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 $B$.


  • $\begingroup$ Do you know that the equation $AX=B$ has a unique solution if, and only if, $A$ is invertible? $\endgroup$
    – Git Gud
    Jul 22, 2013 at 23:14
  • $\begingroup$ No. Why is this so? $\endgroup$
    – user85362
    Jul 22, 2013 at 23:15
  • $\begingroup$ 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). $\endgroup$
    – Billy
    Jul 22, 2013 at 23:16
  • $\begingroup$ @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. $\endgroup$
    – Git Gud
    Jul 22, 2013 at 23:17
  • $\begingroup$ Yeah, I just found it in my book. Thanks. $\endgroup$
    – user85362
    Jul 22, 2013 at 23:24

2 Answers 2


Hence, varying B is equivalent to varying X and will create a new solution for every change made to B.

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 $E_mE_{m-1}\dots E_2E_1=A^{-1}$.


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