# Show that $\det(A) = 0$ [closed]

If $A$ is a $5\times 5$ matrix such that $A^T = - A$, show that $\det(A) = 0$.

I have solved it but I have trouble understanding the significance of the $5\times 5$ in the question.

## closed as off-topic by Namaste, Math Lover, Hans Lundmark, Micah, A. GoodierMar 19 '18 at 16:29

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• Consider $B= \pmatrix{0 & 1 \\ -1 & 0}$. Now $B^t =B$; does your proof show that $\det B = 0$? – John Hughes Mar 19 '18 at 14:54
• It is true for all odd exponents (see the solution of @Math Lover) – Jean Marie Mar 19 '18 at 14:54

Hint: For an $n \times n$ matrix $A$, $\det(A^T) = \det(A)$ and $\det(-A)=(-1)^n \det(A)$.
$det(A)=det(A^t)=det(-A)=(-1)^5det(A)=-det(A)$ this implies that $det(A)=0$.
$$A^t=-A$$ $$\det(A^t)=\det(-A)$$ $$\det(A)=(-1)^n\det(A)$$ $$\det(A)(1-(-1)^n)=0$$
Since $n=5 \implies (-1)^n=-1$
$$2\det(A)=0$$ Therefore $$\det(A)=0$$