Let $A=\left[\begin{array}{cc}1& 2\\ -5& 1\end{array}\right]$ and $A^{-1}=xA+yI$, then the values of $x$ and $y$ are:

Using the Characteristic Equation: $\det (A-\lambda I)=0$ $
$\left|\begin{array}{cc}1-\lambda & 2\\ -5& 1-\lambda \end{array}\right|=0$ $
$$(1-\lambda {)}^2-(-10)=0⟹{\lambda }^2-2\lambda +1+10=0⟹{\lambda }^2-2\lambda +11=0$$
By Cayley-Hamilton Theorem, $A$ satisfies its characteristic equation:
$$A^2-2A+11I=0$$
Multiply by $A^{-1}$:
$$A-2I+11A^{-1}=0$$
$$11A^{-1}=-A+2I⟹A^{-1}=-\frac{1}{11}A+\frac{2}{11}I$$
Comparing with $A^{-1}=xA+yI$:
$$x=-\frac{1}{11},y=\frac{2}{11}$$