Let, there be a matrix of order $N$ with all distinct entries. Using the given entries, is the sum of determinants of all possible matrices zero?

Question

I tried this with a $2 \times 2$ matrix and got the answer zero, is it true for any matrix, or just a coincidence?

Answer 1

Suppose that $A$ is an $n\times n$ matrix with distinct entries in some fixed set $S$ (where $|S|=n^2,\ S\subset \mathbb{C}$), and the $\det(A)>0$. Then by swapping the first and second row of $A$ and naming this matrix $B$, one has $-\det(A)=\det(B)$ (see here). This mapping is a bijection.

So, if we take the sum of all determinants of all $n\times n$ matrices consisting of distinct entries in the set $S$, for every matrix with positive determinant $k$ we also have a matrix with determinant $-k$. The only other matrices are ones with a determinant of zero, but they of course don't contribute to this sum. As such, the sum is indeed zero.