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