I have a problem and a proposed solution. Please tell me if I'm correct.
Problem: Let $A$ be a square matrix. Show that if the system $AX=B$ has a unique solution for some particular column vector B, then it has a unique solution for all $B$.
Solution: If $AX=B$ has a unique solution for some column vector $B$, then $A$ in reduced row echelon form has a pivot in each column and $A$ can be reduced to $I_n$, for $A$,$\\ n \times n$. Since the number of equations = the number of unknowns, we will have column vector $(n \times 1)$ of $x_i$'s = column vector $n \times 1$ of $b_i$'s. Hence, varying $B$ is equivalent to varying $X$ and will create a new solution for every change made to $B$.