If A, B and C are square matrices of same order, then AB=AC always implies that B=C.

Correct Answer False

If $AB = AC = 0$, then it can be possible that B and C are two non-zero matrices such that $B \ne C$.

$\therefore$ $A \cdot B = 0 = A \cdot C$
Let $A = \left[ {\begin{array}{llllllllllllllllllll}1&0\\0&0\end{array}} \right]$, $B = \left[ {\begin{array}{llllllllllllllllllll}0&0\\1&3\end{array}} \right]$

and $C = \left[ {\begin{array}{llllllllllllllllllll}0&0\\3&1\end{array}} \right]$

$\therefore$ $AB = \left[ {\begin{array}{llllllllllllllllllll}1&0\\0&0\end{array}} \right]\left[ {\begin{array}{llllllllllllllllllll}0&0\\1&3\end{array}} \right] = \left[ {\begin{array}{llllllllllllllllllll}0&0\\0&0\end{array}} \right]$

and $AC = \left[ {\begin{array}{llllllllllllllllllll}1&0\\0&0\end{array}} \right] \cdot \left[ {\begin{array}{llllllllllllllllllll}0&0\\3&1\end{array}} \right] = \left[ {\begin{array}{llllllllllllllllllll}0&0\\0&0\end{array}} \right]$

$\Rightarrow$ $AB = AC$but $B \ne C$