If $\;$ $A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$, then show that $\;$ $A^2 - 5 A + 7 I = 0$
$A = \begin{bmatrix}
3 & 1 \\
-1 & 2
\end{bmatrix}$, $\;$ $I = \begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix} = 2 \times 2$ $\;$ identity matrix
$\begin{aligned}
A^2 = A \times A & = \begin{bmatrix}
3 & 1 \\
-1 & 2
\end{bmatrix} \times \begin{bmatrix}
3 & 1 \\
-1 & 2
\end{bmatrix} \\\\
& = \begin{bmatrix}
3 \times 3 + 1 \times \left(-1\right) & 3 \times 1 + 1 \times 2 \\
-1 \times 3 + 2 \times \left(-1\right) & -1 \times 1 + 2 \times 2
\end{bmatrix} \\\\
& = \begin{bmatrix}
8 & 5 \\
-5 & 3
\end{bmatrix}
\end{aligned}$
$\begin{aligned}
5A & = 5 \times \begin{bmatrix}
3 & 1 \\
-1 & 2
\end{bmatrix} \\\\
& = \begin{bmatrix}
15 & 5 \\
-5 & 10
\end{bmatrix}
\end{aligned}$
$\begin{aligned}
7 I & = 7 \times \begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix} \\\\
& = \begin{bmatrix}
7 & 0 \\
0 & 7
\end{bmatrix}
\end{aligned}$
$\begin{aligned}
\therefore \; A^2 - 5 A + 7 I & = \begin{bmatrix}
8 & 5 \\
-5 & 3
\end{bmatrix} - \begin{bmatrix}
15 & 5 \\
-5 & 10
\end{bmatrix} + \begin{bmatrix}
7 & 0 \\
0 & 7
\end{bmatrix} \\\\
& = \begin{bmatrix}
8 - 15 + 7 & 5 - 5 + 0 \\
-5 + 5 + 0 & 3 - 10 + 7
\end{bmatrix} \\\\
& = \begin{bmatrix}
0 & 0 \\
0 & 0
\end{bmatrix} \\\\
& = 0 \;\;\; \left[\text{null matrix}\right]
\end{aligned}$
$\therefore \;$ $A^2 - 5A + 7 I = 0$ $\;\;\;$ Hence proved.