Question

Show that:
$\left(i\right)$ $\left[\begin{array}{cc}5& -1\\ 6& 7\end{array}\right]\left[\begin{array}{cc}2& 1\\ 3& 4\end{array}\right]\ne \left[\begin{array}{cc}2& 1\\ 3& 4\end{array}\right]\left[\begin{array}{cc}5& -1\\ 6& 7\end{array}\right]$

$\left(ii\right)$ $\left[\begin{array}{ccc}1& 2& 3\\ 0& 1& 0\\ 1& 1& 0\end{array}\right]\left[\begin{array}{ccc}-1& 1& 0\\ 0& -1& 1\\ 2& 3& 4\end{array}\right]\ne \left[\begin{array}{ccc}-1& 1& 0\\ 0& -1& 1\\ 2& 3& 4\end{array}\right]\left[\begin{array}{ccc}1& 2& 3\\ 0& 1& 0\\ 1& 1& 0\end{array}\right]$

Answer 1

$\left(i\right)$ Solving $L.H.S.$
${\left[\begin{array}{cc}5& -1\\ 6& 7\end{array}\right]}_{2\times 2}{\left[\begin{array}{cc}2& 1\\ 3& 4\end{array}\right]}_{2\times 2}$
$={\left[\begin{array}{cc}5\times 2+(-1)\times 3& 5\times 1+(-1)\times 4\\ 6\times 2+7\times 3& 6\times 1+7\times 4\end{array}\right]}_{2\times 2}$
$=\left[\begin{array}{cc}10-3& 5-4\\ 12+21& 6+28\end{array}\right]$
$=\left[\begin{array}{cc}7& 1\\ 33& 34\end{array}\right]$

Solving $R.H.S.$
${\left[\begin{array}{cc}2& 1\\ 3& 4\end{array}\right]}_{2\times 2}{\left[\begin{array}{cc}5& -1\\ 6& 7\end{array}\right]}_{2\times 2}$
$=\left[\begin{array}{cc}2\times 5+1\times 6& 2\times (-1)+1\times 7\\ 3\times 5+4\times 6& 3\times (-1)+4\times 7\end{array}\right]$
$=\left[\begin{array}{cc}10+6& -2+7\\ 15+24& -3+28\end{array}\right]$
$=\left[\begin{array}{cc}16& 5\\ 39& 25\end{array}\right]\ne L.H.S$
Thus, $L.H.S\ne R.H.S$
Hence Proved.

$\left(ii\right)$ Solving $L.H.S.$
${\left[\begin{array}{ccc}1& 2& 3\\ 0& 1& 0\\ 1& 1& 0\end{array}\right]}_{3\times 3}{\left[\begin{array}{ccc}-1& 1& 0\\ 0& -1& 1\\ 2& 3& 4\end{array}\right]}_{3\times 3}$
$=\left[\begin{array}{ccc}-1+0+6& 1-2+9& 0+2+12\\ 0+0+0& 0-1+0& 0+1+0\\ -1+0+0& 1-1+0& 0+1+0\end{array}\right]$
$=\left[\begin{array}{ccc}5& 8& 14\\ 0& -1& 1\\ -1& 0& 1\end{array}\right]$

Solving $R.H.S.$
${\left[\begin{array}{ccc}-1& 1& 0\\ 0& -1& 1\\ 2& 3& 4\end{array}\right]}_{3\times 3}{\left[\begin{array}{ccc}1& 2& 3\\ 0& 1& 0\\ 1& 1& 0\end{array}\right]}_{3\times 3}$

$=\left[\begin{array}{ccc}-1+0+0& -2+1+0& -3+0+0\\ 0+0+1& 0-1+1& 0+0+0\\ 2+0+4& 4+3+4& 6+0+0\end{array}\right]$

$=\left[\begin{array}{ccc}-1& -1& -3\\ 1& 0& 0\\ 6& 11& 6\end{array}\right]$
$\ne L.H.S.$
$\therefore L.H.S.\ne R.H.S.$
Hence proved .