For three matrices $A, B$ and $C$ of the same order, if $A$=$B$, then $AC$=$BC$, but converse is not true.

Question

For three matrices $A, B$ and $C$ of the same order, if $A$=$B$, then $AC$=$BC$, but converse is not true.

I guess $A,B,C$ all are square matrices of the same order. $$ A=B $$ multiplying by $C$ from right side, $$ \implies AC=BC $$

But why is the second statement "converse is not true" ?

$AC=BC$, multiplying by $C^{-1}$ from the right $$ ACC^{-1}=BCC^{-1}\implies AI=BI\implies A=B $$ right ?

Answer 1

Hint.

Consider: $$A= \begin{bmatrix} 1&0\\0&2 \end{bmatrix} \qquad B= \begin{bmatrix} 1&0\\0&3 \end{bmatrix} \qquad C= \begin{bmatrix} 0&1\\0&0 \end{bmatrix} $$

and note that $C$ is not invertible