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.

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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 + 2 y)

.
(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{matrix} x \\ y \end{matrix}) \to (\begin{matrix} a & b \\ c & d \end{matrix}) (\begin{matrix} 3 & - 1 \\ 2 & 4 \end{matrix}) = (\begin{matrix} - 1 & 7 \\ 4 & 11 \end{matrix})

3 a + 2 b = - 1 . . . (1)

- a + 4 b = 7 . . . (2)

3 c + 2 d = 4 . . . (3)

- c + 4 d = 11 . . . (4)

(2) \times 3 : - 3 a + 12 b = 21. . . (5)

(1) + (5) : 14 b = 20 ⟹ b = \frac{10}{7}

Substitute for b in (1),

3 a + 2 (\frac{10}{7}) = - 1 ⟹ 3 a = \frac{- 27}{7}

a = \frac{- 9}{7}

Solving for c and d,

(4) \times 3 : - 3 c + 12 d = 33 . . . (6)

(3) + (6) : 14 d = 37 ⟹ d = \frac{37}{14}

- c + 4 (\frac{37}{14}) = 11 ⟹ - c = 11 - \frac{74}{7} = \frac{3}{7}

c = \frac{- 3}{7}

∴ T = (\begin{matrix} \frac{- 9}{7} & \frac{10}{7} \\ \frac{- 3}{7} & \frac{37}{14} \end{matrix})

T : (\begin{matrix} x \\ y \end{matrix}) \to (\begin{matrix} \frac{- 9}{7} & \frac{10}{7} \\ \frac{- 3}{7} & \frac{37}{14} \end{matrix}) (\begin{matrix} x \\ y \end{matrix})

From

P : (x, y) \to (x + y, x + 2 y)

P : (\begin{matrix} x \\ y \end{matrix}) \to (\begin{matrix} x + y \\ x + 2 y \end{matrix})

P : (\begin{matrix} x \\ y \end{matrix}) \to (\begin{matrix} 1 & 1 \\ 1 & 2 \end{matrix}) (\begin{matrix} x \\ y \end{matrix})

Missing \end{pmatrix}

(b)

T P = (\begin{matrix} \frac{- 9}{7} & \frac{10}{7} \\ \frac{- 3}{7} & \frac{37}{14} \end{matrix}) (\begin{matrix} 1 & 1 \\ 1 & 2 \end{matrix})

(\begin{matrix} \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{matrix})

(\begin{matrix} \frac{1}{7} & \frac{11}{7} \\ \frac{31}{14} & \frac{34}{7} \end{matrix})

Missing \end{pmatrix}

(\begin{matrix} \frac{4}{7} + \frac{33}{7} \\ \frac{62}{7} + \frac{102}{7} \end{matrix})

(\begin{matrix} \frac{37}{7} \\ \frac{164}{7} \end{matrix})

The image of X(4, 3) under TP is

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

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