What properties does $A\in M_n(\Bbb F),n\geqslant2\ \ $ have if $\operatorname{rank}(A)=\det(A)$?
My work:
$A$ is either $\text{regular}$ or a $\text{zero-matrix}$.
~For non-trivial case, let $C\in M_n,n\geqslant2$ be regular and $\det(C)=n$ $$d(n):=\{k\in\Bbb N:k\mid n\}$$
Depending on whether $n$ is $\text{prime}$ or $\text{composite}$, we multiplied some number($\leqslant d(k)$) of columns(rows) by some $k$.
To visualise (the simplest LaPlace development or evaluation of a diagonal matrix determinant), let: $$B=\begin{bmatrix}I_{n-1}&0\\0&n\end{bmatrix}$$ $\prod_{i=1}^n a_{ii}=\det(B)=n\cdot\det(I_{n-1})\in \Bbb N$
For $n\in\Bbb N_{2n+1}$ it would work even for $-I_{n-1}$
Is $B\sim C$ s.t. the equivalence realized by an arbitrary number of the $3^{\text{rd}}$ type of elementary transformations and/or an even number of the first two types? Then the matrices $B\in[C]$? in order to $|B|=|C|,\ \ B,C\in M_n(\Bbb Z)$
Such a matrix can be simply build in the following way.
Let $A$ be any non singular square $n \times n$ matrix, over any field $\Bbb F$ extending the rationals ($\Bbb Q \subseteq \Bbb F$).
We have $\det A \neq 0$, while the rank of $A$ is $n$. Let $a_1, \dots , a_n \in \Bbb F$ such that $$a_1 \cdots a_n = n (\det A)^{-1}$$ Then if you multiply for all $j \in \{ 1, \dots , n \}$ the $j$-th row/column of $A$ by $a_j$, you will get a new matrix $\tilde{A}$ whose determinant is $n= r(\tilde{A})$.
In particular, you don't need that the matrix has integer entries, nor integer eigenvalues. In other words, it could be (almost) anything.