Show that $\det(A) = 0$

Question

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.

Answer 1

Hint: For an $n \times n$ matrix $A$, $\det(A^T) = \det(A)$ and $\det(-A)=(-1)^n \det(A)$.

Answer 2

$det(A)=det(A^t)=det(-A)=(-1)^5det(A)=-det(A)$ this implies that $det(A)=0$.

Answer 3

$$A^t=-A$$ $$\det(A^t)=\det(-A)$$ $$\det(A)=(-1)^n\det(A)$$ $$\det(A)(1-(-1)^n)=0$$

Since $n=5 \implies (-1)^n=-1$

$$2\det(A)=0$$ Therefore $$\det(A)=0$$