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If $\quad A=\left[\begin{array}{rrr}1 & 1 & -1 \\ 2 & 0 & 3 \\ 3 & -1 & 2\end{array}\right]$, $B=\left[\begin{array}{rr}1 & 3 \\ 0 & 2 \\ -1 & 4\end{array}\right]$ and $C=\left[\begin{array}{rrrr}1 & 2 & 3 & -4 \\ 2 & 0 & -2 & 1\end{array}\right],$ find $A(B C),(A B) C$ and show that $(A B) C=A(B C)$.
Solution
We have
$AB=$ $\left[ {\begin{array}{*{20}{c}}
1&1&{ – 1} \\
2&0&3 \\
3&{ – 1}&2
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
1&3 \\
0&2 \\
{ – 1}&4
\end{array}} \right]$ $ = \left[ {\begin{array}{*{20}{c}}
{1 + 0 + 1}&{3 + 2 – 4} \\
{2 + 0 – 3}&{6 + 0 + 12} \\
{3 + 0 – 2}&{9 – 2 + 8}
\end{array}} \right]$ $ = \left[ {\begin{array}{*{20}{c}}
2&1 \\
{ – 1}&{18} \\
1&{15}
\end{array}} \right]$
$(AB)$ $(C)=$ $\left[ {\begin{array}{*{20}{c}}
2&1 \\
{ – 1}&{18} \\
1&{15}
\end{array}} \right]$ $\left[\begin{array}{rrrr}1 & 2 & 3 & -4 \\ 2 & 0 & -2 & 1\end{array}\right]=$ $\left[\begin{array}{cccc}2+2 & 4+0 & 6-2 & -8+1 \\ -1+36 & -2+0 & -3-36 & 4+18 \\ 1+30 & 2+0 & 3-30 & -4+15\end{array}\right]$
$=\left[\begin{array}{cccc}4 & 4 & 4 & -7 \\ 35 & -2 & -39 & 22 \\ 31 & 2 & -27 & 11\end{array}\right]$
Now $BC=$ $ \left[ {\begin{array}{*{20}{r}}
1&3 \\
0&2 \\
{ – 1}&4
\end{array}} \right]\left[ {\begin{array}{*{20}{r}}
1&2&3&{ – 4} \\
2&0&{ – 2}&1
\end{array}} \right]$ $=$ $\left[ {\begin{array}{*{20}{r}}
{1 + 6}&{2 + 0}&{3 – 6}&{ – 4 + 3} \\
{0 + 4}&{0 + 0}&{0 – 4}&{0 + 2} \\
{ – 1 + 8}&{ – 2 + 0}&{ – 3 – 8}&{4 + 4}
\end{array}} \right]$
$=$ $\left[ {\begin{array}{*{20}{r}}
7&2&{ – 3}&{ – 1} \\
4&0&{ – 4}&2 \\
7&{ – 2}&{ – 11}&8
\end{array}} \right]$
Therefore $A(BC) = $ $\left[ {\begin{array}{*{20}{r}}
1&1&{ – 1} \\
2&0&3 \\
3&{ – 1}&2
\end{array}} \right]$$\left[ {\begin{array}{*{20}{r}}
7&2&{ – 3}&{ – 1} \\
4&0&{ – 4}&2 \\
7&{ – 2}&{ – 11}&8
\end{array}} \right]$
$=\left[\begin{array}{cccc}7+4-7 & 2+0+2 & -3-4+11 & -1+2-8 \\ 14+0+21 & 4+0-6 & -6+0-33 & -2+0+24 \\ 21-4+14 & 6+0-4 & -9+4-22 & -3-2+16\end{array}\right]$
$=\left[\begin{array}{cccc}4 & 4 & 4 & -7 \\ 35 & -2 & -39 & 22 \\ 31 & 2 & -27 & 11\end{array}\right] .$ Clearly, $(AB)$ $C=A(B C)$