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.
The characteristic polynomial is
$$
\det(\lambda I-A)=
\det\begin{bmatrix}
\lambda-\cos\theta & \sin\theta & 0 \\
-\sin\theta & \lambda-\cos\theta & 0 \\
0 & 0 & \lambda-1
\end{bmatrix}
=
(\lambda-1)
\det\begin{bmatrix}
\lambda-\cos\theta & \sin\theta\\
-\sin\theta & \lambda-\cos\theta
\end{bmatrix}
$$
and, finally,
$$
(\lambda-1)(\lambda^2-2\lambda\cos\theta+1)
$$
and the eigenvalues are $1$, $\cos\theta+i\sin\theta$ and $\cos\theta-i\sin\theta$.
An eigenvector for $1$ is readily computable as $[0\ 0\ 1]^T$, but if you're interested in finding a row echelon form of
$$
\begin{bmatrix}
1-\cos\theta & \sin\theta & 0 \\
-\sin\theta & 1-\cos\theta & 0 \\
0 & 0 & 0
\end{bmatrix}
$$
you can do as follows; set $\theta=2\varphi$, so $1-\cos\theta=2\sin^2\varphi$ and $\sin\theta=2\sin\varphi\cos\varphi$.
The case in which $\sin\varphi=0$ must be dealt with separately; assume $\sin\varphi\ne0$:
\begin{align}
\begin{bmatrix}
2\sin^2\varphi & 2\sin\varphi\cos\varphi & 0 \\
-2\sin\varphi\cos\varphi & 2\sin^2\varphi & 0 \\
0 & 0 & 0
\end{bmatrix}
&\to
\begin{bmatrix}
1 & \cot\varphi & 0 \\
-2\sin\varphi\cos\varphi & 2\sin^2\varphi & 0 \\
0 & 0 & 0
\end{bmatrix}
&&R_1\gets\frac{1}{2\sin^2\varphi}R_1
\\[6px]&\to
\begin{bmatrix}
1 & \cot\varphi & 0 \\
0 & 2 & 0 \\
0 & 0 & 0
\end{bmatrix}
&&R_2\gets R_2+2R_1\sin\varphi\cos\varphi
\\[6px]&\to
\begin{bmatrix}
1 & \cot\varphi & 0 \\
0 & 1 & 0 \\
0 & 0 & 0
\end{bmatrix}
&&R_2\gets \frac{1}{2}R_2
\\[6px]&\to
\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 0
\end{bmatrix}
&&R_1\gets R_1-R_2\cot\varphi
\end{align}
The case $\sin\varphi=0$ corresponds to $\theta=2k\pi$, when the original matrix is the identity.
Best Answer
The matrices you have are in row echelon form for they satisfy the following conditions:
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:
So, as mentioned by NasuSama, the second can be simplified to reduced row echelon form.