Q.39. If the matrix A=\begin{bmatrix} 1 & 2 \\ -5 & 1 \\ \end{bmatrix} and A-1= xA +yI, when I is a uit matrix of order 2, then the value of 2x+3y is
A. \frac{8}{11}
B. \frac{4}{11}
C. \frac{-8}{11}
D. \frac{-4}{11}
Answer: B. \frac{4}{11}
Explanation :-
A = \begin{bmatrix} 1 & 2 \\ -5 & 1 \end{bmatrix} \therefore |\mathrm{A}| = 11 A^{-1} = \frac{1}{|\mathrm{A}|} \begin{bmatrix} 1 & -2 \\ 5 & 1 \end{bmatrix} = \frac{1}{11} \begin{bmatrix} 1 & -2 \\ 5 & 1 \end{bmatrix} \mathrm{A}^{-1} = x \, \mathrm{A} + u \, \mathrm{I \, we \, get} \begin{bmatrix} \frac{1}{11} & \frac{-2}{11} \\ \frac{5}{11} & \frac{1}{11} \end{bmatrix} = \begin{bmatrix} x & 2x \\ -5x & x \end{bmatrix} + \begin{bmatrix} y & 0 \\ 0 & y \end{bmatrix} \begin{bmatrix} \frac{1}{11} & \frac{-2}{11} \\ \frac{5}{11} & \frac{1}{11} \end{bmatrix} = \begin{bmatrix} x + y & 2x \\ -5x & x + y \end{bmatrix} \Rightarrow x = \frac{-1}{11} \text{ and } y = \frac{2}{11} \therefore 2x + 3y = 2 \left( \frac{-1}{11} \right) + 3 \left( \frac{2}{11} \right) = \frac{4}{11}