Yes, the matrix can still be in row-reduced echelon form (RREF) even if it corresponds to a system with no solution, such as when you encounter a row like $0 \ 0 \ 0 \ | \ 1$ (which represents the equation $0 = 1$).

So in your case, while the system of equations has no solution, the matrix could still be row-reduced. The condition for being in row-reduced form is about the structure of the matrix itself, not the solvability of the system.

Do you want more details or have any other questions?

Here are some related questions:

1. What are the specific criteria for a matrix to be in row-reduced echelon form?
2. How can I identify when a system of linear equations has no solution using RREF?
3. Can a matrix in RREF have infinite solutions? If so, what does that look like?
4. How do I perform row operations to put a matrix into row-reduced echelon form?
5. What is the difference between row echelon form and row-reduced echelon form?

Tip: Always check if a system is consistent or inconsistent after reducing the matrix to RREF. Inconsistent systems will have a row like $0 \ 0 \ 0 \ | \ 1$.