WAEC 2011 · Paper 2 · Q10

(a)
If $\mathbf A = \begin{pmatrix} -2 & 5 \\ 4 & 3 \end{pmatrix}$ and $\mathbf B = \begin{pmatrix} 3 & 1 \\ 2 & 3 \end{pmatrix}$ , find the values of $x$ and $y$ such that $\mathbf{BA} = 2\begin{pmatrix} 3 & 7 \\ -2 & x \end{pmatrix} + \begin{pmatrix} y & 4 \\ 12 & -3 \end{pmatrix}$ .
Separate values with commas, e.g. 3, −2
(b)
Two functions $f$ and $g$ are defined by $f: x \to \frac12x + 1$ and $g: x \to \frac{5x - 1}{3}$ . Find: (i) $g^{-1}$ ; (ii) $g^{-1}\circ f$ .

Worked solution (try it first)

Multiply in the order asked, $\mathbf{BA}$ : row of $\mathbf B$ times column of $\mathbf A$ .
Top row: $3(-2) + 1(4) = -2$ and $3(5) + 1(3) = 18$ .
Bottom row: $2(-2) + 3(4) = 8$ and $2(5) + 3(3) = 19$ .
So $\mathbf{BA} = \begin{pmatrix} -2 & 18 \\ 8 & 19 \end{pmatrix}$ .
The right side: $\begin{pmatrix} 6 & 14 \\ -4 & 2x \end{pmatrix} + \begin{pmatrix} y & 4 \\ 12 & -3 \end{pmatrix} = \begin{pmatrix} 6 + y & 18 \\ 8 & 2x - 3 \end{pmatrix}$ .
Match the top-left entries: $6 + y = -2$ , so $y = -8$ .
Match the bottom-right entries: $2x - 3 = 19$ , so $x = 11$ .

$f$ acts first: $g^{-1} \circ f(x) = g^{-1}\left(\frac12x + 1\right)$
$= \dfrac{3\left(\frac12x + 1\right) + 1}{5}$ .
Multiply out the top: $\dfrac{\frac32x + 4}{5}$ .
Multiply top and bottom by 2: $\dfrac{3x + 8}{10}$ .