If then verify that

Given, $A=\left[\begin{array}{ccc}1 & 0 & -1 \\ 2 & 1 & 3 \\ 0 & 1 & 1\end{array}\right]$

So, $A^{2}=A \cdot A$

Thus, $\quad=\left[\begin{array}{ccc}1 & 0 & -1 \\ 2 & 1 & 3 \\ 0 & 1 & 1\end{array}\right]\left[\begin{array}{ccc}1 & 0 & -1 \\ 2 & 1 & 3 \\ 0 & 1 & 1\end{array}\right]$

$=\left[\begin{array}{ccc}1+0+0 & 0+0-1 & -1+0-1 \\ 2+2+0 & 0+1+3 & -2+3+3 \\ 0+2+0 & 0+1+1 & 0+3+1\end{array}\right]=\left[\begin{array}{ccc}1 & -1 & -2 \\ 4 & 4 & 4 \\ 2 & 2 & 4\end{array}\right]$

$A^{2}+A=\left[\begin{array}{ccc}1 & -1 & -2 \\ 4 & 4 & 4 \\ 2 & 2 & 4\end{array}\right]+\left[\begin{array}{ccc}1 & 0 & -1 \\ 2 & 1 & 3 \\ 0 & 1 & 1\end{array}\right]=\left[\begin{array}{ccc}2 & -1 & -3 \\ 6 & 5 & 7 \\ 2 & 3 & 5\end{array}\right]$ $\ldots(\mathrm{i})$

Now, $\quad A+I=\left[\begin{array}{ccc}1 & 0 & -1 \\ 2 & 1 & 3 \\ 0 & 1 & 1\end{array}\right]+\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}2 & 0 & -1 \\ 2 & 2 & 3 \\ 0 & 1 & 2\end{array}\right]$

So,

$A(A+I)=\left[\begin{array}{ccc}1 & 0 & -1 \\ 2 & 1 & 3 \\ 0 & 1 & 1\end{array}\right]\left[\begin{array}{ccc}2 & 0 & -1 \\ 2 & 2 & 3 \\ 0 & 1 & 2\end{array}\right]$

$=\left[\begin{array}{ccc}2+0+0 & 0+0-1 & -1+0-2 \\ 4+2+0 & 0+2+3 & -2+3+6 \\ 0+2+0 & 0+2+1 & 0+3+2\end{array}\right]=\left[\begin{array}{ccc}2 & -1 & -3 \\ 6 & 5 & 7 \\ 2 & 3 & 5\end{array}\right]$ $\ldots$ (ii)

From (i) and (ii), we get

$\mathbf{A}^{2}+\mathbf{A}=\mathbf{A}(\mathbf{A}+\mathbf{I})$