Suppose that $A \in M_{5\times5}$, and suppose that $RREF_A = B$. Find $\text{rank}(A)$, if we know that $\det A = \det B + 1$
I'm not really sure how to approach this question. I know that $n = 5$ ($A$ has $5$ columns), so $\text{rank}(A) + \text{nullity}(A) = 5$. I also know that $\det(A) \neq \det(B) \neq 0$ because that would just be nonsense. However I don't know what other steps to take to find an answer. I believe it is the case that if $\det(A) \neq 0$, then $\text{rank}(A) = n = 5$, but I'm not really sure if that's true or (if it is true) how to justify that it is the case.
Any guidance would be greatly appreciated.
If $\det A = 0$, then $\det B = 0$ as well because $B$ is obtained from $A$ through either swapping rows or multiplying a row by a non-zero number** (hence $\det B = \alpha \det A$ for some $\alpha \in \Bbb R$). You can also argue that if $\det A = 0$, then $A$ has two linearly dependent rows, hence $B$ must have at least a zero row, and so has determinant zero as well.
Therefore $\det A \neq 0$ and rank$(A) = 5$.
** There is another possible row operation, but it doesn't affect the determinant.