The proof for this is straightforward, but I didn’t find it explicitly stated elsewhere on the web – it may just be one of those propositions that’s hard to phrase as a search query. In words: the negative determinant of an odd square matrix is the determinant of the negative matrix.

Proposition

If \(A\) is an \(n \times n\) square matrix and \(n\) is odd, then \(\det(-A) = -\det(A)\).

Proof

Negating all elements of a row of the matrix \(A\) negates the determinant of the matrix (proof – think about it as multiplying a row by the scalar \(-1\)).

\(-A\) is equivalent to \(A\) with each of its rows negated. Because \(n\) is odd:

\[\det(-A) = (-1)^n \det(A) = -\det(A)\]

Context

This fact was needed to prove that if \(A\) is a special orthogonal matrix and \(n\) is odd, then \(A\) has at least one eigenvector with eigenvalue \(1\).

Note that if \(n\) is even, we can prove that \(\det(A) = \det(-A)\) using the same technique.