The matrices you have are in row echelon form for they satisfy the following conditions:
- All nonzero rows (rows with at least one nonzero element) are above any rows of all zeroes (all zero rows, if any, belong at the bottom of the matrix).
- The leading coefficient (the first nonzero number from the left, also called the pivot) of a nonzero row is always strictly to the right of the leading coefficient of the row above it (some texts add the condition that the leading coefficient must be $1$.).
- All entries in a column below a leading entry are zeroes (implied by the first two criteria.).
However, the second matrix is not in reduced row echelon form because on top of being in row echelon form it also needs to satisfy:
- Every leading coefficient is 1 and is the only nonzero entry in its column.
So, as mentioned by NasuSama, the second can be simplified to reduced row echelon form.
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
In a logical sense, yes. The zero matrix is vacuously in RREF as it satisfies:
All zero rows are at the bottom of the matrix
The leading entry of each nonzero row subsequently to the first is right of the leading entry of the preceding row.
The leading entry in any nonzero row is a 1.
All entries in the column above and below a leading 1 are zero.