leading 1s in reduced echelon form

Question

I was following a tutorial here; the matrix in echelon form is as follows:

In the video he says that that would be further processed to be in its reduced echelon form and that each row should be a leading $1$. Starting with $-32$ in $R_3$ using a Scale operation so $\frac{1}{-32R_3}\rightarrow R_3$ to suffice a leading $1$ on every row.

While further down the video there was another example here and the reduced echelon form of the matrix he was solving was this:

Why did he not work to produce a leading $1$ in the second matrix; why did he just stop at $R_3$ and let $R_4$ have $0$s and $3$ and no leading one? While in the first matrix he converted the $-32$ entry to become a leading $1$? I understand scaling does not change the solution set but why in this example he worked to have leading $1$s in all rows in the first matrix while that was not the second for the second matrix?

Answer 1

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.