Prove that for a $n\times n$ matrix $A\in M_n(\mathbb{Z})$ ($n$ is odd), $det(A)$ is even if $a_{ii}$ is even and $a_{ij}+a_{ji}$ is even ($i\neq j$)?

Question

This is a linear algebra problem regarding a sufficient condition for the determinant of an odd-order square matrix to be even. As said above, if all elements along the main diagonal are even, and the sum of any pair of symmetric elements is even, how can we prove that the determinant is even? I've learned the methods of calculating determinants and the properties of matrices, but I struggle to prove it. I tried induction along with Laplace expansion, but it doesn't seem to help. Or regarding the definition of determinant, I thought the key point might be a product is even as long as one factor is even, but I find it challeging as the statement only provides that the sum of symmetric entries is even. Can anyone help? Thanks.

Answer 1

$\det(A)$ is a polynomial over the entries of $A$. Thus, if $A \equiv B \pmod 2$, then it follows that $\det(A) \equiv \det(B) \pmod 2$. So, let $B$ be the matrix obtained by replacing each entry $a_{i,j}$ with $0$ if $a_{i,j}$ is even, with $1$ if $a_{i,j}$ is odd and $i < j$, and with $-1$ if $a_{i,j}$ is odd and $i > j$.

Note that $B$ is an antisymmetric matrix with an odd size. It follows that $\det(B) = 0$. Thus, we have $$ \det(A) \equiv \det(B) = 0 \pmod 2. $$ So, $\det(A)$ is equal to $0$ modulo $2$, which is to say that $\det(A)$ is even, which is what we wanted.