Prove that determinant of the matrix is non-zero

Question

Given a square matrix $A$ of order $2n$ such that $a_{ii}=0$ and $a_{ij}\in\{-1,1\},\space i\neq j$, prove that $\det(A)\neq0$.

Answer 1

$\det A = \sum_{\sigma\in D_{2n}} (\pm 1)$, where the sum is over all fixed point free permutation $\sigma\in S_{2n}$ ("derangements"). So it suffices to show that the number $!(2n)$ of derangements of an even number of objects is odd. This follows from the recursion $$ !k = (k-1)(!(k-1)+!(k-2)) $$ by induction; in fact, it's immediate from this that $!k$ alternates between even and odd values. See here for background.

Answer 2

a suitably-chosen cofactor expansion shows that if a single non-zero element changes sign then the odd-even parity of the determinant is unaltered.

consider a matrix $M=[m_{ij}]$ with zeroes on the diagonal, $1$ elsewhere.

in the determinant mod(2) terms cancel in pairs, except for the case of the "codiagonal" which contributes $$ \prod_{j=1}^{2n}m_{j,2n+1-j} \equiv_2 1 $$