If \(A=\left[\begin{array}{cc}1 & 2 \\ -5 & 1\end{array}\right]\) and \(A^{-1}=x A+y I\), where \(I\) is unit matrix of order 2 , then the values of \(x\) and \(y\) are respectively

\(|\mathrm{A}|=\left|\begin{array}{cc}1 & 2 \\ -5 & 1\end{array}\right|=1+10=11\) and adj \(\mathrm{A}=\left[\begin{array}{cc}1 & -2 \\ 5 & 1\end{array}\right]\)
\(\therefore \mathrm{A}^{-1}=\frac{1}{11}\left[\begin{array}{cc}1 & -2 \\ 5 & 1\end{array}\right]\)
Given \(\mathrm{A}^{-1}=\mathrm{x} \mathrm{A}+\mathrm{yI}\)
\(
\frac{1}{11}\left[\begin{array}{cc}1 & -2 \\ 5 & 1\end{array}\right]=\left[\begin{array}{cc}\mathrm{x} & 2 \mathrm{x} \\ -5 \mathrm{x} & \mathrm{x}\end{array}\right]+\left[\begin{array}{ll}\mathrm{y} & 0 \\ 0 & \mathrm{y}\end{array}\right]
\)
\(\therefore \frac{1}{11}\left[\begin{array}{cc}1 & -2 \\ 5 & 1\end{array}\right] \quad=\left[\begin{array}{cc}\mathrm{x}+\mathrm{y} & 2 \mathrm{x} \\ -5 \mathrm{x} & \mathrm{x}+\mathrm{y}\end{array}\right]\)
\(\therefore 2 \mathrm{x}=\frac{-2}{11} \Rightarrow \mathrm{x}=\frac{-1}{11}\) and \(\mathrm{x}+\mathrm{y}=\frac{1}{11} \Rightarrow \mathrm{y}=\frac{2}{11}\)