Solution
= Solution
\b[Yes, in even dimension.] For example, choose the <skew-symmetric matrix>
$$
\boxed{B=\begin{pmatrix}0&-1\\1&0\end{pmatrix},\qquad C=\begin{pmatrix}0&1\\-1&0\end{pmatrix}=-B.}
$$
Here $B^2=-I$, so $(I+B)^{-1}=(I-B)/2$ and $C=(I-B)^2/2=-B$. It is an <orthogonal skew-symmetric matrix>. More generally the skew-symmetry condition $C=-C^{-1}$ gives $(I-B)^2=-(I+B)^2$, or $B^2=-I$; no real odd-dimensional matrix can satisfy that condition.