WAEC 2018 · Paper 2 · Q11

(a)
Given the matrix $A = \begin{pmatrix} 3 & 8 \\ 5 & -2 \end{pmatrix}$ , find its inverse.
Show the answer
$A^{-1} = \begin{pmatrix} \frac{1}{23} & \frac{4}{23} \\ \frac{5}{46} & -\frac{3}{46} \end{pmatrix}$
(b)
Solve the simultaneous equations $\log(x - 2) + \log 2 = 2\log y$ and $\log(x - 3y + 3) = 0$ .
Show the answer
$x = 4, y = 2$ or $x = 10, y = 4$

Worked solution (try it first)

(a)

The determinant is $|A| = 3(-2) - 8(5) = -6 - 40 = -46$ .
Swap the leading diagonal and change the signs of the other two entries: $\begin{pmatrix} -2 & -8 \\ -5 & 3 \end{pmatrix}$ .
Divide by the determinant: $A^{-1} = -\dfrac{1}{46}\begin{pmatrix} -2 & -8 \\ -5 & 3 \end{pmatrix}$
$= \begin{pmatrix} \frac{1}{23} & \frac{4}{23} \\ \frac{5}{46} & -\frac{3}{46} \end{pmatrix}$ .

(b)

First equation: $\log[2(x - 2)] = \log y^2$ , so $2(x - 2) = y^2$ .
Second equation: $\log(x - 3y + 3) = 0$ means $x - 3y + 3 = 10^0 = 1$ , so $x = 3y - 2$ .
Substitute: $2(3y - 4) = y^2$ , so $y^2 - 6y + 8 = 0$ and $(y - 2)(y - 4) = 0$ .
$y = 2$ gives $x = 4$ , and $y = 4$ gives $x = 10$ .
Both keep every log positive, so $x = 4, y = 2$ or $x = 10, y = 4$ .

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