Question

Show that AB ≠ BA in each of the following cases:
(i) $A=\left[\begin{array}{ccc}1& 3& -1\\ 2& -1& -1\\ 3& 0& -1\end{array}\right]\mathrm{and}B=\left[\begin{array}{ccc}-2& 3& -1\\ -1& 2& -1\\ -6& 9& -4\end{array}\right]$

(ii) $A=\left[\begin{array}{ccc}10& -4& -1\\ -11& 5& 0\\ 9& -5& 1\end{array}\right]\mathrm{and}B=\left[\begin{array}{ccc}1& 2& 1\\ 3& 4& 2\\ 1& 3& 2\end{array}\right]$

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right) \\ AB=\left[\begin{array}{ccc}1& 3& -1\\ 2& -1& -1\\ 3& 0& -1\end{array}\right]\left[\begin{array}{ccc}-2& 3& -1\\ -1& 2& -1\\ -6& 9& -4\end{array}\right] \\ \Rightarrow AB=\left[\begin{array}{ccc}-2-3+6& 3+6-9& -1-3+4\\ -4+1+6& 6-2-9& -2+1+4\\ -6-0+6& 9+0-9& -3-0+4\end{array}\right] \\ \Rightarrow AB=\left[\begin{array}{ccc}1& 0& 0\\ 3& -5& 3\\ 0& 0& 1\end{array}\right]...\left(1\right) \\ \mathrm{Also}, \\ BA=\left[\begin{array}{ccc}-2& 3& -1\\ -1& 2& -1\\ -6& 9& -4\end{array}\right]\left[\begin{array}{ccc}1& 3& -1\\ 2& -1& -1\\ 3& 0& -1\end{array}\right] \\ \Rightarrow BA=\left[\begin{array}{ccc}-2+6-3& -6-3+0& 2-3+1\\ -1+4-3& -3-2+0& 1-2+1\\ -6+18-12& -18-9+0& 6-9+4\end{array}\right] \\ \Rightarrow BA=\left[\begin{array}{ccc}1& -9& 0\\ 0& -5& 0\\ 0& -27& 1\end{array}\right]...\left(2\right) \\ \therefore \mathrm{AB\; \ne \; BA}\left(\mathrm{From}\mathrm{eqs}.\left(1\right)\mathrm{and}\left(2\right)\right) \end{gathered}$$