Let $B$ and $C$ be any matrices such that $B \neq C$. For an arbitrary square matrix $A$, $B \neq C$ does not necessarily imply that $AB \neq AC$. This is clear from looking at the trivial case where $A$ is the zero matrix.
However, if $A$ is invertible then (I believe) $B \neq C \implies AB \neq AC$. I'm not sure how to show this. My attempt was $B \neq C = I C = A^{-1} A C$, thus $B \neq A^{-1} A C$. The final step would then be multiplying both sides by $A$, but that would require assuming the premise.
Best Answer
Thank you, Michael Burr!
Given that $B \neq C$, if $A$ is invertible then $AB \neq AC$. We take the contrapositive: given that $B \neq C$, if $AB = AC$ then $A$ is not invertible.
Assume $A$ is invertible with inverse $A^{-1}$, then $AB = AC \implies A^{-1}AB = A^{-1}AC \implies I B = I C \implies B=C$. This is a contradiction, therefore $A$ must not be invertible. This proves the contrapositive and the initial statement.