Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How does one prove that if 2 systems of linear equations $Ax=c_1$ and $Bx=c_2$ have the same solution set, then they are row equivalent?

I see that it's true from wikipedia:

[...] two linear systems are equivalent if and only if they have the same solution set.

Original question before edits:

For fixed $m$ and $n$, is it possible to have two $m$ by $n$ matrices $A$ and $B$ that are not row equivalent but have the same solution set? What about if $A$ and $B$ are allowed to be of differing dimensions?

share|cite|improve this question
A matrix doesn't have a solution set; rather, an equation has a solution set. I assume you mean the solution set of the homogeneous matrix equations $Ax=0$ and $Bx=0$, that is, the null spaces of $A$ and $B$. – Greg Martin May 11 '12 at 1:19
If $A$ and $B$ have different dimensions then it's very possible: for example, $A$ could be a single row and $B$ could be multiple copies of that row. – Greg Martin May 11 '12 at 1:23
@Greg has answered your question for the differing-dimension case, e.g., the system $x+y=2$ is not row-equivalent to the system $x+y=2,x+y=2$, but it has the same solutions. – Gerry Myerson May 11 '12 at 3:40
The solution set to $Ax=c_1$ is just an affine shift of the solution set to $Ax=0$. So if $Ax=c_1$ and $Bx=c_2$ have the same solution sets, then so do $Ax=0$ and $Bx=0$, and the previous discussion applies. – Greg Martin May 11 '12 at 20:57
Two consistent systems $A\mathbf{x}=\mathbf{c}$ and $B\mathbf{x}=\mathbf{d}$ have the same set of solutions if and only if the rowspace of $A$ equals the rowspace of $B$. See this previous question. – Arturo Magidin May 14 '12 at 19:39

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.