Question

For the matrices $A$ and $B$, verify that $(AB{)}^{\prime }=B^{\prime }{A}^{\prime }$ where:
$\left(i\right)$ $A=\left[\begin{array}{c}1\\ -4\\ 3\end{array}\right],B=\left[\begin{array}{ccc}-1& 2& 1\end{array}\right]$
$\left(ii\right)$ $A=\left[\begin{array}{c}0\\ 1\\ 2\end{array}\right],B=\left[\begin{array}{ccc}1& 5& 7\end{array}\right]$

Answer 1

$\left(i\right)$ Given: $A=\left[\begin{array}{c}1\\ -4\\ 3\end{array}\right],B=\left[\begin{array}{ccc}-1& 2& 1\end{array}\right]$
To verify: $(AB{)}^{\prime }=B^{\prime }{A}^{\prime }$
Solving $L.H.S.$
$(AB{)}^{\prime }={\left({\left[\begin{array}{c}1\\ -4\\ 3\end{array}\right]}_{3\times 1}{\left[\begin{array}{ccc}-1& 2& 1\end{array}\right]}_{1\times 3}\right)}^{\prime }$
$\Rightarrow (AB{)}^{\prime }={\left({\left[\begin{array}{ccc}1\times (-1)& 1\times 2& 1\times 1\\ -4\times (-1)& -4\times 2& -4\times 1\\ 3\times (-1)& 3\times 2& 3\times 1\end{array}\right]}_{3\times 3}\right)}^{\prime }$
$\Rightarrow (AB{)}^{\prime }={\left(\left[\begin{array}{ccc}-1& 2& 1\\ 4& -8& -4\\ -3& 6& 3\end{array}\right]\right)}^{\prime }$
$\Rightarrow (AB{)}^{\prime }=\left[\begin{array}{ccc}-1& 4& -3\\ 2& -8& 6\\ 1& -4& 3\end{array}\right]$

Solving $R.H.S.$
$B^{\prime }{A}^{\prime }={\left[\begin{array}{ccc}-1& 2& 1\end{array}\right]}^{\prime }{\left[\begin{array}{c}1\\ -4\\ 3\end{array}\right]}^{\prime }$
$\Rightarrow B^{\prime }{A}^{\prime }=\left[\begin{array}{c}-1\\ 2\\ 1\end{array}\right]\left[\begin{array}{ccc}1& -4& 3\end{array}\right]$
$\Rightarrow B^{\prime }{A}^{\prime }={\left[\begin{array}{ccc}-1\times 1& -1\times (-4)& -1\times 3\\ 2\times 1& 2\times (-4)& 2\times 3\\ 1\times 1& 1\times (-4)& 1\times 3\end{array}\right]}_{3\times 3}$
$\Rightarrow B^{\prime }{A}^{\prime }=\left[\begin{array}{ccc}-1& 4& -3\\ 2& -8& 6\\ 1& -4& 3\end{array}\right]$
Hence, $L.H.S.=R.H.S.$
Hence verified.

$\left(ii\right)$ Given :
$A=\left[\begin{array}{c}0\\ 1\\ 2\end{array}\right],B=\left[\begin{array}{ccc}1& 5& 7\end{array}\right]$
To verify: $(AB{)}^{\prime }=B^{\prime }{A}^{\prime }$
Solving $L.H.S.$
$(AB{)}^{\prime }={\left({\left[\begin{array}{c}0\\ 1\\ 2\end{array}\right]}_{3\times 1}{\left[\begin{array}{ccc}1& 5& 7\end{array}\right]}_{1\times 3}\right)}^{\prime }$
$\Rightarrow (AB{)}^{\prime }={\left({\left[\begin{array}{ccc}0\times 1& 0\times 5& 0\times 7\\ 1\times 1& 1\times 5& 1\times 7\\ 2\times 1& 2\times 5& 2\times 7\end{array}\right]}_{3\times 3}\right)}^{\prime }$
$\Rightarrow (AB{)}^{\prime }={\left(\left[\begin{array}{ccc}0& 0& 0\\ 1& 5& 7\\ 2& 10& 14\end{array}\right]\right)}^{\prime }$
$\Rightarrow (AB{)}^{\prime }=\left[\begin{array}{ccc}0& 1& 2\\ 0& 5& 10\\ 0& 7& 14\end{array}\right]$

Solving $R.H.S.$
$B^{\prime }{A}^{\prime }={\left[\begin{array}{ccc}1& 5& 7\end{array}\right]}^{\prime }{\left[\begin{array}{c}0\\ 1\\ 2\end{array}\right]}^{\prime }$
$\Rightarrow B^{\prime }{A}^{\prime }={\left[\begin{array}{c}1\\ 5\\ 7\end{array}\right]}_{3\times 1}{\left[\begin{array}{ccc}0& 1& 2\end{array}\right]}_{1\times 3}$
$\Rightarrow B^{\prime }{A}^{\prime }={\left[\begin{array}{ccc}1\times 0& 1\times 1& 1\times 2\\ 5\times 0& 5\times 1& 5\times 2\\ 7\times 0& 7\times 1& 7\times 2\end{array}\right]}_{3\times 3}$
$\Rightarrow B^{\prime }{A}^{\prime }=\left[\begin{array}{ccc}0& 1& 2\\ 0& 5& 10\\ 0& 7& 14\end{array}\right]$
Hence, $L.H.S.=R.H.S.$
Hence verified.