Prove that a linear system $A\bf{x} = \bf{b}$ is consistent if and only if the rank of ($A\;|\;\bf{b}$) equals the rank of A.
I can see that it's impossible for the rank of ($A\;|\;\bf{b}$) to be less than the rank of $A$, and if the rank of ($A\;|\;\bf{b}$) is greater than the rank of $A$, then in echelon form we'll have something like:
$(A\;|\;\bf{b}) = \left[\begin{array}{rrrr|r}1 & 2 & 3 & 4 & 5 \\ 0 & 1 & 2 & 3 & 7 \\ 0 & 0 & 0 & 1 & 3 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0\end{array}\right]$
which is clearly inconsistent, but I don't know how to express it. I think I need to say that $\bf{b}$ is not in the column space of A.