Q.40. If the matrix A = \begin{bmatrix}1 & 2 \\ -5 & 1\end{bmatrix} and A^{-1} = x A + yI, where I is the unit 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} |A| = 1(1) - (-5)(2) = 1 + 10 = 11 A^{-1} = \frac{1}{|A|} \begin{bmatrix}1 & -2 \\ 5 & 1\end{bmatrix} = \frac{1}{11} \begin{bmatrix}1 & -2 \\ 5 & 1\end{bmatrix}Since A^{-1} = xA + yI, 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}Comparing corresponding elements:
x + y = \frac{1}{11} 2x = \frac{-2}{11} \Rightarrow x = \frac{-1}{11} -5x = \frac{5}{11} \Rightarrow x = \frac{-1}{11} x + y = \frac{1}{11} \Rightarrow \frac{-1}{11} + y = \frac{1}{11} \Rightarrow y = \frac{2}{11}Now, calculating 2x + 3y:
2x + 3y = 2\left(\frac{-1}{11}\right) + 3\left(\frac{2}{11}\right) = \frac{-2}{11} + \frac{6}{11} = \frac{4}{11}