If \(A=\left[\begin{array}{ll}3 & 2 \\ 1 & 1\end{array}\right]\), then \(A^{2}+x A+y I=0\) for \((x, y)\) is

\[
A=\left[\begin{array}{ll}
3 & 2 \\
1 & 1
\end{array}\right]
\]
The characteristic equation of ' \(A\) ' is
\[
\begin{array}{r}
|\mathrm{A}-\lambda \mathrm{I}|=0 \\
\left|\begin{array}{cc}
3-\lambda & 2 \\
1 & 1-\lambda
\end{array}\right|=0 \\
(3-\lambda)(1-\lambda)-2=0 \\
3-\lambda-3 \lambda+\lambda^{2}-2=0 \\
\left(\lambda^{2}-4 \lambda+1\right)=0 \quad \text{...(i)}
\end{array}
\]
By Caylay-Hamilton theorem : Every square matrix satisfied its characteristic equation, then put \((\lambda=\mathrm{A})\) is in Eq. (i)
\[
\begin{aligned}
&\mathrm{A}^{2}-4 \mathrm{~A}+\mathrm{I}=0 \\
&\text { On comparing with } \mathrm{A}^{2}+\mathrm{xA}+\mathrm{yI}=0 \\
&\Rightarrow \quad \mathrm{x}=-4, \mathrm{y}=1
\end{aligned}
\]