# Row rank and solutions to $Ax=b$

Let $A$ be an $n\times m$ matrix, $x$ is an $m\times 1$ vector, and $b$ is an $n\times 1$ vector.

The system of equations $Ax=b$ has a solution iff the row rank of $A$ equals the row rank of $A|b$ ($A$ augmented with $b$).

My attempt at a proof: Let $A'|b'$ represent the row reduced form of $A|b$. Suppose that $Ax=b$ has a solution. Then $A'|b'$ has a solution if and only if the nonzero rows in $A'$ are nonzero rows in $b'$. Since row reduction doesn't affect rank, we have that $rank(A)=rank(A|b)$.

I'm stuck on this problem working backwards.

• is it a Rouché–Capelli theorem? – D F Feb 1 '18 at 18:42
• Not sure. It certainly wasn't named in the book. – emka Feb 1 '18 at 19:33
• – D F Feb 1 '18 at 19:34

Hint: This a proof by contradiction. Suppose for contradiction that $Ax = b$ does not have a single solution. You want to contradict the fact that $\text{rank}(A) = \text{rank}(A|b)$. How would you do that?
• @emka You misunderstood the premise. If $Ax=b$ has no solutions, then $rank(A)< rank(A|b)$. This contradicts the assumption that $rank(A)=rank(A|b)$. Hence we’re done. – user513685 Feb 2 '18 at 1:04