here are the rrefs of $3 \times 3:$
of rank $3:$ $\quad \pmatrix{1&0&0\\0&1&0\\0&0&1}$
of rank $2: \quad\pmatrix{1&0&x\\0&1&x\\0&0&0} , \pmatrix{1&x&0\\0&0&1\\0&0&0}, \pmatrix{0&1&0\\0&0&1\\0&0&0}$
of rank $1: \quad\pmatrix{1&x&x\\0&0&0\\0&0&0}, \pmatrix{0&1&x\\0&0&0\\0&0&0}, \pmatrix{0&0&1\\0&0&0\\0&0&0}$
of rank $0: \quad \pmatrix{0&0&0\\0&0&0\\0&0&0}$
where $x$ represents an arbitrary number. hopefully this is complete.
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
There is "row echelon form" and "reduced row echelon form"; the stricter of the two is the last.
The linked entry contains the criteria for each, first the row echelon form, then additional criteria for reduced row echelon form. So using the term "echelon form", or even the casual use of the terms "row echelon form", can be ambiguous, depending on context, and depending on textbooks, to some degree, unless an explicit distinction is made.
Also, as an aside, consistency isn't a property of a matrix, per se, but of the system of equations that an augmented coefficient matrix may represent. For your matrices, the first reveals that there exist two linearly independent row (and column) vectors. Your second matrix reveals that the row vectors (and column vectors) are linearly dependent. We can also determine the rank of a matrix from a matrix in echelon form: the rank of a matrix is equal to the number of non-zero rows in the matrix when reduced to row echelon form.