Proving a linear system $A\bf{x} = \bf{b}$ is consistent iff the rank of ($A\;|\;\bf{b}$) equals the rank of $A$ 
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.
 A: I would first show, or appeal to,  the following:
$\ \ 1)$ The rank of a matrix is the number of nonzero rows in an echelon form of the matrix. 
$\ \ 2)$ An echelon form of $[A\,|\,\bf b]$ also gives an echelon form of A. 
So the rank of $[A\,|\,\bf b]$ as at least the rank of $A$ and if the rank of $[A\,|\,\bf b]$ exceeds the rank of $A$, then there is a row of $[A\,|\,\bf b]$ with zeroes in all entries except for the last entry (that is you have an echelon form as in your post). 
A: I'll assume the definition of the rank of a matrix as the dimension of the space spanned by its columns, which space is the image of the linear map $\mathbb x\mapsto A\,\mathbb x$ that is represented by $A$. Clearly the rank of $(A~b)$ cannot be less than that of $A$. Now obviously $A\,\mathbf x=\mathbf b$ has a solution if and only if $\mathbb b$ is in the image of the linear map represented by $A$, and this means that adding $\mathbb b$ as an additional column to $A$ does not increase the rank, so it stays equal. Conversely, if  $A\,\mathbf x=\mathbf b$ does not have a solution then $\mathbb b$ is not in the mentioned image, so adding $\mathbf b$ strictly increases the span of the columns of the matrix, and therefore the rank (by $1$).
