If, prove that

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

$=\left[\begin{array}{ccc}1+0+4 & 0+0+0 & 2+0+6 \\ 0+0+2 & 0+4+0 & 0+2+3 \\ 2+0+6 & 0+0+0 & 4+0+9\end{array}\right]=\left[\begin{array}{ccc}5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13\end{array}\right]$

Now $A^{3}=A^{2} \cdot A$

$=\left[\begin{array}{ccc}5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13\end{array}\right]\left[\begin{array}{lll}1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3\end{array}\right]$

$=\left[\begin{array}{lll}5+0+16 & 0+0+0 & 10+0+24 \\ 2+0+10 & 0+8+0 & 4+4+15 \\ 8+0+26 & 0+0+0 & 16+0+39\end{array}\right]$

$=\left[\begin{array}{lll}21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55\end{array}\right]$

$\therefore A^{3}-6 A^{2}+7 A+2 I$

$=\left[\begin{array}{ccc}21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55\end{array}\right]-6\left[\begin{array}{ccc}5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13\end{array}\right]+7\left[\begin{array}{ccc}1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3\end{array}\right]+2\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$

$=\left[\begin{array}{lll}21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55\end{array}\right]-\left[\begin{array}{lll}30 & 0 & 48 \\ 12 & 24 & 30 \\ 48 & 0 & 78\end{array}\right]+\left[\begin{array}{lll}7 & 0 & 14 \\ 0 & 14 & 7 \\ 14 & 0 & 21\end{array}\right]+\left[\begin{array}{lll}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{array}\right]$

$=\left[\begin{array}{lll}21+7+2 & 0+0+0 & 34+14+0 \\ 12+0+0 & 8+14+2 & 23+7+0 \\ 34+14+0 & 0+0+0 & 55+21+2\end{array}\right]-\left[\begin{array}{lll}30 & 0 & 48 \\ 12 & 24 & 30 \\ 48 & 0 & 78\end{array}\right]$

$=\left[\begin{array}{lll}30 & 0 & 48 \\ 12 & 24 & 30 \\ 48 & 0 & 78\end{array}\right]-\left[\begin{array}{lll}30 & 0 & 48 \\ 12 & 24 & 30 \\ 48 & 0 & 78\end{array}\right]$

$=\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]=O$

$\therefore A^{3}-6 A^{2}+7 A+2 I=O$