WAEC 2011 · Paper 2 · Q11

Matrices & linear transformations

The images of points $(2, 3)$ and $(4, 5)$ under a linear transformation $\mathbf A$ are $(3, 4)$ and $(5, 6)$ respectively. Find the:

(a)
matrix $\mathbf A$ ;
Separate values with commas, e.g. 3, −2
(b)
inverse of $\mathbf A$ ;
Separate values with commas, e.g. 3, −2
(c)
point whose image is $(-1, 1)$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

Let $\mathbf A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ .
Its images of $(2, 3)$ and $(4, 5)$ give four equations.
First rows: $2a + 3b = 3$ and $4a + 5b = 5$ .
Double the first and subtract: $b = 1$ , so $a = 0$ .
Second rows: $2c + 3d = 4$ and $4c + 5d = 6$ .
Likewise $d = 2$ , so $c = -1$ .
So $\mathbf A = \begin{pmatrix} 0 & 1 \\ -1 & 2 \end{pmatrix}$ .

(b)

$|\mathbf A| = 0(2) - 1(-1) = 1$ , so $\mathbf A^{-1} = \begin{pmatrix} 2 & -1 \\ 1 & 0 \end{pmatrix}$ .

(c)

$\mathbf A^{-1}\begin{pmatrix} -1 \\ 1 \end{pmatrix} = \begin{pmatrix} -2 - 1 \\ -1 + 0 \end{pmatrix}$
$= \begin{pmatrix} -3 \\ -1 \end{pmatrix}$ : the point $(-3, -1)$ .

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