Exact same solutions implies same row-reduced echelon form? In Hoffman and Kunze they have two exercises where they ask to show that if two homogeneous linear systems have the exact same solutions then they have the same row-reduced echelon form.
They first ask to prove it in the case of $2\times 2$ (Exercise 1.2.6) and then they ask to prove it in the case $2\times 3$ (Exercise 1.4.10).  I am able to prove it in both of these special cases, but as far as I can tell Hoffman and Kunze never tell us whether or not this is true in general. 
So that's my question, is this true in general?  And if not, can anybody provide a counter-example?  Thank you!
 A: Solutions to the homogeneous system associated with a matrix is the same as determining the null space of the relevant matrix. The row space of a matrix is complementary to the null space. This is true not only for inner product spaces, and can be proved using the theory of non-degenerate symmetric bilinear forms. 
So if two matrices of the same order have exactly the same null space, they must also have exactly the same row space. In the row reduced echelon form the nonzero rows form a basis for the row space of the original matrix, and hence two matrices with the same row space will have the same row reduced echelon form.
A: I am (very much) unsure if this counts as a rigorous proof, but given my limited chapter 1 knowledge I have the following argument:
Let A, B be the RREFs of 2 consistent linear systems -(1) with $n$ variables. We claim that the 2 linear systems have the same solution -(2), yet $A\neq B$ -(3). Denote $a_i, b_i$ as the ith row of matrices A and B respectively. Now let $i$ be the smallest integer such that $a_i\neq b_i$.
There are 3 cases where this might be true.
Case 1: The leading variables are different. Let the leading variables be $A_{ij_i }, B_{ik_i} $. Then WLOG, let $j_i<k_i$. This implies that $j_i$ is a pivot column in $A$ but not in $B$. Now using (1), we know that the solution sets for A and B are non-empty. Using (2), we are able to choose some n-tuple of $(x_1,\dots,x_n)$ that is a common solution to $A$ and $B$. Consider A, since $x_{j_i}$ is a leading variable, should we fix, $x_\lambda, \lambda\neq j_i$, there is only 1 possible value of $x_{j_i}$ in $A$. Whereas for $B$, since $x_{j_i}$ is a free variable, it can take any value in the field $\mathbb{F}$.
Case 2: Same leading variable,i.e., $j_i=k_i$ but the free variables are different. WLOG, there must exist some index $l$ such that $A_{i\alpha}=0$ but $B_{i\alpha}\neq 0$. Again, consider an n-tuple particular solution of both $A$ and $B$. Then if we vary $x_l$, and keeping all other free variables and $x_{j_i}$ fixed, the equality represented by $b_i$ would fail, but the equality represented by $a_i$ would still hold.
Case 3: $b_i$ is a zero row or non-existent (i.e. $i$ > no. of rows of B), where $a_i$ is a non-zero row. This would mean $A$ has less free variables than $B$, and it is easy to show that a difference in the number of free variables implies difference in solution sets.
