A2 June 2025 Paper 2 Q13
13 The matrix \(\mathbf{M}\) is defined by \(\mathbf{M} = \begin{bmatrix} 1 & -\sqrt{3} \\ \sqrt{3} & 1 \end{bmatrix}\)
(a) The matrix \(\mathbf{M}\) represents an anticlockwise rotation about the origin through an angle \(\theta\), where \(0 \leqslant \theta \leqslant 2\pi\), followed by an enlargement, scale factor \(r\), with centre at the origin where \(r\) is a positive integer.
Find the value of \(r\) and the value of \(\theta\) [3 marks]
(b) It is given that \(\begin{bmatrix} u \\ v \end{bmatrix} = \mathbf{M}\begin{bmatrix} x \\ y \end{bmatrix}\)
Using the value of \(r\) and the value of \(\theta\) which you obtained in part (a), verify that
\[r\mathrm{e}^{\mathrm{i}\theta}(x + \mathrm{i}y) = u + \mathrm{i}v\] [3 marks](c) Hence, find the value of \(x\) and the value of \(y\) such that\[\mathbf{M}^8\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 2 \end{bmatrix}\]
Give your answers in an exact form. [4 marks]
| Scheme | Marks | AO |
|---|---|---|
| Forms a matrix equation Or Forms equations in \(r\) and \(\theta\) PI by \(r = 2\) or \(\theta = \dfrac{\pi}{3}\) | M1 | 3.1a |
| Deduces \(r = 2\) or \(\theta = \dfrac{\pi}{3}\) | A1 | 2.2a |
| Obtains \(r = 2\) and \(\theta = \dfrac{\pi}{3}\) | A1 | 1.1b |
| (3) |
Typical solution
\[\begin{bmatrix} r & 0 \\ 0 & r \end{bmatrix}\begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix} = \begin{bmatrix} 1 & -\sqrt{3} \\ \sqrt{3} & 1 \end{bmatrix}\]\[r\cos\theta = 1\]\[r\sin\theta = \sqrt{3}\]\[r = 2\]\[\theta = \frac{\pi}{3}\]| Scheme | Marks | AO |
|---|---|---|
| Uses matrix multiplication to deduce \(u = x - \sqrt{3}y\) and \(v = \sqrt{3}x + y\) | B1 | 2.2a |
| Uses their values of \(r\) and \(\theta\) to obtain an expression for \(r\mathrm{e}^{\mathrm{i}\theta}(x + \mathrm{i}y)\) with no trigonometric or exponential terms. | M1 | 1.1a |
| Uses correct reasoning to obtain \(x - \sqrt{3}y + \mathrm{i}(\sqrt{3}x + y)\) and to conclude that \(r\mathrm{e}^{\mathrm{i}\theta}(x + \mathrm{i}y) = u + \mathrm{i}v\) or \(2\mathrm{e}^{\frac{\mathrm{i}\pi}{3}}(x + \mathrm{i}y) = u + \mathrm{i}v\) | R1 | 2.1 |
| (3) |
Typical solution
\[\begin{bmatrix} u \\ v \end{bmatrix} = \begin{bmatrix} 1 & -\sqrt{3} \\ \sqrt{3} & 1 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} x - \sqrt{3}y \\ \sqrt{3}x + y \end{bmatrix}\]\[u = x - \sqrt{3}y,\ v = \sqrt{3}x + y\]\[\begin{aligned} r\mathrm{e}^{\mathrm{i}\theta}(x + \mathrm{i}y) &= 2\mathrm{e}^{\frac{\mathrm{i}\pi}{3}}(x + \mathrm{i}y) \\ &= 2\left(\cos\frac{\pi}{3} + \mathrm{i}\sin\frac{\pi}{3}\right)(x + \mathrm{i}y) \\ &= (1 + \mathrm{i}\sqrt{3})(x + \mathrm{i}y) \\ &= x - \sqrt{3}y + \mathrm{i}(\sqrt{3}x + y) \\ &= u + \mathrm{i}v \end{aligned}\]| Scheme | Marks | AO |
|---|---|---|
| Deduces \(\left(2\mathrm{e}^{\frac{\mathrm{i}\pi}{3}}\right)^8(x + \mathrm{i}y) = 2\mathrm{i}\) OE | B1 | 2.2a |
| Forms an equation including \(2^8\) or \(2^{-8}\), and \(\mathrm{e}^{\pm\frac{8\pi\mathrm{i}}{3}}\) or \(\mathrm{e}^{\pm\frac{2\pi\mathrm{i}}{3}}\) OE Or Obtains the modulus and argument of \(x + \mathrm{i}y\) | M1 | 3.1a |
| Obtains values of \(x\) and \(y\), including no trigonometric or exponential terms, from an equation including \(\mathrm{e}^{\pm\frac{8\pi\mathrm{i}}{3}}\) or \(\mathrm{e}^{\pm\frac{2\pi\mathrm{i}}{3}}\) | M1 | 1.1a |
| Obtains \(x = \dfrac{\sqrt{3}}{256}\), \(y = \dfrac{-1}{256}\) | A1 | 1.1b |
| (4) | ||
| (10 marks) |