Let

Here,$A=\left[\begin{array}{lll}1 & 1 & 1 \\ 3 & 3 & 3\end{array}\right]$ $B=\left[\begin{array}{cc}3 & 1 \\ 5 & 2 \\ -2 & 4\end{array}\right]$ and $C=\left[\begin{array}{cc}4 & 2 \\ -3 & 5 \\ 5 & 0\end{array}\right]$

Now,

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

$\Rightarrow A B=\left[\begin{array}{cc}3+5-2 & 1+2+4 \\ 9+15-6 & 3+6+12\end{array}\right]$

$\Rightarrow A B=\left[\begin{array}{cc}6 & 7 \\ 18 & 21\end{array}\right]$

$A C=\left[\begin{array}{lll}1 & 1 & 1 \\ 3 & 3 & 3\end{array}\right]\left[\begin{array}{cc}4 & 2 \\ -3 & 5 \\ 5 & 0\end{array}\right]$

$\Rightarrow A C=\left[\begin{array}{cc}4-3+5 & 2+5+0 \\ 12-9+15 & 6+15+0\end{array}\right]$

$\Rightarrow A C=\left[\begin{array}{cc}6 & 7 \\ 18 & 21\end{array}\right]$

So, $A B=A C$ though $B \neq C, A \neq O$.