# Same solution implying row equivalence?

Suppose $R$ and $R ‘$ are $2 \times 3$ row-reduced echelon matrices and that the system $Rx=0$ and $R’x=0$ have exactly the same solutions. Prove that $R=R’$.

In general, is it true that any 2 $m \times n$ matrices that have the same solution must be row-equivalent?

#### Solutions Collecting From Web of "Same solution implying row equivalence?"

There are seven possible $2\times3$ row-reduced echelon matrices:

R_1=\left[\begin{array}{ccc}
0 &0 &0 \\
0 & 0& 0
\end{array}\right]
\label{m1}

R_2=\left[\begin{array}{ccc}
1 &0 &a \\
0 &1 & b
\end{array}\right]
\label{m2}

R_3=\left[\begin{array}{ccc}
1 & a& 0\\
0 &0 &1
\end{array}\right]
\label{m3}

R_4=\left[\begin{array}{ccc}
1 &a & b\\
0& 0& 0
\end{array}\right]
\label{m4}

R_5=\left[\begin{array}{ccc}
0& 1& a\\
0&0 &0
\end{array}\right]
\label{m5}

R_6=\left[\begin{array}{ccc}
0&1 &0 \\
0&0 &1
\end{array}\right]
\label{m6}

R_7=\left[\begin{array}{ccc}
0& 0&1 \\
0& 0&0
\end{array}\right]
\label{m7}

We must show that no two of these have exactly the same solutions. For the first one $R_1$, any $(x,y,z)$ is a solution and that’s not the case for any of the other $R_i$’s. Consider next $R_7$. In this case $z=0$ and $x$ and $y$ can be anything. We can have $z\not=0$ for $R_2$, $R_3$ and $R_5$. So the only ones $R_7$ could share solutions with are $R_3$ or $R_6$. But both of those have restrictions on $x$ and/or $y$ so the solutions cannot be the same. Also $R_3$ and $R_6$ cannot have the same solutions since $R_6$ forces $y=0$ while $R_3$ does not.

Thus we have shown that if two $R_i$’s share the same solutions then they must be among $R_2$, $R_4$, and $R_5$.

The solutions for $R_2$ are $(-az, -bz, z)$, for $z$ arbitrary. The solutions for $R_4$ are $(-a’y-b’z,y,z)$ for $y,z$ arbitrary. Thus $(-b’,0,1)$ is a solution for $R_4$. Suppose this is also a solution for $R_2$. Then $z=1$ so it is of the form $(-a,-b,1)$ and it must be that $(-b’,0,1)=(-a,-b,1)$. Comparing the second component implies $b=0$. But if $b=0$ then $R_2$ implies $y=0$. But $R_4$ allows for arbitrary $y$. Thus $R_2$ and $R_4$ cannot share the same solutions.

The solutions for $R_2$ are $(-az, -bz, z)$, for $z$ arbitrary. The solutions for $R_5$ are $(x,-a’z,z)$ for $x,z$ arbitrary. Thus $(0,-a’,1)$ is a solution for $R_5$. As before if this is a solution of $R_2$ then $a=0$. But if $a=0$ then $R_2$ forces $x=0$ while in $R_5$ $x$ can be arbitrary. Thus $R_2$ and $R_5$ cannot share the same solutions.

The solutions for $R_4$ are $(-ay-bz,y,z)$ for $y,z$ arbitrary. The solutions for $R_5$ are $(x,-a’z,z)$ for $x,z$ arbitrary. Thus setting $x=1$, $z=0$ gives $(1,0,0)$ is a solution for $R_5$. But this cannot be a solution for $R_4$ since if $y=z=0$ then first component must also be zero.

Thus we have shown that no two $R_i$ and $R_j$ have the same solutions unless $i=j$.

We can actually write out all the $2\times 3$ row-reduced echelon matrices, and check that if $RX=0$ and $R’X=0$ have the same solution, then $R=R’$, though this is very inelegant and does not help the $(m\times n)$ case.