WAEC 2016 · Paper 2 · Q9

(a)
Without using mathematical tables or a calculator, evaluate $\dfrac{\frac32\log27 - 3\log5\sqrt5}{\log0.6}$ .
(b)
Two linear transformations $A$ and $B$ in the $Oxy$ plane are defined by $A : (x, y) \to (x + 2y, -x + y)$ and $B : (x, y) \to (2x + 3y, x + 2y)$ . (i) Write down the matrices $A$ and $B$ . (ii) Find the image of the point $P(-2, 2)$ under the linear transformation $A$ followed by $B$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

Write the numbers as prime powers: $27 = 3^3$ and $5\sqrt5 = 5^{\frac32}$ .
So $\frac32\log 27 = \frac32 \times 3\log 3$
$= \frac92\log 3$ and $3\log 5\sqrt5 = 3 \times \frac32\log 5$
$= \frac92\log 5$ .
The top is $\frac92(\log 3 - \log 5)$ .
Write the bottom the same way: $0.6 = \frac35$ , so $\log 0.6 = \log 3 - \log 5$ .
The bracket cancels, so the value is $\dfrac92 = 4\frac12$ .

(b)(i)

Read the coefficients off each rule: $A = \begin{pmatrix} 1 & 2 \\ -1 & 1 \end{pmatrix}$ and $B = \begin{pmatrix} 2 & 3 \\ 1 & 2 \end{pmatrix}$ .

(ii)

$A$ followed by $B$ is the matrix $BA$ (the first transformation goes on the right): $BA = \begin{pmatrix} 2 - 3 & 4 + 3 \\ 1 - 2 & 2 + 2 \end{pmatrix}$
$= \begin{pmatrix} -1 & 7 \\ -1 & 4 \end{pmatrix}$ .
Apply it to $P(-2, 2)$ : $\begin{pmatrix} -1 & 7 \\ -1 & 4 \end{pmatrix}\begin{pmatrix} -2 \\ 2 \end{pmatrix} = \begin{pmatrix} 2 + 14 \\ 2 + 8 \end{pmatrix}$
$= \begin{pmatrix} 16 \\ 10 \end{pmatrix}$ .
The image is $(16, 10)$ .

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