det(-A) = -det(A) for Odd Square Matrix

Published on 6 November 2014

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.


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


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:

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


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.