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.
When the matrix is in echelon form, for each non-zero row $R_i$, you can divide the row by its leading non-zero value. that makes the leading value $1$. Next for each row $R_n$ above $R_i$, you can subtract $R_i$ multiplied by the entry in row $R_n$ in the same column as that leading $1$ from $R_n$. This results in $R_n$ having $0$ in that column. If you follow this procedure starting with the first row and going down, by the time you are done, the matrix will be reduced echelon form, and guess what! Those leading $1$s that define the pivot points are exactly the locations of the leading non-zero values in each row back before it was reduced.
I.e., When a matrix is in echelon form, the pivot points are exactly the leading non-zero values in each row. Quite frankly, if I had written the definition, that's how I would have defined it, since the two are equivalent, and you need to know them before you get in reduced echelon form.
For example, in your matrix, I marked the leading non-zero entries in red:
$$\begin{bmatrix}
\color{red}1 &4 &5 &-9 &7 \\
0 &\color{red}2 &4 &-6 &-6 \\
0 &0 &0 &\color{red}{-5} &0 \\
0 &0 &0 &0 &0 \\
\end{bmatrix}$$
First divide each row by its leading non-zero value
$$\begin{bmatrix}
\color{red}1 &4 &5 &-9 &7 \\
0 &\color{red}1 &2 &-3 &-3 \\
0 &0 &0 &\color{red}1 &0 \\
0 &0 &0 &0 &0 \\
\end{bmatrix}$$
Subtract $4$ times Row 2 from Row 1:
$$\begin{bmatrix}
\color{red}1 &0 &-3 &3 &19 \\
0 &\color{red}1 &2 &-3 &-3 \\
0 &0 &0 &\color{red}1 &0 \\
0 &0 &0 &0 &0 \\
\end{bmatrix}$$
Subtract $3$ times row 3 from row 1, and add 3 times row 3 to row 2:
$$\begin{bmatrix}
\color{red}1 &0 &-3 &0 &19 \\
0 &\color{red}1 &2 &0 &-3 \\
0 &0 &0 &\color{red}1 &0 \\
0 &0 &0 &0 &0 \\
\end{bmatrix}$$
And now, we are in reduced echelon form. See that the the pivot points from the definition are in the same locations as the echelon from leading non-zero values.
Best Answer
Row-echelon form doesn’t require that the leftmost nonzero element of a row be $1$. It simply has to be strictly farther right than the leftmost nonzero element of the preceding row (and, of course, that all zero rows come after all nonzero rows). That said, some texts do insist on a leading $1$. The text you’ve quoted isn’t one of them.