WAEC 2008 · Paper 2 · Q12

(a)
The point $P(3, -5)$ is rotated through an angle of $60^\circ$ anticlockwise about the origin. (i) Obtain the matrix for the rotation. (ii) Find the image $P_1$ of the point $P$ under the rotation.
Separate values with commas, e.g. 3, −2
(b)
A linear transformation is given by $N : \begin{pmatrix} x \\ y \end{pmatrix} \to \begin{pmatrix} 2x + 3y \\ 3x - y \end{pmatrix}$ . (i) Write down the matrix $N$ of the transformation. (ii) If $N^2 + aN + bI = 0$ , where $a, b \in \mathbb{R}$ , $I$ is the $2 \times 2$ unit matrix and $0$ is the $2 \times 2$ null matrix, find the values of $a$ and $b$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

An anticlockwise rotation through $\theta$ about the origin has matrix $\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$ .
With $\theta = 60^\circ$ : $\cos 60^\circ = \frac12$ and $\sin 60^\circ = \frac{\sqrt3}{2}$ , giving $\begin{pmatrix} \frac12 & -\frac{\sqrt3}{2} \\ \frac{\sqrt3}{2} & \frac12 \end{pmatrix}$ .

Multiply the matrix by $\begin{pmatrix} 3 \\ -5 \end{pmatrix}$ .
Top row: $\frac32 + \frac{5\sqrt3}{2} = \frac{3 + 5\sqrt3}{2}$ .
Bottom row: $\frac{3\sqrt3}{2} - \frac52 = \frac{3\sqrt3 - 5}{2}$ .
So $P_1\left(\frac{3 + 5\sqrt3}{2}, \frac{3\sqrt3 - 5}{2}\right)$ , about $(5.83, 0.10)$ .

Read the coefficients of $x$ and $y$ row by row: $N = \begin{pmatrix} 2 & 3 \\ 3 & -1 \end{pmatrix}$ .

$N^2 = N \times N$ .
Top row: $(4 + 9,\ 6 - 3) = (13, 3)$ .
Bottom row: $(6 - 3,\ 9 + 1) = (3, 10)$ .
So $\begin{pmatrix} 13 & 3 \\ 3 & 10 \end{pmatrix} + a\begin{pmatrix} 2 & 3 \\ 3 & -1 \end{pmatrix} + b\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$ .
Top-right entries: $3 + 3a = 0$ , so $a = -1$ .
Top-left entries: $13 + 2a + b = 0$ , so $13 - 2 + b = 0$ and $b = -11$ .
Check with the bottom-right entries: $10 - a + b = 10 + 1 - 11 = 0$ ✓.
So $a = -1$ and $b = -11$ .