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...

Answer Details

(a) Let \(T = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\). Using the images \((3,2)\to(-1,4)\) and \((-1,4)\to(7,11)\):

From the \(x\)-outputs: \(3a + 2b = -1\) and \(-a + 4b = 7\). Solving gives \(b = \tfrac{10}{7},\ a = -\tfrac{9}{7}\).

From the \(y\)-outputs: \(3c + 2d = 4\) and \(-c + 4d = 11\). Solving gives \(d = \tfrac{37}{14},\ c = -\tfrac{3}{7}\).

\[T = \frac{1}{14}\begin{pmatrix} -18 & 20 \\ -6 & 37 \end{pmatrix}.\]

For \(P:(x,y)\to(x+y,\,x+2y)\), \(P = \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}\).

(b) \(\displaystyle TP = \frac{1}{14}\begin{pmatrix} -18 & 20 \\ -6 & 37 \end{pmatrix}\begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix} = \frac{1}{14}\begin{pmatrix} 2 & 22 \\ 31 & 68 \end{pmatrix}.\)

(c) Image of \(X(4,3)\) under \(TP\):

\[\frac{1}{14}\begin{pmatrix} 2 & 22 \\ 31 & 68 \end{pmatrix}\begin{pmatrix} 4 \\ 3 \end{pmatrix} = \frac{1}{14}\begin{pmatrix} 74 \\ 328 \end{pmatrix} = \left(\frac{37}{7},\ \frac{164}{7}\right).\]