Answer
$AX=B$ has a unique solution $A^{-1}B$
Work Step by Step
Let $A$ be an invertible $n$$\times$$n$ matrix and let $B$ be an $n$$\times$$p$ matrix.
Since $A$ is invertible, $A^{-1}B$ satisfies $AX=B$
$A(A^{-1}B)=AA^{-1}B=IB=B$
Let $X$ be a solution of $AX=B$. Then:
$A^{-1}(AX)=A^{-1}B$
$IX=A^{-1}B$
$X=A^{-1}B$
