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# From the matrix equation $AB=AC$ we can conclude $B=C$ provided.

$\begin{array}{1 1}(A)\;\text{A is singular}\\(B)\;\text{A is non singular}\\(C)\;\text{A is symmetric}\\(D)\;\text{A is square}\end{array}$

Now $|A|$ is non zero
$\therefore A^{-1}$ exists
As $AB=AC$
$\Rightarrow A^{-1}(AB)=A^{-1}(AC)$
$\Rightarrow B=C$