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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.

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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?

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  • $\begingroup$ Are you saying Suppose that rank(A)=rank(A|b) and Ax=b has no solution....? In this case, the number of linearly independent rows of A are the same as those in A|b. The only other fact that I know is that a system of linear equations is consistent iff the number of row reduced nonzero rows of A is the same as the number of row reduced nonzero rows of A|b $\endgroup$ – emka Feb 1 '18 at 18:57
  • $\begingroup$ @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. $\endgroup$ – user513685 Feb 2 '18 at 1:04

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