Can two singular matrices ever be row equivalent?

Question

Since they are singular, they have a zero row when theyre in rref form, so they will not have any solutions (inconsistent linear system). Thats the extent to which i understand. I am not sure if they can be considered row equivalent when they dont have solution(s).

Answer 1

To find an example, start with a singular matrix that's already in rref form and do a row operation on it to make another singular matrix that is row equivalent to it.

Answer 2

Take $A=\begin{bmatrix}1&0&0\\0&0&0\end{bmatrix}$ and $B=\begin{bmatrix}1&0&0\\2&0&0\end{bmatrix}$. Then both $A$ and $B$ are singular and they are row-equivalent as well because their rref will be the same.

Answer 3

Note that a row operation on a matrix $A$ is the same as multiplying it with an elementary matrix $E$ in the front. ($E$ being the matrix obtained from the identity matrix by the same row operation).

And this matrix $E$ is always invertible (the row operation is reversible/can be undone with opposite row operation).

So $A$ and $EA$ are row-equivalent. This is irrespective of whether $A$ is singular or not or whether $Ax=b$ is inconsistent or not (for whatever vector $b$).

If $A$ were singular $EA$ would be singular too.

Answer 4

Yes, of course. Start with any singular matrix and perform row operations on it to get it in RREF. At each step along the way, you produce a different matrix that is row-equivalent to the original one.