Row Equivalent Matrix Systems have the same Solution Set
# Statement
Let $F$ be a Field, let $m,n \in \mathbb{N}$ and let $A \mathbf{x} = \mathbf{b}$ and $B \mathbf{x} = \mathbf{c}$ be two Row Equivalent Matrix Systemss. Then $${\mathbf{x} \in F^{n} : A \mathbf{x} = \mathbf{b}} = {\mathbf{y} \in F^{n} : B \mathbf{y} = \mathbf{c}}.$$ That is, the two Matrix Equation Systems have the same Solution Set.
# Proof
Since $\begin{pmatrix}A &\Big| &\mathbf{b}\end{pmatrix} \sim_{R} \begin{pmatrix}B &\Big| &\mathbf{c}\end{pmatrix}$, there exists $e_{1}, \dots, e_{k} \in \text{Row}F^{m \times \cdot}$ so that $$e_{1} \circ \cdots \circ e_{k}(\begin{pmatrix}A &\Big| &\mathbf{b}\end{pmatrix}) = \begin{pmatrix}B &\Big| &\mathbf{c}\end{pmatrix}.$$ Since row operations can be represented as matrices, we can rewrite this chain of Function Composition as $$E_{1}\cdots E_{k} \begin{pmatrix}A &\Big| &\mathbf{b}\end{pmatrix} = \begin{pmatrix}B &\Big| &\mathbf{c}\end{pmatrix},$$ where $E_{i}$ is the matrix representation of $e_{i}$ for $i \in [k]$. Since Matrix Multiplication Distributes over Columns we have that $$\begin{align*} E_{1} \cdots E_{k} \mathbf{b} &= \mathbf{c}\\ E_{1} \cdots E_{k} A &= B. \end{align*}$$ Now suppose $\mathbf{x} \in F^{n}$ so that $A \mathbf{x} = \mathbf{b}$. Then $$\mathbf{c} = E_{1} \cdots E_{k} \mathbf{b} = E_{1} \cdots E_{k} A = B$$ so $\mathbf{x} \in {\mathbf{y} \in F^{n} : B \mathbf{y} = \mathbf{0}}$ and $${\mathbf{x} \in F^{n} : A \mathbf{x} = \mathbf{b}} \subset {\mathbf{y} \in F^{n} : B \mathbf{y} = \mathbf{c}}.$$ But since $\begin{pmatrix}A &\Big| &\mathbf{b}\end{pmatrix} \sim_{R} \begin{pmatrix}B &\Big| &\mathbf{c}\end{pmatrix}$ means $\begin{pmatrix}B &\Big| &\mathbf{c}\end{pmatrix} \sim_{R} \begin{pmatrix}A &\Big| &\mathbf{b}\end{pmatrix}$, the same argument gives us that $${\mathbf{x} \in F^{n} : A \mathbf{x} = \mathbf{b}} \supset {\mathbf{y} \in F^{n} : B \mathbf{y} = \mathbf{c}}.$$ Therefore $${\mathbf{x} \in F^{n} : A \mathbf{x} = \mathbf{b}} = {\mathbf{y} \in F^{n} : B \mathbf{y} = \mathbf{c}}.$$ $\blacksquare$
# Remarks
- TODO this could be simplified by looking at invertible matrices as a Group. Then establish that Elementary Row Operations are just invertible matrices.