Question

If $A=\left[\begin{array}{c}-2\\ 4\\ 5\end{array}\right],B=\left[\begin{array}{ccc}1& 3& -6\end{array}\right]$, verify that $(AB{)}^{\prime }=B^{\prime }{A}^{\prime }.$

Answer 1

Given : $A=\left[\begin{array}{c}-2\\ 4\\ 5\end{array}\right],B=\left[\begin{array}{ccc}1& 3& -6\end{array}\right]$
To verify: $(AB{)}^{\prime }=B^{\prime }{A}^{\prime }.$
Solving $L.H.S.$
$\Rightarrow (AB{)}^{\prime }={\left({\left[\begin{array}{ccc}-2\times 1& -2\times 3& -2\times (-6)\\ 4\times 1& 4\times 3& 4\times (-6)\\ 5\times 1& 5\times 3& 5\times (-6)\end{array}\right]}_{3\times 3}\right)}^{\prime }$
$\Rightarrow (AB{)}^{\prime }={\left(\left[\begin{array}{ccc}-2& -6& 12\\ 4& 12& -24\\ 5& 15& -30\end{array}\right]\right)}^{\prime }$
$\Rightarrow (AB{)}^{\prime }=\left[\begin{array}{ccc}-2& 4& 5\\ -6& 12& 15\\ 12& -24& -30\end{array}\right]$

Solving $R.H.S.$
$B^{\prime }{A}^{\prime }={\left[\begin{array}{ccc}1& 3& -6\end{array}\right]}^{\prime }{\left[\begin{array}{c}-2\\ 4\\ 5\end{array}\right]}^{\prime }$

$\Rightarrow B^{\prime }{A}^{\prime }={\left[\begin{array}{c}1\\ 3\\ -6\end{array}\right]}_{3\times 1}{\left[\begin{array}{ccc}-2& 4& 5\end{array}\right]}_{1\times 3}$
$\Rightarrow B^{\prime }{A}^{\prime }={\left[\begin{array}{ccc}1\times (-2)& 1\times 4& 1\times 5\\ 3\times (-2)& 3\times 4& 3\times 5\\ -6\times (-2)& -6\times 4& -6\times 5\end{array}\right]}_{3\times 3}$
$\Rightarrow B^{\prime }{A}^{\prime }=\left[\begin{array}{ccc}-2& 4& 5\\ -6& 12& 15\\ 12& -24& -30\end{array}\right]$
So, $L.H.S.=R.H.S.$
Hence, verified.