If the determinants of two matrices, $A$ and $B$, of order $n$, $= 0$, are they row-equivalent?

Question

Question is in the title. I'm thinking no, they aren't, and a counter-example below:

Let's say the reduced row-echelon forms of matrices $A$ and $B$ are below:

$$\text{rref}(A) = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix}\text{,} \quad \text{rref}(B) = \begin{pmatrix} 1 & 0 & 2 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix}$$

Hence, it is clear that although $\text{det}(A) = \text{det}(B) = 0$, since the reduced row-echelon forms of matrices are unique, A and B are not row-equivalent.

Is this a sufficient and logical proof?

Thanks in advance.

Answer 1

Yes, it is fine. A simpler example would be$$A=\begin{pmatrix}0&0&0\\0&0&0\\0&0&0\end{pmatrix}\text{ and }B=\begin{pmatrix}1&0&0\\0&0&0\\0&0&0\end{pmatrix}.$$