The images of (3, 2) and (-1, 4) under a linear transformation T are (-1, 4) and (7, 11) respectively. P is another transformation where \(P : (x, y) \to (x + y, x + 2y)\). (a) Find the matrices T and P of the linear transformations T and P; (b) Calculate TP. (c) Find the image of the point X(4, 3) under TP.
Explanation
Let the linear transformation T be represented by the following: \(T : \begin{pmatrix} x \\ y \end{pmatrix} \to \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} 3 & -1 \\ 2 & 4 \end{pmatrix} = \begin{pmatrix} -1 & 7 \\ 4 & 11 \end{pmatrix}\). \(3a + 2b = -1 ... (1)\) \(-a + 4b = 7 ... (2)\) \(3c + 2d = 4 ... (3)\) \(-c + 4d = 11 ... (4)\) \((2) \times 3 : -3a + 12b = 21... (5)\) \((1) + (5) : 14b = 20 \implies b = \frac{10}{7}\) Substitute for b in (1), \(3a + 2(\frac{10}{7}) = -1 \implies 3a = \frac{-27}{7}\) \(a = \frac{-9}{7}\) Solving for c and d, \((4) \times 3 : -3c + 12d = 33 ... (6)\) \((3) + (6) : 14d = 37 \implies d = \frac{37}{14}\) \(-c + 4(\frac{37}{14}) = 11 \implies -c = 11 - \frac{74}{7} = \frac{3}{7}\) \(c = \frac{-3}{7}\) \(\therefore T = \begin{pmatrix} \frac{-9}{7} & \frac{10}{7} \\ \frac{-3}{7} & \frac{37}{14} \end{pmatrix}\) \(T : \begin{pmatrix} x \\ y \end{pmatrix} \to \begin{pmatrix} \frac{-9}{7} & \frac{10}{7} \\ \frac{-3}{7} & \frac{37}{14} \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}\) From \(P : (x, y) \to (x + y, x + 2y)\) \(P : \begin{pmatrix} x \\ y \end{pmatrix} \to \begin{pmatrix} x + y \\ x + 2y \end{pmatrix}\) \(P : \begin{pmatrix} x \\ y \end{pmatrix} \to \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}\) \(\therefore P = \begin{pmatrix} 1 & 1 \\ 1 & 2 \\end{pmatrix}\) (b) \(TP = \begin{pmatrix} \frac{-9}{7} & \frac{10}{7} \\ \frac{-3}{7} & \frac{37}{14} \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}\) = \(\begin{pmatrix} \frac{-9}{7} + \frac{10}{7} & \frac{-9}{7} + \frac{20}{7} \\ \frac{-3}{7} + \frac{37}{14} & \frac{-3}{7} + \frac{74}{14} \end{pmatrix}\) = \(\begin{pmatrix} \frac{1}{7} & \frac{11}{7} \\ \frac{31}{14} & \frac{34}{7} \end{pmatrix}\) (c) The image of X(4, 3) under TP: \(TP : \begin{pmatrix} 4 \\ 3 \end{pmatrix} \to \begin{pmatrix} \frac{1}{7} & \frac{11}{7} \\ \frac{31}{17} & \frac{34}{7} \endd{pmatrix} \begin{pmatrix} 4 \\ 3 \end{pmatrix}\) = \(\begin{pmatrix} \frac{4}{7} + \frac{33}{7} \\ \frac{62}{7} + \frac{102}{7} \end{pmatrix}\) = \(\begin{pmatrix} \frac{37}{7} \\ \frac{164}{7} \end{pmatrix}\) The image of X(4, 3) under TP is \((\frac{37}{7}, \frac{164}{7})\).
