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.
$\begin{bmatrix}1&-1\\1&3\end{bmatrix}$ has eigenvalues $\lambda_1 = 2$ and $\lambda_2 = 2$.
$\begin{bmatrix}1&-1\\0&4\end{bmatrix}$ has eigenvalues $\lambda_1 = 4$ and $\lambda_2 = 1$
On the other hand, note that $2\times 2 = 4 \times 1$.
Best Answer
Assume that a matrix $A$ is in row echelon form. Let $r$ be the number of nonzero rows in $A$. Let $a_{ij_{i}}$ denote the leading $1$ in row $i$.
Notice that $1\leq j_{1} < j_{2} <\ ...<\ j_{r-1}< j_{r} \ $and so $n\leq j_n$ for $n=1,2,...,r$.
Also, for rows $i\leq r,\ $ $a_{ik} = 0$ for $k<j_{i} \implies a_{ik} = 0 \ $ for $k < i\ $ since $i\leq j_i$.
Lastly, rows $i>r$ are zero rows, so $a_{ik} = 0$ for $k<i$.
$\therefore A$ is an upper triangular matrix.