# Showing that if an equation has a unique solution for one variable, then it has unique solutions for all.

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 $B$.

Thanks!

• Do you know that the equation $AX=B$ has a unique solution if, and only if, $A$ is invertible? Jul 22, 2013 at 23:14
• No. Why is this so?
– user85362
Jul 22, 2013 at 23:15
• Your solution looks more or less fine to me (though it needs some patching up). The point is that, regardless of what B is, you solve the equation by row-reducing A, and you can either hit the identity (in which case AX=B has exactly one solution) or you can't (in which case it has no solution or many solutions). Jul 22, 2013 at 23:16
• @AbhishekMallela Too long for a comment, but Jim is mentioning that in his answer. Maybe he'll elaborate on that. But you can find that on any linear algebra book. Jul 22, 2013 at 23:17
• Yeah, I just found it in my book. Thanks.
– user85362
Jul 22, 2013 at 23:24

This statement is not precise. There are several ways to fix it, depending on how much you know. Do you know what non-singular matrices are? Do you know that they are invertible and that if $A$ can be reduced to $I_n$ then it is nonsingular? If you know that $A$ is invertible then from $AX = B$ you can write $X = A^{-1}B$ so there is only one choice for $X$ no matter what $B$ is.
Another way of seeing that the solution is unique (that doesn't use non-singularity explicitly) is the following. As $A$ reduces to $I_n$, when you reduce the augmented matrix $[A \ | \ B]$ do any of the choices you make depend on $B$? Try arguing that no matter what that last column is, reducing the augmented matrix will always yield something with a pivot in each of the first $n$ columns. Thus there will be no independent variables in your solution. I suspect that this is what you had in mind with what you wrote, but you should explain it a little further.
Your argument is essentially correct, but the end of it is a bit vague. What you’ve shown is that if $AX=B$ has a unique solution for some $B$, then $A$ can be row-reduced to $I_n$. This row-reduction is independent of the augmentation column in the augmented matrix, so for any $n\times 1$ column vector $C$ we can row-reduce the augmented matrix $[A\mid C]$ to some $[I_n\mid Y]$, and $Y$ will be the unique solution to $AX=C$. Thus, $AX=C$ has a unique solution for each $C$.
There are many ways to see why this shows that $A$ is invertible (non-singular). If you know that when $[A\mid I_n]$ can be row-reduced to something of the form $[I_n\mid D]$, then $D=A^{-1}$, the result is clear. If you know that row-reduction can be accomplished by premultiplying by elementary matrices, then you can argue that row-reducibility of $A$ to $I_n$ means that there are elementary matrices $E_1,\dots,E_m$ such that $E_mE_{m-1}\dots E_2E_1A=I_n$ and hence $E_mE_{m-1}\dots E_2E_1=A^{-1}$.