WAEC 2022 · Paper 2 · Q11

(a)
Write down the matrix $M$ of the linear transformation defined by $T : (x, y) \to (3x + y, -2x + 3y)$ .
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$M = \begin{pmatrix} 3 & 1 \\ -2 & 3 \end{pmatrix}$
(b)
Find the: (i) inverse of $M$ ; (ii) coordinates of the point whose image under $M$ is $(9, 5)$ .
Separate values with commas, e.g. 3, −2
(c)
A line passes through the centres of the circles $x^2 + y^2 - 10x - 8y + 28 = 0$ and $3x^2 + 3y^2 + 6x - 9y + 1 = 0$ . Find the equation of the line.
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$12y - 5x - 23 = 0$

Worked solution (try it first)

$|M| = 9 - (-2) = 11$ , so $M^{-1} = \frac{1}{11}\begin{pmatrix} 3 & -1 \\ 2 & 3 \end{pmatrix}$ .

$\begin{pmatrix} x \\ y \end{pmatrix} = M^{-1}\begin{pmatrix} 9 \\ 5 \end{pmatrix}$
$= \frac{1}{11}\begin{pmatrix} 22 \\ 33 \end{pmatrix}$
$= \begin{pmatrix} 2 \\ 3 \end{pmatrix}$ : the point $(2, 3)$ .

The first circle has centre $(5, 4)$ .
Divide the second by 3: $x^2 + y^2 + 2x - 3y + \frac13 = 0$ , with centre $\left(-1, \frac32\right)$ .
Gradient: $\dfrac{4 - \frac32}{5 + 1} = \dfrac{5}{12}$ , so $y - 4 = \frac{5}{12}(x - 5)$ .
$12y - 48 = 5x - 25$ , so $12y - 5x - 23 = 0$ .