If a matrix is not all 0's, then why the same is true for any matrix that A row-reduces to?

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$\space \begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix}$

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 $A = 0$.