Find if

We have $A^{2}=A \times A$

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

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

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

$=\left[\begin{array}{rrr}5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2\end{array}\right]$

$\therefore A^{2}-5 A+6 I$

$=\left[\begin{array}{rrr}5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2\end{array}\right]-5\left[\begin{array}{rrr}2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0\end{array}\right]+6\left[\begin{array}{rrr}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$

$=\left[\begin{array}{rrr}5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2\end{array}\right]-\left[\begin{array}{rrr}10 & 0 & 5 \\ 10 & 5 & 15 \\ 5 & -5 & 0\end{array}\right]+\left[\begin{array}{rrr}6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6\end{array}\right]$

$=\left[\begin{array}{ccc}5-10 & -1-0 & 2-5 \\ 9-10 & -2-5 & 5-15 \\ 0-5 & -1+5 & -2-0\end{array}\right]+\left[\begin{array}{ccc}6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6\end{array}\right]$

$=\left[\begin{array}{ccc}-5 & -1 & -3 \\ -1 & -7 & -10 \\ -5 & 4 & -2\end{array}\right]+\left[\begin{array}{ccc}6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6\end{array}\right]$

$=\left[\begin{array}{lll}-5+6 & -1+0 & -3+0 \\ -1+0 & -7+6 & -10+0 \\ -5+0 & 4+0 & -2+6\end{array}\right]$

$=\left[\begin{array}{rrr}1 & -1 & -3 \\ -1 & -1 & -10 \\ -5 & 4 & 4\end{array}\right]$