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

Question:

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

(i) $A=\left[\begin{array}{rrr}1 & 3 & -1 \\ 2 & -1 & -1 \\ 3 & 0 & -1\end{array}\right]$ and $B=\left[\begin{array}{lll}-2 & 3 & -1 \\ -1 & 2 & -1 \\ -6 & 9 & -4\end{array}\right]$

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

Solution:

(i)

$A B=\left[\begin{array}{ccc}1 & 3 & -1 \\ 2 & -1 & -1 \\ 3 & 0 & -1\end{array}\right]\left[\begin{array}{ccc}-2 & 3 & -1 \\ -1 & 2 & -1 \\ -6 & 9 & -4\end{array}\right]$

$\Rightarrow A B=\left[\begin{array}{ccc}-2-3+6 & 3+6-9 & -1-3+4 \\ -4+1+6 & 6-2-9 & -2+1+4 \\ -6-0+6 & 9+0-9 & -3-0+4\end{array}\right]$

$\Rightarrow A B=\left[\begin{array}{ccc}1 & 0 & 0 \\ 3 & -5 & 3 \\ 0 & 0 & 1\end{array}\right]$                 $\ldots(1)$

Also,

$B A=\left[\begin{array}{lll}-2 & 3 & -1 \\ -1 & 2 & -1 \\ -6 & 9 & -4\end{array}\right]\left[\begin{array}{ccc}1 & 3 & -1 \\ 2 & -1 & -1 \\ 3 & 0 & -1\end{array}\right]$

$\Rightarrow B A=\left[\begin{array}{ccc}-2+6-3 & -6-3+0 & 2-3+1 \\ -1+4-3 & -3-2+0 & 1-2+1 \\ -6+18-12 & -18-9+0 & 6-9+4\end{array}\right]$

$\Rightarrow B A=\left[\begin{array}{ccc}1 & -9 & 0 \\ 0 & -5 & 0 \\ 0 & -27 & 1\end{array}\right]$              $\ldots(2)$

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