(a) \(A(x, y) \to (2x - y , -5x + 3y)\)
Matrix \(A = \begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix}\)
(b) \(A = \begin{pmatrix} 2 & -1 \\ -5 & 2 \end{pmatrix} ; B = \begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix}\)
(i) \(A^{2} - B^{2} = \begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix} \begin{pmatrix} 2 & -1 \\ -5 & 2 \end{pmatrix} - \begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix} \begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix}\)
= \(\begin{pmatrix} 9 & -5 \\ -25 & 14 \end{pamtrix} - \begin{pmatrix} 14 & 5 \\ 25 & 9 \end{pmatrix}\)
= \(\begin{pmatrix} -5 & -10 \\ -50 & 5 \end{pmatrix}\)
= \(-5 \begin{pmatrix} 1 & 2 \\ 10 & -1 \end{pmatrix}\)
(ii) \(C = B^{2} A\)
= \(\begin{pmatrix} 14 & 5 \\ 25 & 9 \end{pmatrix} \begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix}\)
= \(\begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix}\)
(iii) \(\begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 10 \\ 18 \end{pmatrix}\)
\(3x + y = 10 ... (1)\)
\(5x + 2y = 18 ... (2)\)
Multiply (1) by 2 :
\(6x + 2y = 20 ... (3)\)
\((3) - (2) : (6x + 2y) - (5x + 2y) = (20 - 18)\)
\(x = 2\)
Put x = 2 in (1) :
\(3x + y = 10 \implies 3(2) + y = 10\)
\(6 + y = 10 \implies y = 10 - 6 = 4\)
\(M(2, 4)\).
(c) C is the inverse of matrix A and A is the inverse of matrix C.