(a) Write down the matrix A of the linear transformation (A(x, y) to (2x -y, -5x + 3y)).
(b) If (B = begin{pmatrix} 3 & 1 \ 5 & 2 end{pmatrix}), find :
(i) (A^{2} – B^{2}) ; (ii) matrix (C = B^{2} A) ; (iii) the point (M(x, y)) whose image under the linear transformation (C) is (M’ (10, 18)).
(c) What is the relationship between matrix A and matrix C?
Explanation
(a) (A(x, y) to (2x – y , -5x + 3y))
Matrix (A = begin{pmatrix} 2 & -1 \ -5 & 3 end{pmatrix})
(b) (A = begin{pmatrix} 2 & -1 \ -5 & 2 end{pmatrix} ; B = begin{pmatrix} 3 & 1 \ 5 & 2 end{pmatrix})
(i) (A^{2} – B^{2} = begin{pmatrix} 2 & -1 \ -5 & 3 end{pmatrix} begin{pmatrix} 2 & -1 \ -5 & 2 end{pmatrix} – begin{pmatrix} 3 & 1 \ 5 & 2 end{pmatrix} begin{pmatrix} 3 & 1 \ 5 & 2 end{pmatrix})
= (begin{pmatrix} 9 & -5 \ -25 & 14 end{pamtrix} – begin{pmatrix} 14 & 5 \ 25 & 9 end{pmatrix})
= (begin{pmatrix} -5 & -10 \ -50 & 5 end{pmatrix})
= (-5 begin{pmatrix} 1 & 2 \ 10 & -1 end{pmatrix})
(ii) (C = B^{2} A)
= (begin{pmatrix} 14 & 5 \ 25 & 9 end{pmatrix} begin{pmatrix} 2 & -1 \ -5 & 3 end{pmatrix})
= (begin{pmatrix} 3 & 1 \ 5 & 2 end{pmatrix})
(iii) (begin{pmatrix} 3 & 1 \ 5 & 2 end{pmatrix} begin{pmatrix} x \ y end{pmatrix} = begin{pmatrix} 10 \ 18 end{pmatrix})
(3x + y = 10 … (1))
(5x + 2y = 18 … (2))
Multiply (1) by 2 :
(6x + 2y = 20 … (3))
((3) – (2) : (6x + 2y) – (5x + 2y) = (20 – 18))
(x = 2)
Put x = 2 in (1) :
(3x + y = 10 implies 3(2) + y = 10)
(6 + y = 10 implies y = 10 – 6 = 4)
(M(2, 4)).
(c) C is the inverse of matrix A and A is the inverse of matrix C.