(a) If (A = begin{pmatrix} -2 & 5 \ 4 & 3 end{pmatrix}) and (B = begin{pmatrix} 3 & 1 \ 2 & 3 end{pmatrix}), find the values of x and y such that (BA = 2begin{pmatrix} 3 & 7 \ -2 & x end{pmatrix} + begin{pmatrix} y & 4 \ 12 & -3 end{pmatrix}).
(b) Two functions, f and g are defined by (f : x to frac{1}{2}x + 1) and (g : x to frac{5x – 1}{3}). Find :
(i) (g^{-1}) ; (ii) (g^{-1} circ f).
Explanation
(A = begin{pmatrix} -2 & 5 \ 4 & 3 end{pmatrix} ; B = begin{pmatrix} 3 & 1 \ 2 & 3 end{pmatrix})
(BA = begin{pmatrix} 3 & 1 \ 2 & 3 end{pmatrix} begin{pmatrix} -2 & 5 \ 4 & 3 end{pmatrix})
= (begin{pmatrix} -6 + 4 & 15 + 3 \ -4 + 12 & 10 + 9 end{pmatrix})
= (begin{pmatrix} -2 & 18 \ 8 & 19 end{pmatrix}).
(BA = 2 begin{pmatrix} 3 & 7 \ -2 & x end{pmatrix} + begin{pmatrix} y & 4 \ 12 & -3 end{pmatrix})
= (begin{pmatrix} 6 + y & 18 \ 8 & 2x – 3 end{pmatrix})
(therefore 6 + y = -2 implies y = -8)
(therefore 2x – 3 = 19 implies x = 11)
(b) (f(x) = frac{1}{2}x + 1 ; g(x) = frac{5x – 1}{3})
(i) (g^{-1} (x) )
Let (y = g(x))
(y = frac{5x – 1}{3} implies 5x = 3y + 1)
(x = frac{3y + 1}{5})
(therefore g^{-1} (x) = frac{3x + 1}{5})
(ii) (g^{-1} circ f = g^{-1} [f(x)])
= (frac{3(frac{1}{2}x + 1) + 1}{5})
= (frac{frac{3}{2}x + 3 + 1}{5})
= ((frac{3}{2}x + 4) times frac{1}{5})
= (frac{3x + 8}{10})