonal infection, each person is either healthy or infected at the start of each week. Each week, 90% of healthy people stay healthy and 10% become infected; 30% of infected people recover and 70% stay infected. The state at the start of a week is \(\binom{h}{i}\), where \(h\) and \(i\) are the numbers of healthy and infected people, and the next week's state is \(T\binom{h}{i}\), where \(T = \begin{pmatrix} 0.9 & 0.3 \\ 0.1 & 0.7 \end{pmatrix}\).

(a)

Show that the eigenvalues of \(T\) are 1 and 0.6.

(3)
(b)

Find an eigenvector of \(T\) corresponding to the eigenvalue 1.

(2)
(c)

Hence find the long-term proportion of the population that is infected.

(2)
(d)

The other eigenvalue is 0.6. Explain what this tells you about how the population approaches its long-term state.

(2)
(e)

At the start of week 1, 1000 people are all healthy. Find the expected number of infected people at the start of week 3.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 attempt to solve \(\det(T - \lambda I) = 0\)✓ 1
A1 \((0.9-\lambda)(0.7-\lambda) - 0.03 = 0\), leading to \(\lambda^2 - 1.6\lambda + 0.6 = 0\)✓ 1
A1 \((\lambda - 1)(\lambda - 0.6) = 0\), so \(\lambda = 1\) or \(\lambda = 0.6\) (AG)✓ 1
Part (b)
M1 attempt to solve \((T - I)\mathbf{v} = \mathbf{0}\), for example \(-0.1h + 0.3i = 0\)✓ 1
A1 any non-zero multiple of \(\binom{3}{1}\)✓ 1
Part (c)
M1 attempt to scale the eigenvector so that its entries sum to 1, for example \(\frac{1}{3+1}\)✓ 1
A1 \(0.25\) (25%)✓ 1
Part (d)
R1 the part of the state vector that differs from the long-term state is multiplied by 0.6 each week✓ 1
R1 so this difference shrinks geometrically towards zero, and the proportions settle to the long-term state✓ 1
Part (e)
M1 attempt to evaluate \(T^2\binom{1000}{0}\) (or two successive applications of \(T\))✓ 1
A1 \(160\) people✓ 1

Common error: An eigenvector is only defined up to a non-zero multiple; scale it so the entries sum to 1 before reading off proportions.

509IBMI3H-1-05r48-08Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Hard11 marks

Calculator

In a bioreactor producing biofuel, algae cells are either young or mature. In each day, half of the young cells stay young and half become mature, and each mature cell is replaced by 2 new young cells. Let \(y_n\) and \(m_n\) be the numbers of young and mature cells at the start of day \(n\), so that \(\binom{y_{n+1}}{m_{n+1}} = L\binom{y_n}{m_n}\).

(a)

Write down the matrix \(L\).

(1)
(b)

Find the eigenvalues of \(L\), correct to 3 significant figures.

(4)
(c)

State which eigenvalue determines the long-term behaviour of the cell population, and find the long-term percentage increase in the total number of cells per day, correct to 3 significant figures.

(2)
(d)

Find an eigenvector of \(L\) for this eigenvalue, giving it in the form \(\binom{1}{k}\) with \(k\) to 3 significant figures.

(2)
(e)

Hence find the long-term proportion of cells that are young, correct to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(L = \begin{pmatrix} 0.5 & 2 \\ 0.5 & 0 \end{pmatrix}\)✓ 1
Part (b)
M1 attempt to solve \(\det(L - \lambda I) = 0\)✓ 1
A1 \((0.5 - \lambda)(-\lambda) - 1 = 0\), that is \(\lambda^2 - 0.5\lambda - 1 = 0\)✓ 1
M1 attempt to solve the quadratic equation✓ 1
A1 \(\lambda = 1.28\) and \(\lambda = -0.781\)✓ 1
Part (c)
R1 the eigenvalue of largest modulus, \(1.28\), because its term dominates after many days✓ 1
A1 \(28.1\%\) (using the unrounded eigenvalue)✓ 1
Part (d)
M1 use of the second row, \(0.5y = \lambda m\), with \(\lambda = 1.2808\ldots\)✓ 1
A1 \(\binom{1}{0.390}\)✓ 1
Part (e)
M1 attempt to find \(\frac{1}{1 + k}\) using the unrounded value of \(k\)✓ 1
A1 \(0.719\)✓ 1

Exam tip: The long-term growth factor is the eigenvalue of largest modulus, and its eigenvector gives the long-term mix of the groups.

510IBMI3H-1-05r48-14Complex numbers in Cartesian, polar and exponential form; matrices, inverses and systems; eigenvalues and eigenvectors in context Paper 2Hard10 marks

Calculator

Two alternating voltages in a solar-array circuit are represented by the complex numbers \(V_1 = 5e^{i\pi/3}\) and \(V_2 = 4e^{-i\pi/6}\), both in volts. The total voltage is \(V = V_1 + V_2\).

(a)

Write \(V_1\) and \(V_2\) in the form \(a + bi\), giving \(a\) and \(b\) to 3 significant figures.

(2)
(b)

Find \(V\) in the form \(a + bi\), giving \(a\) and \(b\) to 3 significant figures.

(2)
(c)

Find the modulus and argument of \(V\), giving the argument in radians. Give each answer to 3 significant figures.

(3)
(d)

Show that the exact value of \(|V|\) is \(\sqrt{41}\), by considering the difference between the arguments of \(V_1\) and \(V_2\).

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 \(V_1 = 2.50 + 4.33i\)✓ 1
A1 \(V_2 = 3.46 - 2.00i\)✓ 1
Part (b)
M1 attempt to add real parts and imaginary parts✓ 1
A1 \(V = 5.96 + 2.33i\)✓ 1
Part (c)
A1 \(|V| = 6.40\) V✓ 1
M1 attempt to use \(\tan^{-1}\left(\frac{2.3301\ldots}{5.9641\ldots}\right)\)✓ 1
A1 \(\arg V = 0.372\) rad✓ 1
Part (d)
R1 the arguments differ by \(\frac{\pi}{3} - \left(-\frac{\pi}{6}\right) = \frac{\pi}{2}\), so the phasors are perpendicular✓ 1
M1 attempt to use Pythagoras with sides of length 5 and 4✓ 1
A1 \(|V| = \sqrt{25 + 16} = \sqrt{41}\) (AG)✓ 1

Exam tip: Complex numbers in exponential form are easiest to multiply, but to add them convert to Cartesian form first.