If the matrix \(\mathrm{A}=\left[\begin{array}{cc}1 & 2 \\ -5 & 1\end{array}\right]\) and \(\mathrm{A}^{-1}=x \mathrm{~A}+y \mathrm{I}\), when \(I\) is a unit matrix of order 2 , then the value of \(2 x+3 y\) is

\(\begin{aligned} & \mathrm{A}=\left[\begin{array}{cc}1 & 2 \\ -5 & 1\end{array}\right] \\ \therefore \quad & \mathrm{A} \mid=11 \\ & \mathrm{~A}^{-1}=\frac{1}{|\mathrm{~A}|}\left[\begin{array}{cc}1 & -2 \\ 5 & 1\end{array}\right]=\frac{1}{11}\left[\begin{array}{cc}1 & -2 \\ 5 & 1\end{array}\right] \\ & \mathrm{A}^{-1}=x \mathrm{~A}+y \mathrm{I}, \text { we get } \\ & {\left[\begin{array}{cc}\frac{1}{11} & \frac{-2}{11} \\ \frac{5}{11} & \frac{1}{11}\end{array}\right]=\left[\begin{array}{cc}x & 2 x \\ -5 x & x\end{array}\right]+\left[\begin{array}{ll}y & 0 \\ 0 & y\end{array}\right] }\end{aligned}\)
\(\begin{array}{ll}\therefore & {\left[\begin{array}{cc}\frac{1}{11} & \frac{-2}{11} \\ \frac{5}{11} & \frac{1}{11}\end{array}\right]=\left[\begin{array}{cc}x+y & 2 x \\ -5 x & x+y\end{array}\right]} \\ \Rightarrow & x=\frac{-1}{11} \text { and } y=\frac{2}{11} \\ \therefore & 2 x+3 y=2\left(\frac{-1}{11}\right)+3\left(\frac{2}{11}\right)=\frac{4}{11}\end{array}\)