# 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.

## 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.