Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2010 / Paper 2 / 14 / b / i / Solution

The transpose rule gives $(BC{)}^{\mathsf{T}}=C^{\mathsf{T}}{B}^{\mathsf{T}}$. If $C$ is orthogonal, $C^{-1}=C^{\mathsf{T}}$, and the identity becomesMultiplying on the left by $C$ yields $\boxed{B^{\mathsf{T}}=-B}$. This is precisely the defining condition for an antisymmetric matrix.