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(a) Without using mathematical tables or calculator, evaluate $\frac{\frac{3}{2} \log 27-3 \log 5 \sqrt{5}}{\log 0.6}$

(b) Two linear transformations $\mathrm{A}$ and $\mathrm{B}$ in the $O_{x y}$ plane, are defined by :
\begin{aligned} &A:(x, y)(x+2 y,-x+y) \\ &B:(x, y)(2 x+3 y, x+2 y) \end{aligned}
(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.
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(a) $\frac{\frac{3}{2} \log 27-3 \log 5 \sqrt{5}}{\log 0.6}$
\begin{aligned} &=\frac{\frac{3}{2} \log 3^{3}-3 \log 5^{\frac{3}{2}}}{\log \left(\frac{3}{5}\right)} \\ &=\frac{\frac{2}{3} \times 3 \log ^{3}-3 \times \frac{3}{2} \log 5}{\log 3-\log 5} \\ &=\frac{\frac{9}{2}(\log 3-\log 5)}{\log 3-\log 5} \\ &=\frac{9}{2}=4 \frac{1}{2} \end{aligned}
(b)(i) $A=\left(\begin{array}{cc}1 & 2 \\ -1 & 1\end{array}\right), B=\left(\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right)$
(ii) Transformation of A followed by $B$ is $B A$.
\begin{aligned} &B A=\left(\begin{array}{cc} 2 & 3 \\ 1 & 2 \end{array}\right)\left(\begin{array}{cc} 1 & 2 \\ -1 & 1 \end{array}\right) \\ &=\left(\begin{array}{ll} -1 & 7 \\ -1 & 4 \end{array}\right) \end{aligned}
Image of $(-2,2)$ is given by

$\left(\begin{array}{ll}-1 & 7 \\ -1 & 4\end{array}\right)\left(\begin{array}{c}-2 \\ 2\end{array}\right)$
$=\left(\begin{array}{l}16 \\ 10\end{array}\right)$
by Platinum (119,120 points)

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