Proof that $A^{-1} + A = 0$ holds only for $n \times n$ matrices, with n even

Question

Let $A$ be an invertible $n \times n$ matrix, then: $A^{-1} + A = 0$ holds only for $n$ even.

How can I proof that the above holds?

Answer 1

As DonAntonio hinted: $$ \det A^{-1} = (\det A)^{-1} = \det (-A) = (-1)^n \det A $$

Now, $\det A$ and $(\det A)^{-1}$ have the same sign (and are non zero since $A$ is invertible) so $(-1)^n \geq 0$ and $n$ is even.

Answer 2

Hint:

Observe that from the given equality we have

$$\left(\det A\right)^{-1}=\det A^{-1}=\det(-A)\ldots$$