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

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

$2(2)+0(2)+1(1)  2(0)+0(1)+1(-1)  2(1)+0(3)+1(0)$

$2(2)+1(2)+3(1)  2(0)+1(1)+3(-1)  2(1)+1(3)+3(0)$

$1(2)+(-1)(2)+0(1)  1(0)+(-1)(1)+0(-1)  1(1)+(-1)(3)+0(0)$

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

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

Substituting the matrices in the given equation $: A^{2}-5 A+6 I$

$ = \left[ {\begin{array}{*{20}{c}}
  5&{ – 1}&2 \\ 
  9&{ – 2}&5 \\ 
  0&{ – 1}&{ – 2} 
\end{array}} \right]$ $ – 5\left[ {\begin{array}{*{20}{c}}
  2&0&1 \\ 
  2&1&3 \\ 
  1&{ – 1}&0 
\end{array}} \right]$ $ + 6\left[ {\begin{array}{*{20}{c}}
  1&0&0 \\ 
  0&1&0 \\ 
  0&0&1 
\end{array}} \right]$

$ = \left[ {\begin{array}{*{20}{c}}
  5&{ – 1}&2 \\ 
  9&{ – 2}&5 \\ 
  0&{ – 1}&{ – 2} 
\end{array}} \right]$  $ – \left[ {\begin{array}{*{20}{c}}
  {10}&0&5 \\ 
  {10}&5&{15} \\ 
  5&{ – 5}&0 
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
  6&0&0 \\ 
  0&6&0 \\ 
  0&0&6 
\end{array}} \right]$

$ = \left[ {\begin{array}{*{20}{c}}
  {5 – 10}&{ – 1 – 0}&{2 – 5} \\ 
  {9 – 10}&{ – 2 – 5}&{5 – 15} \\ 
  {0 – 5}&{ – 1 + 5}&{ – 2 – 0} 
\end{array}} \right]$ $ + \left[ {\begin{array}{*{20}{l}}
  6&0&0 \\ 
  0&6&0 \\ 
  0&0&6 
\end{array}} \right]$

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

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

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