The images of points (2, -3) and (4, 5) under a linear transformation A are (3, 4) and (5, 6) respectively. Find the : (a) matrix A ; (b) inverse of A ; (c) point whose image is (-1, 1).
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).