In short, row reduced echelon form(RREF) of a matrix $A$ is such that
i) Every leading entry is 1
ii) Any nonzero rows are above zero rows
iii) any leading entry is strictly to the right of any leading entries above that row
iv) any other entry in a column containing a leading entry is 0 except for the leading entry.
So it does not have to be put in augmented matrix $[A|b]$ to get a RRE form. You are comparing RRE form of matrix $A$ and $[A|b]$.
To see why the statement is true, suppose that you put the matrix $[A|b]$ into RRE form, so you have a matrix E. If E contains a leading entry in its last column, in terms of system of equations, what does it say? And what is the condition for E to not have any leading entry in last column?
Note: If RRE form of $[A|b]$ does contain a leading entry, then it is different from that of $A$. Also, note that RRE form of $[A|b]$ is m by n+1 whereas that of $A$ is m by n.
Solve:
$x+y=1$
$x+y=2$
Then we have
$\
A =
\left( {\begin{array}{cc}
1 & 1 \\
1 & 1
\end{array} } \right)
$
$\
b =
\left( {\begin{array}{cc}
1 \\
2
\end{array} } \right)
$
and $Ax=b$
If we turn A into RREF, we get
$\
E =
\left( {\begin{array}{cc}
1 & 1 \\
0 & 0
\end{array} } \right)
$
So A has rank 1
and if we put $[A|b]$ into RRE form, we get
$\
E' =
\left( {\begin{array}{cc}
1 & 1 & 0 \\
0 & 0 & 1
\end{array} } \right)
$
So augmented matrix has rank 2. Observe what last row says in terms of equations.
For the resolution of linear systems, Gaussian elimination is preferred over Gauss-Jordan (these parallel the echelon and reduced echelon forms) as the former involves less operations (roughly $n^3/3$ vs $n^3/2$), for a similar numerical stability.
Anyway, Gauss-Jordan does not require a backsubstitution step, which makes the code a little more compact. On the other hand, the reduced form is not compatible with LU decomposition.
From a theoretical point of view, the reduced form has the advantage of being uniquely defined for a given matrix.
Anyway, it is worth to note that for an invertible matrix, the reduced form is simply a unit matrix, totally uniniteresting. The concept is only useful for degenerate cases.
Best Answer
Row echelon form is simply a matrix such that all nonzero rows are above rows of all zereos, and the leading coefficient is a nonzero row is strictly to the right of the leading coefficient of the row above it.
Reduced row echelon form is a matrix that is in row echelon form but adds the condition that the leading coefficient is the only nonzero element in a column.
It may be tedious to go from row echelon form to reduced row echelon form but there are equivalent by a finite number of steps. Thus since they are equivalently matrix then the rank of the matrix must be the same.