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 \((\frac{37}{7}, \frac{164}{7})\).