Solution
= Solution
Direct <matrix multiplication> gives
$$
A^2=\begin{pmatrix}13&12\\12&13\end{pmatrix},\qquad \operatorname{tr}A=6,\qquad\det A=5.
$$
Consequently $\boxed{A^2-6A+5I=0}$, the <Cayley-Hamilton theorem> for its <characteristic polynomial> $t^2-6t+5$. Rearranging gives $A(6I-A)=5I$, so
$$
\boxed{A^{-1}=\frac{6I-A}5=\frac15\begin{pmatrix}3&-2\\-2&3\end{pmatrix}.}
$$