Question

If A=$\left[\begin{array}{ccc}0& 6& 7\\ -6& 0& 8\\ 7& -8& 0\end{array}\right]$,B=$\left[\begin{array}{ccc}0& 1& 1\\ 1& 0& 1\\ 1& 2& 0\end{array}\right]$ ,and C$\left[\begin{array}{c}2\\ -2\\ 3\end{array}\right]$ calculate AC, BC and (A + B) C, also verify (A+B)C = AC + BC

Answer 1

$AC=\left[\begin{array}{ccc}0& 6& 7\\ -6& 0& 8\\ 7& -8& 0\end{array}\right]\left[\begin{array}{c}2\\ -2\\ 3\end{array}\right]=\left[\begin{array}{c}0\times 2+6\times (-2)+7\times 3\\ -6\times 2+0\times -2+8\times 3\\ 7\times 2+(-8)\times (-2)+0\times 3\end{array}\right]=\left[\begin{array}{c}9\\ 12\\ 30\end{array}\right]AC=\left[\begin{array}{ccc}0& 1& 1\\ 1& 0& 1\\ 1& 2& 0\end{array}\right]\left[\begin{array}{c}2\\ -2\\ 3\end{array}\right]=\left[\begin{array}{c}0\times 2+1\times (-2)+1\times 3\\ 1\times 2+0\times (-2)+1\times 3\\ 1\times 2+2\times (-2)+0\times 3\end{array}\right]=\left[\begin{array}{c}1\\ 5\\ -2\end{array}\right](A+B)C=\left[\begin{array}{ccc}0+0& 6+1& 7+1\\ -6+1& 0+0& 8+1\\ 7+1& -8+2& 0+0\end{array}\right]\left[\begin{array}{c}2\\ -2\\ 3\end{array}\right]=\left[\begin{array}{ccc}0& 7& 8\\ -5& 0& 9\\ 8& -6& 0\end{array}\right]\left[\begin{array}{c}2\\ -2\\ 3\end{array}\right]=\left[\begin{array}{c}0\times 2+7\times (-2)+8\times 3\\ -5\times 2+0\times (-2)+9\times 3\\ 8\times 2+(-6)\times (-2)+0\times 3\end{array}\right]=\left[\begin{array}{c}10\\ 17\\ 28\end{array}\right]$
$AC+BC=\left[\begin{array}{c}9\\ 12\\ 30\end{array}\right]+\left[\begin{array}{c}1\\ 5\\ -2\end{array}\right]=\left[\begin{array}{c}10\\ 17\\ 28\end{array}\right]$
As we can see (A+B) C = AC + BC
Note that this denotes the distributive property of matrix multiplication.