Find a quadratic integer polynomial that annihilates a given $2\times2$ integer matrix Suppose $A=\begin{bmatrix}3&2\\2&3\end{bmatrix}$. How can we find integers $b$ and $c$ such that $A^2+bA+cI_2=0$?
 A: $A=\begin{pmatrix}3&2\\ 2&3\end{pmatrix}\implies A^2= \begin{pmatrix}3&2\\ 2&3\end{pmatrix}\begin{pmatrix}3&2\\ 2&3\end{pmatrix} =\begin{pmatrix}13&12\\ 12&13\end{pmatrix}$
Then $$A^2+bA+cI_2=0\implies \begin{pmatrix}13&12\\ 12&13\end{pmatrix}+b\begin{pmatrix}3&2\\ 2&3\end{pmatrix}+c\begin{pmatrix}1&0\\ 0&1\end{pmatrix} =\begin{pmatrix}0&0\\ 0&0\end{pmatrix}$$
Then we get the following equations:
$$13+3b+c=0\tag{1}$$
$$12+2b=0\tag{2}$$
$(2)\implies b=-6\quad\text{then}\quad (1)\implies 13+3\cdot -6+c=0\implies c=5$
A: This is an application of the Cayley-Hamilton Theorem.

Every square matrix satisfies its own characteristic function.

Compute the characteristic polynomial of 
$$\mathbf{A}=
\left(
\begin{array}{cc}
 3 & 2 \\
 2 & 3 \\
\end{array}
\right)
$$
$$ \det \mathbf{A} = 5, \qquad \text{tr }\mathbf{A} = 6$$
The characteristic polynomial is
$$
 \color{blue}{p(\lambda)} = \lambda^{2} - \lambda \text{tr }\mathbf{A} +  \det \mathbf{A} = \color{blue}{\lambda^{2} - 6 \lambda + 5}
$$

By the Cayley-Hamilton theorem
$$
\boxed{
\color{blue}{\mathbf{A}^{2} - 6 \mathbf{A} + 5 \mathbf{I}_{2}} = \mathbf{0}
}
$$

This is the characteristic polynomial with the target matrix as the argument: $\color{blue}{p \left( \mathbf{A} \right)} $

Confirmation:

$$
%
\begin{align}
%
\mathbf{A}^{2} - 6 \mathbf{A} + 5 \mathbf{I}_{2} =
%
\left(
\begin{array}{cc}
 13 & 12 \\
 12 & 13 \\
\end{array}
\right)
%
-
\left(
\begin{array}{cc}
 18 & 12 \\
 12 & 18 \\
\end{array}
\right)
+
\left(
\begin{array}{cc}
 5 & 0 \\
 0 & 5 \\
\end{array}
\right) =
\left(
\begin{array}{cc}
 0 & 0 \\
 0 & 0 \\
\end{array}
\right)
%
\end{align}
%
$$

(Thanks to @amd for helpful suggestions.)
A: You can find characteristic polynome of 
$\ A=\begin{bmatrix}3&2\\2&3\end{bmatrix}$
$\ det(A- \lambda I )=0=> \lambda^2-6\lambda+5=0=>b=-6,c=5$
A: Here's a hint: A = 3I + 2(sqrt(I)) ...
