# left multiplication by invertible matrix doesn't change reduced row echelon form

How to proof that for the matrices:

$A=\begin{pmatrix} 1 & 0 \\ a & 1 \end{pmatrix}$ and $B=\begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix}$, $X\in M_{23}(\mathbb R)$

$A(BX)$ has the same reduced row echelon form as $X$ ?

Of course I know: $A(BX) <=> (AB)X$, A is an elementary matrix and B is the product of an elementary matrix. I also know, that the reduced row echelon form is distinct. But how to show that ?

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Hint: $A=\begin{pmatrix} 1 & 0 \\ a & 1 \end{pmatrix}$, $B_1=\begin{pmatrix} 2 & 0 \\ 0 & 1 \end{pmatrix}$ and $B_2=\begin{pmatrix} 1 & 0 \\ 0 & 3 \end{pmatrix}$ are elemanter matrices and $B=B_1B_2$
i know there are finite elementary matrices required to transform X to the reduced row echelon form. but i don't know how to show that the multiplication of some arbitrary elementary matrices to X transforms X to X and X can be transformed to the same reduced row echelon form as X –  fast-forward Oct 27 '13 at 21:02
$T_1, T_2 \in M_{23}(\mathbb R)$ are matrices in reduced row echelon form. $E=E_s,E_{s-1}...E_1$ and $F=F_t,F_{t-1}....F_1$ is the product of elementary matrices so that $EX=T_1$ and $F(AB)X=T_2$. $E$, $F$ and $AB$ are invertible because they are a product of elementary matrices: $X=E^{-1}T_1=F^{-1}(AB)^{-1}T_2 => (FABE^{-1})T_1=T_2$ But we have to show that $T_1=T_2$ !? –  fast-forward Oct 27 '13 at 21:55