Question

If $A=(\begin{array}{ccc}-1& 2& 1\\ 1& 2& 3\end{array}),B=(\begin{array}{c}0\\ 1\\ 2\end{array})$ and $C=(\begin{array}{cc}2& 1\end{array})$ verify (AB)C = A(BC).

Answer 1

$\Rightarrow A=\left[\begin{array}{ccc}-1& 2& 1\\ 1& 2& 3\end{array}\right]B=\left[\begin{array}{c}0\\ 1\\ 2\end{array}\right]C=\left[\begin{array}{cc}2& 1\end{array}\right]$

To verify : $\left(AB\right)C=A\left(BC\right)$

$\Rightarrow$ LHS $=\left(AB\right)C\left[\left[\begin{array}{ccc}-1& 2& 1\\ 1& 2& 3\end{array}\right]\left[\begin{array}{c}0\\ 1\\ 2\end{array}\right]\right]\left[\begin{array}{cc}2& 1\end{array}\right]$

$=\left[\begin{array}{c}0+2+2\\ 0+2+6\end{array}\right]\left[\begin{array}{cc}2& 1\end{array}\right]$

$={\left[\begin{array}{c}4\\ 8\end{array}\right]}_{2\times 1}{\left[\begin{array}{cc}2& 1\end{array}\right]}_{1\times 2}$

$=\left[\begin{array}{cc}8& 4\\ 16& 8\end{array}\right]$

$\Rightarrow$ RHS$=A\left(BC\right)=\left[\begin{array}{ccc}-1& 2& 1\\ 1& 2& 3\end{array}\right]\left[\left[\begin{array}{c}0\\ 1\\ 2\end{array}\right]\left[\begin{array}{cc}2& 1\end{array}\right]\right]$

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

$=\left[\begin{array}{cc}0+4+4& 0+2+2\\ 0+4+12& 0+2+6\end{array}\right]$

$=\left[\begin{array}{cc}8& 4\\ 16& 8\end{array}\right]$

Hence, verified LHS $=$ RHS.