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enter image description here Source: https://math.colorado.edu/~monkd/m3130/rowred.pdf

It seems the author uses A as a row-reduced echelon form at first in the second line "so that A has only one...", and then a matrix before it's not row-reduced from the fourth line "Clearly if A is all 0's...".

But I don't understand that if A is not all $0's$, then clearly the same is true for any matrix row-equivalent to A.
For example let A be $3×1 $ matrix as follows:
$\begin{bmatrix} 1\\ 0\\ 1 \end{bmatrix}$

Then we can make $R_1$ and $R_2$ 0 by subtracting the entries in the rows each other. So if A is not all 0's, a row-equivalent matrix can be all 0's.

[EDIT I checked again] So
$\begin{bmatrix} 1\\ 0\\ 1 \end{bmatrix} \space becomes\space by $R_3-R_1$\space \begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix} $

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You can't perform both row operations at the same time, so your example doesn't quite work. If for instance you first replace the last row with itself minus the first row, the last row becomes zero, but then it's impossible to make the first row zero.

In general, it helps to recall that a row operation simply corresponds to left-multiplication by an elementary matrix $E$, and that an elementary matrix is always invertible. Therefore if $A \neq 0$, we must have $EA \neq 0$, because otherwise we could just left-multiply by $E^{-1}$ to obtain $A = 0$.

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