Proof that determinant rank equals row/column rank
If the matrix $A$ has rank $k$, then it has $k$ linearly independent lines. Those form an $k\times n$ submatrix, which of course also has rank $k$. But if it has rank $k$, then it has $k$ linearly independent columns. Those form a $k\times k$ submatrix of $A$, which of course also has rank $k$. But a $k\times k$ submatrix with rank $k$ is a full-rank square matrix, therefore invertible, thus is has a non-zero determinant. And therefore the determinant rank has to be at least $k$.