Let \(\mathrm{A}=\left[\begin{array}{cc}1 & 2 \\ -5 & 1\end{array}\right]\) and \(\mathrm{A}^{-1}=x \mathrm{~A}+y \mathrm{I}_2\), (where \(\mathrm{I}_2\) is unit matrix of order 2), then

\(\begin{array}{ll}
& |A|=\left|\begin{array}{cc}
1 & 2 \\
-5 & 1
\end{array}\right|=11 \neq 0 \\
\therefore & A^{-1}=\frac{1}{11}\left[\begin{array}{cc}
1 & -2 \\
5 & 1
\end{array}\right]
\end{array}\)
\(\ldots\left[\right.\) If \(A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]\) and \(a d-b c \neq 0\), then
\(\left.A^{-1}=\frac{1}{(a d-b c)}\left[\begin{array}{cc}
d & -b \\
-c & a
\end{array}\right] .\right]\)
\(\begin{aligned}
& \mathrm{A}^{-1}=x \mathrm{~A}+y \mathrm{I} \\
& \therefore \quad \frac{1}{11}\left[\begin{array}{cc}
1 & -2 \\
5 & 1
\end{array}\right]=\left[\begin{array}{cc}
x & 2 x \\
-5 x & x
\end{array}\right]+\left[\begin{array}{ll}
y & 0 \\
0 & y
\end{array}\right] \\
& \therefore \quad\left[\begin{array}{cc}
\frac{1}{11} & \frac{-2}{11} \\
\frac{5}{11} & \frac{1}{11}
\end{array}\right]=\left[\begin{array}{cc}
x+y & 2 x \\
-5 x & x+y
\end{array}\right]
\end{aligned}\)
\(\therefore \quad\) By the equality of matrices,
\(\begin{array}{ll}
& 2 x=\frac{-2}{11} \text { and } x+y=\frac{1}{11} \\
\therefore & x=\frac{-1}{11} \text { and } \frac{-1}{11}+y=\frac{1}{11} \\
\therefore & x=\frac{-1}{11} \text { and } y=\frac{2}{11}
\end{array}\)