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})).