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.
When you perform elementary row operations, then linear relations between columns don't change. More precisely, if $A=[a_1\;a_2\;\dots\;a_n]$ is a matrix (given with its columns $a_i$) and $B$ is obtained by $A$ by performing elementary row operations, if $1\le i_1,i_2,\dots,i_j,p\le n$ are column indices, then
$$
b_p=\alpha_1b_{i_1}+\alpha_2b_{i_2}+\dots+\alpha_jb_{i_j}
\quad\text{if and only if}\quad
a_p=\alpha_1a_{i_1}+\alpha_2a_{i_2}+\dots+\alpha_ja_{i_j}
$$
This follows from the fact that performing row operations is the same as multiplying $A$ on the left by invertible matrices. So $B=FA$ for some invertible matrix $F$ and proving the above results is easy.
In particular, a set of columns of $B$ is linearly independent if and only if the corresponding set of columns of $A$ is linearly independent.
If $U=[u_1\;u_2\;\dots\;u_n]$ is a reduced row echelon form for $A$, let $u_{i_1},u_{i_2},\dots,u_{i_j}$ be the pivot columns, with $i_1<i_2<\dots<i_j$.
Note that if $i_j<p<i_{j+1}$, then $u_p$ is a linear combination of $u_{i_1},\dots,u_{i_j}$ and the coefficients of the linear combination are determined by $u_p$: if
$$
u_p=\begin{bmatrix}\alpha_1\\\alpha_2\\\vdots\\\alpha_m\end{bmatrix}
$$
then $\alpha_i=0$ if $i>i_j$ and
$$
u_p=\alpha_1u_{i_1}+\alpha_2u_{i_2}+\dots+\alpha_{i_j}u_{i_j}
$$
Therefore
$$
a_p=\alpha_1a_{i_1}+\alpha_2a_{i_2}+\dots+\alpha_{i_j}a_{i_j}
$$
and this linear combination is unique, because the set of columns
$$
\{a_{i_1},a_{i_2},\dots,a_{i_j}\}
$$
is linearly independent.
Thus the position of the pivot columns in $U$ is uniquely determined by the columns of $A$ and the coefficients on the non-pivot columns are likewise determined by the linear relations between the columns of $A$.
Best Answer
As you pointed out, there's no change in the solution set if he changes $R_3$ to $-\frac{1}{32}R_3$ in the first example. The same can be said for the second example; $[0\text{ }0\text{ }0\text{ }|\text{ }3]$ describes the same set as $[0\text{ }0\text{ }0\text{ }|\text{ }1]$, which, as he (and the above commenter) pointed out, doesn't make any sense.
However, the reason why the youtube guy did this, if you continue watching on, is because he wanted to get the matrices to row reduced echelon form. In other words, he divided by $-32$ so that he might simplify the matrix further. The second matrix, you'll note, is already $0$'ed and $1$'ed for all other rows bar $R_4$, and we can see that $R_4$ is inconsistent, so there's really no need to do anything to it.
tl;dr: Simplifying the inconsistent row in the first matrix helped us gain more info about the matrix, while doing that in the second matrix would not have.