Now,  $A+B=$ $\left[\begin{array}{rrr}0 & 7 & 8 \\ -5 & 0 & 10 \\ 8 & -6 & 0\end{array}\right]$

So        $(A+B)C=$ $\left[ {\begin{array}{*{20}{r}}
  0&7&8 \\ 
  { – 5}&0&{10} \\ 
  8&{ – 6}&0 
\end{array}} \right]\left[ {\begin{array}{*{20}{r}}
  2 \\ 
  { – 2} \\ 
  3 
\end{array}} \right]$ $ = \left[ {\begin{array}{*{20}{r}}
  {0 – 14 + 24} \\ 
  { – 10 + 0 + 30} \\ 
  {16 + 12 + 0} 
\end{array}} \right] = \left[ {\begin{array}{*{20}{l}}
  {10} \\ 
  {20} \\ 
  {28} 
\end{array}} \right]$

Further     $AC=$ $\left[ {\begin{array}{*{20}{c}}
  0&6&7 \\ 
  { – 6}&0&8 \\ 
  7&{ – 8}&0 
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  2 \\ 
  { – 2} \\ 
  3 
\end{array}} \right]$ $=$ $\left[ {\begin{array}{*{20}{c}}
  {0 – 12 + 21} \\ 
  { – 12 + 0 + 24} \\ 
  {14 + 16 + 0} 
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  9 \\ 
  {12} \\ 
  {30} 
\end{array}} \right]$

and    $BC=$  $\left[ {\begin{array}{*{20}{l}}
  0&1&1 \\ 
  1&0&2 \\ 
  1&2&0 
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  2 \\ 
  { – 2} \\ 
  3 
\end{array}} \right]$ $ = \left[ {\begin{array}{*{20}{c}}
  {0 – 2 + 3} \\ 
  {2 + 0 + 6} \\ 
  {2 – 4 + 0} 
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  1 \\ 
  8 \\ 
  { – 2} 
\end{array}} \right]$

So       $AC+BC=$  $\left[\begin{array}{l}9 \\ 12 \\ 30\end{array}\right]+\left[\begin{array}{l}1 \\ 8 \\ -2\end{array}\right]=\left[\begin{array}{l}10 \\ 20 \\ 28\end{array}\right]$

Clearly,          $(A+B) C=A C+B C$