Question

Show that AB ≠ BA in each of the following cases:
(i) $A=\left[\begin{array}{cc}5& -1\\ 6& 7\end{array}\right]\mathrm{and}B=\left[\begin{array}{cc}2& 1\\ 3& 4\end{array}\right]$

(ii) $A=\left[\begin{array}{ccc}-1& 1& 0\\ 0& -1& 1\\ 2& 3& 4\end{array}\right]\mathrm{and}B=\left[\begin{array}{ccc}1& 2& 3\\ 0& 1& 0\\ 1& 1& 0\end{array}\right]$

(iii) $A=\left[\begin{array}{ccc}1& 3& 0\\ 1& 1& 0\\ 4& 1& 0\end{array}\right]\mathrm{and}B=\left[\begin{array}{ccc}0& 1& 0\\ 1& 0& 0\\ 0& 5& 1\end{array}\right]$

Answer 1

$\left(i\right)$
$$\begin{gathered} AB=\left[\begin{array}{cc}5& -1\\ 6& 7\end{array}\right]\left[\begin{array}{cc}2& 1\\ 3& 4\end{array}\right] \\ \Rightarrow AB=\left[\begin{array}{cc}10-3& 5-4\\ 12+21& 6+28\end{array}\right] \\ \Rightarrow AB=\left[\begin{array}{cc}7& 1\\ 33& 34\end{array}\right]...\left(1\right) \\ \mathrm{Also}, \\ BA=\left[\begin{array}{cc}2& 1\\ 3& 4\end{array}\right]\left[\begin{array}{cc}5& -1\\ 6& 7\end{array}\right] \\ \Rightarrow BA=\left[\begin{array}{cc}10+6& -2+7\\ 15+24& -3+28\end{array}\right] \\ \Rightarrow BA=\left[\begin{array}{cc}16& 5\\ 39& 25\end{array}\right]...\left(2\right) \\ \therefore \mathrm{AB\; \ne \; BA}\left(\mathrm{From}\mathrm{eqs}.\left(1\right)\mathrm{and}\left(2\right)\right) \end{gathered}$$


$$\begin{gathered} \left(\mathrm{ii}\right) \\ AB=\left[\begin{array}{ccc}-1& 1& 0\\ 0& -1& 1\\ 2& 3& 4\end{array}\right]\left[\begin{array}{ccc}1& 2& 3\\ 0& 1& 0\\ 1& 1& 0\end{array}\right] \\ \Rightarrow AB=\left[\begin{array}{ccc}-1+0+0& -2+1+0& -3+0+0\\ 0+0+1& 0-1+1& 0+0+0\\ 2+0+4& 4+3+4& 6+0+0\end{array}\right] \\ \Rightarrow AB=\left[\begin{array}{ccc}-1& -1& -3\\ 1& 0& 0\\ 6& 11& 6\end{array}\right]...\left(1\right) \\ \mathrm{Also}, \\ BA=\left[\begin{array}{ccc}1& 2& 3\\ 0& 1& 0\\ 1& 1& 0\end{array}\right]\left[\begin{array}{ccc}-1& 1& 0\\ 0& -1& 1\\ 2& 3& 4\end{array}\right] \\ \Rightarrow BA=\left[\begin{array}{ccc}-1+0+6& 1-2+9& 0+2+12\\ 0+0+0& 0-1+0& 0+1+0\\ -1+0+0& 1-1+0& 0+1+0\end{array}\right] \\ \Rightarrow BA=\left[\begin{array}{ccc}5& 8& 14\\ 0& -1& 1\\ -1& 0& 1\end{array}\right]...\left(2\right) \\ \therefore \mathrm{AB\; \ne \; BA}\left(\mathrm{From}\mathrm{eqs}.\left(1\right)\mathrm{and}\left(2\right)\right) \end{gathered}$$


$$\begin{gathered} \left(\mathrm{iii}\right) \\ AB=\left[\begin{array}{ccc}1& 3& 0\\ 1& 1& 0\\ 4& 1& 0\end{array}\right]\left[\begin{array}{ccc}0& 1& 0\\ 1& 0& 0\\ 0& 5& 1\end{array}\right] \\ \Rightarrow AB=\left[\begin{array}{ccc}0+3+0& 1+0+0& 0+0+0\\ 0+1+0& 1+0+0& 0+0+0\\ 0+1+0& 4+0+0& 0+0+0\end{array}\right] \\ \Rightarrow AB=\left[\begin{array}{ccc}3& 1& 0\\ 1& 1& 0\\ 1& 4& 0\end{array}\right]...\left(1\right) \\ \mathrm{Also}, \\ BA=\left[\begin{array}{ccc}0& 1& 0\\ 1& 0& 0\\ 0& 5& 1\end{array}\right]\left[\begin{array}{ccc}1& 3& 0\\ 1& 1& 0\\ 4& 1& 0\end{array}\right] \\ \Rightarrow BA=\left[\begin{array}{ccc}0+1+0& 0+1+0& 0+0+0\\ 1+0+0& 3+0+0& 0+0+0\\ 0+5+4& 0+5+1& 0+0+0\end{array}\right] \\ \Rightarrow BA=\left[\begin{array}{ccc}1& 1& 0\\ 1& 3& 0\\ 9& 6& 0\end{array}\right]...\left(2\right) \\ \therefore \mathrm{AB\; \ne \; BA}\left(\mathrm{From}\mathrm{eqs}.\left(1\right)\mathrm{and}\left(2\right)\right) \end{gathered}$$