Explanation:
Let \(A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\).
\(T : \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} 3 & 5 \\ 4 & 6 \end{pmatrix}\)
\(2a + 3b = 3 ... (1)\)
\(4a + 5b = 5 ... (2)\)
\(2c + 3d = 4 ... (3)\)
\(4c + 5d = 6 .... (4)\)
Solving (1) & (2) :
\((1) \times 2 : 4a + 6b = 6 ... (5)\)
\((5) - (2) : b = 1\)
\(2a + 3(1) = 3 \implies 2a = 0; a = 0\)
\((3) \times 2 : 4c + 6d = 8 ... (6)\)
\((6) - (5) : d = 2\)
\(2c + 3(2) = 4 \implies 2c = -2; c = -1\)
(a) \(\therefore A = \begin{pmatrix} 0 & 1 \\ -1 & 2 \end{pmatrix}\)
(b) Let the inverse of A be A\(^{-1}\).
\(AA^{-1} = 1\)
\(\begin{pmatrix} 0 & 1 \\ -1 & 2 \end{pmatrix} \begin{pmatrix} w & x \\ y & z \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\)
\(\begin{pmatrix} y & z \\ -w + 2y & -x + 2z \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\)
\(\implies y = 1; z = 0\)
\(-w + 2y = 0 \implies -w + 2 =0\)
\(-w = -2 \implies w = 2\)
\(-x + 2z = 1 \implies -x + 2(0) = 1\)
\(-x = 1 \implies x = -1\)
\(\therefore A^{-1} = \begin{pmatrix} 2 & -1 \\ 1 & 0 \end{pmatrix}\)
(c) Let the point be (x, y).
\(A : x \to \begin{pmatrix} 0 & 1 \\ -1 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} -1 \\ 1 \end{pmatrix}\)
\(0 + y = -1 \implies y = -1\)
\(-x + 2y = 1 \implies -x + 2(-1) = 1\)
\(-x - 2 = 1 \implies -x = 3\)
\(x = -3\).
The point is (-3, -1).