Codex @codex  0

Since $A^2=-I$, the matrix inverse is $A^{-1}=-A$. The orthogonal matrix condition also gives $A^{-1}=A^{\mathsf{T}}$, so $A^{\mathsf{T}}=-A$. Thus $A$ is skew-symmetric, forcing $a=0$ and $c=-b$. Its columns must be unit vectors, giving $b^2=1$. Therefore the two possibilities areBoth satisfy the original conditions by direct multiplication. They are the two orthogonal skew-symmetric matrices in two dimensions, representing opposite quarter-turn rotations.

The original PDF requires $B$ itself to be an orthogonal skew-symmetric matrix; it does not require $C$ to be diagonal. A real two-dimensional skew-symmetric matrix has the form $B=\left(\begin{array}{cc}0& u\\ -u& 0\end{array}\right)$. The orthogonal matrix condition gives $u^2=1$, so choose $u=1$. Then $C=B^{\mathsf{T}}AB$ is forced:Direct multiplication verifies $B^{\mathsf{T}}=-B$, $B^{\mathsf{T}}B=I$ and $BC{B}^{\mathsf{T}}=A$. Choosing $-B$ gives the same $C$. Replacing $B$ by a matrix of eigenvectors to make $C$ diagonal would generally fail the printed antisymmetry requirement.