In addition to lessening the workload involved in computing a determinant of a square matrix, e.g.,
- one can confirm or rule out whether a square matrix is invertible by stopping before full reduced row-echelon form,
- we can determine the rank of a matrix without needing to go through the tedious work sometimes involved in obtaining full reduced form row-echelon form, and
- we can determine whether the row or column vectors of a matrix are linearly dependent or independent with just plain old row-echelon form.
Certainly, each form has its uses, and the fact that we can sometimes avoid extensive and tedious computations (which also leave room for introducing simple errors along the way), that's not to say that it is not important to know how and when to obtain reduced row echelon form: but we can "pick-and-choose" to some extent "how far" we need to row-reduce.
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.
Best Answer
Row-echelon form (REF):
(i) Leading nonzero entry of each row is 1.
(ii) The leading 1 of a row is strictly to the right of the leading 1 of the row above it.
(iii) Any all-zero rows are at the bottom of the matrix.
$ $
Reduced row-echelon form (RREF):
(i) REF.
(ii) The column of any row-leading 1 is cleared (all other entries are 0).
$ $
Note: A given matrix (generally) has more than one row-echelon form; however, for any matrix, the reduced row-echelon form is unique. This uniqueness allows one to determine if two matrices are row equivalent (can one be transformed to the other by a sequence of elementary row operation).