If (B = begin{pmatrix} 2 & 5 \ 1 & 3 end{pmatrix}), find (B^{-1}).
-
A.
(A = begin{pmatrix} -3 & -5 \ 1 & 2 end{pmatrix}) -
B.
(A = begin{pmatrix} 3 & -5 \ 1 & 2 end{pmatrix}) -
C.
(A = begin{pmatrix} 3 & -5 \ -1 & 2 end{pmatrix}) -
D.
(A = begin{pmatrix} -3 & 5 \ 1 & -2 end{pmatrix})
Correct Answer: Option C
Explanation
(B.B^{-1} = 1), let (B^{-1} = begin{pmatrix} a & b \ c & d end{pmatrix})
(begin{pmatrix} 2 & 5 \ 1 & 3 end{pmatrix})(begin{pmatrix} a & b \ c & d end{pmatrix}) = (begin{pmatrix} 1 & 0 \ 0 & 1 end{pmatrix})
Multiplying (B times B^{-1}), we have the following equations:
(2a+5c = 1……… (1)); (a+3c = 0 ……..(2))
(2b+5d = 0 ……….(3)); (b+3d = 1……….(4))
Solving the equations simultaneously, we have
(a = 3; b = -5; c = -1; d = 2 implies B^{-1} = begin{pmatrix} 3 & -5 \ -1 & 2 end{pmatrix})