If $A$ is an invertible $n \times n$ matrix, show that $AX=B$ has a unique solution for any $n \times k$ matrix $B$.
I'm not sure where to start. What I have is that, if $A$ is invertible then there exists a matrix $C$ such that $AC=CA=I$. I am having trouble understanding how this relates to their existing a unique solution for any matrix $B$.
Thanks
Best Answer
Let $A$ be an $n\times n$ invertible matrix. Then by definition there exists a matrix $A^{-1}$ such that $A^{-1}A=A^{-1}A=I_n$. Now consider the equation $AX=B$. Left Multiplying both sides by $A^{-1}$ yields $A^{-1}AX=I_nX=X$ on the left hand side and $A^{-1}B$ on right hand side. Thus $X=A^{-1}B$ is a solution of the equation $AX=B$, as desired. Now why is this the unique solution? Well if $X_2$ is another solution then $AX_2=B$, then by left multiplying by $A^{-1}$ we get $X_2=A^{-1}B=X$, so $X$ is the unique solution.