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Suppose $R$ and $R'$ are $2\times3$ row-reduced echelon matrices and that the systems $RX=0$ and $R'X=0$ have exactly the same solutions. Prove that $R=R'$.

Any hints? I'm tring to use the fact that the matrices are echelon row-reduced but I'm struggling on it because we don't know the entries of both matrices, and analysing each case seperately seems too much work.

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The actual size of $R$ and $R'$ doesn't matter, as long as they are the same size and echelon row-reduced. We then have $$ R=\begin{bmatrix}I & B\\ 0&0\end{bmatrix},\ \ \ R'=\begin{bmatrix}I & C\\ 0&0\end{bmatrix} $$ for appropriate matrices $B$ and $C$. The assertion $Rx=0$ if and only if $R'x=0$ then says \begin{align} x_1+Bx_2=0& \iff \begin{bmatrix}I & B\\ 0&0\end{bmatrix}\begin{bmatrix}x_1\\ x_2\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix} \iff Rx=0 \\ \ \\ &\iff R'x=0 \iff \begin{bmatrix}I & C\\ 0&0\end{bmatrix}\begin{bmatrix}x_1\\ x_2\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}\\ \ \\ &\iff x_1+Cx_2=0. \end{align} So, if we fix $x_2$ and put $x_1=-Bx_2$, we get $x_1+Bx_2=0$, and so $x_1+Cx_2=0$. That is, $$ Bx_2=-x_1=Cx_2. $$ As we can do this for all $x_2$, it follows that $B=C$. Thus $R=R'$.

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