Past exam of the year 1 of the computer science course of the University of Oxford / 2016 / Digital systems, linear algebra, introduction to formal proof / 5 / e

For a skew-symmetric matrix, $A^T=-A$. By parts (c) and (d),If $n$ is odd, $\det (A)=-\det (A)$, hence $2\det (A)=0$ over $\mathbb{R}$. Therefore $\det (A)=0$, so $A$ is a singular matrix. The conclusion depends on odd dimension: in dimension two, $\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$ is skew-symmetric and has determinant one.