Show that AB ≠ BA in each of the following cases:

Solution:

(i)

$A B=\left[\begin{array}{cc}5 & -1 \\ 6 & 7\end{array}\right]\left[\begin{array}{ll}2 & 1 \\ 3 & 4\end{array}\right]$

$\Rightarrow A B=\left[\begin{array}{cc}10-3 & 5-4 \\ 12+21 & 6+28\end{array}\right]$

$\Rightarrow A B=\left[\begin{array}{cc}7 & 1 \\ 33 & 34\end{array}\right]$               ...(1)

Also,

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

$\Rightarrow B A=\left[\begin{array}{ccc}-1+0+6 & 1-2+9 & 0+2+12 \\ 0+0+0 & 0-1+0 & 0+1+0 \\ -1+0+0 & 1-1+0 & 0+1+0\end{array}\right]$

$\Rightarrow B A=\left[\begin{array}{ccc}5 & 8 & 14 \\ 0 & -1 & 1 \\ -1 & 0 & 1\end{array}\right]$

$\therefore \mathrm{AB} \neq \mathrm{BA}$

(iii)

$A B=\left[\begin{array}{lll}1 & 3 & 0 \\ 1 & 1 & 0 \\ 4 & 1 & 0\end{array}\right]\left[\begin{array}{lll}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 5 & 1\end{array}\right]$

$\Rightarrow A B=\left[\begin{array}{lll}0+3+0 & 1+0+0 & 0+0+0 \\ 0+1+0 & 1+0+0 & 0+0+0 \\ 0+1+0 & 4+0+0 & 0+0+0\end{array}\right]$

$\Rightarrow A B=\left[\begin{array}{lll}3 & 1 & 0 \\ 1 & 1 & 0 \\ 1 & 4 & 0\end{array}\right]$ $\ldots(1)$

Also,

$B A=\left[\begin{array}{lll}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 5 & 1\end{array}\right]\left[\begin{array}{lll}1 & 3 & 0 \\ 1 & 1 & 0 \\ 4 & 1 & 0\end{array}\right]$

$\Rightarrow B A=\left[\begin{array}{ccc}0+1+0 & 0+1+0 & 0+0+0 \\ 1+0+0 & 3+0+0 & 0+0+0 \\ 0+5+4 & 0+5+1 & 0+0+0\end{array}\right]$

$\Rightarrow B A=\left[\begin{array}{lll}1 & 1 & 0 \\ 1 & 3 & 0 \\ 9 & 6 & 0\end{array}\right]$            ...(2)

$\therefore \mathrm{AB} \neq \mathrm{BA}$     (From eqs. (1) and (2))