\(A = \begin{pmatrix} -2 & 5 \\ 4 & 3 \end{pmatrix} ; B = \begin{pmatrix} 3 & 1 \\ 2 & 3 \end{pmatrix}\)
\(BA = \begin{pmatrix} 3 & 1 \\ 2 & 3 \end{pmatrix} \begin{pmatrix} -2 & 5 \\ 4 & 3 \end{pmatrix}\)
= \(\begin{pmatrix} -6 + 4 & 15 + 3 \\ -4 + 12 & 10 + 9 \end{pmatrix}\)
= \(\begin{pmatrix} -2 & 18 \\ 8 & 19 \end{pmatrix}\).
\(BA = 2 \begin{pmatrix} 3 & 7 \\ -2 & x \end{pmatrix} + \begin{pmatrix} y & 4 \\ 12 & -3 \end{pmatrix}\)
= \(\begin{pmatrix} 6 + y & 18 \\ 8 & 2x - 3 \end{pmatrix}\)
\(\therefore 6 + y = -2 \implies y = -8\)
\(\therefore 2x - 3 = 19 \implies x = 11\)
(b) \(f(x) = \frac{1}{2}x + 1 ; g(x) = \frac{5x - 1}{3}\)
(i) \(g^{-1} (x) \)
Let \(y = g(x)\)
\(y = \frac{5x - 1}{3} \implies 5x = 3y + 1\)
\(x = \frac{3y + 1}{5}\)
\(\therefore g^{-1} (x) = \frac{3x + 1}{5}\)
(ii) \(g^{-1} \circ f = g^{-1} [f(x)]\)
= \(\frac{3(\frac{1}{2}x + 1) + 1}{5}\)
= \(\frac{\frac{3}{2}x + 3 + 1}{5}\)
= \((\frac{3}{2}x + 4) \times \frac{1}{5}\)
= \(\frac{3x + 8}{10}\)