WAEC 2019 · Paper 2 · Q11

Given that $M : (x, y) \to (7y, 3x - y)$ and $N : (x, y) \to (2x - y, 5x + 3y)$ ,

(a)
write down the matrices $M$ and $N$ of the linear transformations;
Show the answer
$M = \begin{pmatrix} 0 & 7 \\ 3 & -1 \end{pmatrix}$ , $N = \begin{pmatrix} 2 & -1 \\ 5 & 3 \end{pmatrix}$
(b)
find the image of $P(2, -3)$ under the linear transformation $N$ followed by $M$ ;
Separate values with commas, e.g. 3, −2
(c)
find the coordinates of the point $Q$ whose image is $Q'(2, 4)$ under the linear transformation $N$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

Read the coefficients: $M = \begin{pmatrix} 0 & 7 \\ 3 & -1 \end{pmatrix}$ and $N = \begin{pmatrix} 2 & -1 \\ 5 & 3 \end{pmatrix}$ .

$N$ followed by $M$ is $MN = \begin{pmatrix} 0 + 35 & 0 + 21 \\ 6 - 5 & -3 - 3 \end{pmatrix}$
$= \begin{pmatrix} 35 & 21 \\ 1 & -6 \end{pmatrix}$ .
$MN\begin{pmatrix} 2 \\ -3 \end{pmatrix} = \begin{pmatrix} 70 - 63 \\ 2 + 18 \end{pmatrix}$
$= \begin{pmatrix} 7 \\ 20 \end{pmatrix}$ : the image is $(7, 20)$ .

Solve $N\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 2 \\ 4 \end{pmatrix}$ : $2x - y = 2$ and $5x + 3y = 4$ .
Multiply the first by 3 and add: $11x = 10$ , so $x = \frac{10}{11}$ and $y = 2x - 2 = -\frac{2}{11}$ .
So $Q\left(\frac{10}{11}, -\frac{2}{11}\right)$ .