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.
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.
Best Answer
The key point is that two vectors like
can't be linearly dependent for $a_1\neq 0$ because we can't never obtain the zero vector by linear combinations.
Therefore in the RREF we can show that row vectors are lineraly independent.