(a) Without using mathematical tables or calculator, evaluate \(\frac{\frac{3}{2}\log 27 - 3\log 5\sqrt{5}}{\log 0.6}\) (b)...
(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 A and B in the \(O_{xy}\) plane, are defined by :
\(A : (x, y) (x + 2y, -x + y)\)
\(B : (x, y) (2x + 3y, x + 2y)\).
(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.
Explanation
(a) \(\frac{\frac{3}{2} \log 27 - 3 \log 5\sqrt{5}}{\log 0.6}\)
= \(\frac{\frac{3}{2} \log 3^{3} - 3 \log 5^{\frac{3}{2}}}{\log (\frac{3}{5})}\)
= \(\frac{\frac{2}{3} \times 3 \log ^3 - 3 \times \frac{3}{2} \log 5}{\log 3 - log 5}\)
= \(\frac{\frac{9}{2}(\log 3 - \log 5)}{\log 3 - \log 5}\)
= \(\frac{9}{2} = 4\frac{1}{2}\)
(b)(i) \(A = \begin{pmatrix} 1 & 2 \\ -1 & 1 \end{pmatrix}, B = \begin{pmatrix} 2 & 3 \\ 1 & 2 \end{pmatrix}\)
(ii) Transformation of A followed by B is BA.
\(BA = \begin{pmatrix} 2 & 3 \\ 1 & 2 \end{pmatrix} \begin{pmatrix} 1 & 2 \\ -1 & 1 \end{pmatrix}\)
= \(\begin{pmatrix} -1 & 7 \\ -1 & 4 \end{pmatrix}\)
Image of (-2, 2) is given by
\(\begin{pmatrix} -1 & 7 \\ -1 & 4 \end{pmatrix} \begin{pmatrix} -2 \\ 2 \end{pmatrix}\)
= \(\begin{pmatrix} 16 \\ 10 \end{pmatrix}\)