How to show that $AX=B$ has unique solution for invertible matrix $A$

Question

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

Answer 1

Let $A$ be an $n\times n$ invertible matrix. 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 $X_1:=A^{-1}B$ is a solution of the equation $AX=B$, as desired.

Now, why is $X_1=A^{-1}B$ the only solution? Well, suppose $X_2$ is another solution, meaning $AX_2=B$. Then, left-multiplying by $A^{-1}$ produces $X_2=A^{-1}B$, but we have already shown $X_1 = A^{-1}B$, so $X_2 = X_1$. Therefore, the solution is unique.

Answer 2

If $A$ is invertible then

$$AX=B\iff X=A^{-1}B$$ and the unicity of $X$ comes from the unicity of the inverse matrix $A^{-1}$.