WAEC 2012 · Paper 2 · Q12

(a)
Write down the matrices $P$ and $Q$ of the transformations $P : (x, y) \to (3x - 4y, -x)$ and $Q : (x, y) \to (y, -2x + y)$ .
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$P = \begin{pmatrix} 3 & -4 \\ -1 & 0 \end{pmatrix}$ , $Q = \begin{pmatrix} 0 & 1 \\ -2 & 1 \end{pmatrix}$
(b)
Calculate the matrix $PQ - 2Q$ .
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$\begin{pmatrix} 8 & -3 \\ 4 & -3 \end{pmatrix}$
(c)
Find the image of the point $(1, -2)$ under the linear transformation $PQ - 2Q$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

Read the coefficients of $x$ and $y$ in each new coordinate: $P = \begin{pmatrix} 3 & -4 \\ -1 & 0 \end{pmatrix}$ and $Q = \begin{pmatrix} 0 & 1 \\ -2 & 1 \end{pmatrix}$ .

Multiply rows of $P$ by columns of $Q$ : $PQ = \begin{pmatrix} 0 + 8 & 3 - 4 \\ 0 + 0 & -1 + 0 \end{pmatrix}$
$= \begin{pmatrix} 8 & -1 \\ 0 & -1 \end{pmatrix}$ .
Double $Q$ : $2Q = \begin{pmatrix} 0 & 2 \\ -4 & 2 \end{pmatrix}$ .
Subtract: $PQ - 2Q = \begin{pmatrix} 8 & -3 \\ 4 & -3 \end{pmatrix}$ .

Multiply: $\begin{pmatrix} 8 & -3 \\ 4 & -3 \end{pmatrix}\begin{pmatrix} 1 \\ -2 \end{pmatrix} = \begin{pmatrix} 8 + 6 \\ 4 + 6 \end{pmatrix}$
$= \begin{pmatrix} 14 \\ 10 \end{pmatrix}$ .
The image is the point $(14, 10)$ .