Question

If $A=\left[\begin{array}{ccc}1& -2& 3\\ -4& 2& 5\end{array}\right]$ and $B=\left[\begin{array}{cc}1& 3\\ -1& 0\\ 2& 4\end{array}\right]$, then show that $(AB{)}^{\prime }=B^{\prime }{A}^{\prime }$

Answer 1

$AB=\left[\begin{array}{ccc}1& -2& 3\\ -4& 2& 5\end{array}\right]\left[\begin{array}{cc}1& 3\\ -1& 0\\ 2& 4\end{array}\right]=\left[\begin{array}{cc}1+2+6& 3+0+12\\ -4-2+10& -12+0+20\end{array}\right]=\left[\begin{array}{cc}9& 15\\ 4& 8\end{array}\right]A^{\prime }=\left[\begin{array}{cc}1& -4\\ -2& 2\\ 3& 5\end{array}\right]B^{\prime }=\left[\begin{array}{ccc}1& -1& 2\\ 3& 0& 4\end{array}\right]B^{\prime }{A}^{\prime }=\left[\begin{array}{ccc}1& -1& 2\\ 3& 0& 4\end{array}\right]\left[\begin{array}{cc}1& -4\\ -2& 2\\ 3& 5\end{array}\right]=\left[\begin{array}{cc}1+2+6& -4-2+1\\ 3+0+12& -12+0+2\end{array}\right]=\left[\begin{array}{cc}9& 4\\ 15& 8\end{array}\right]{\left(AB\right)}^{\prime }=\left[\begin{array}{cc}9& 4\\ 15& 8\end{array}\right]$

$\therefore {\left(AB\right)}^{\prime }=B^{\prime }{A}^{\prime }$