What properties does $A\in M_n(\Bbb F),n\geqslant2\ \$ have if $\operatorname{rank}(A)=\det(A)$?

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 $$|B|=|C|,\ \ B,C\in M_n(\Bbb Z)$$

• I cannot understand what you mean by $r(A)$ (is it the rank of the matrix?), and by "regular matrix" (is it maybe a non-singular matrix?). Then, I dont' understand why you pick any "regular matrix", and state that its determinant is necessarily an integer (it could be any element of the field $\mathbb F$). Then I don't understand what "multiply $d(k)$ columns", if $d(k)$ denotes the set of divisors of $k$ (for some $k$?). As it is stated, this question is unclear. – Crostul Jan 15 at 17:43
• $r(A)\iff\operatorname{rank} (A)$ Regular matrix $B\in M_{mn}(\mathbb F)$ has a full rank $r(B)=\min\{m,n\}$ $A\in M_n(\mathbb F)$ is a square matrix and is regular. It is necessarily a positive integer because $r(A)\in\mathbb N$ We can multiply, if we want so, by distinct divisors of $n$, which isn't important. – Invisible Jan 15 at 17:53

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 $$n= r(\tilde{A})$$.