 Solution

Take determinants of $(CB{)}^{\mathsf{T}}=-B^{-1}C$. In order three, multiplication of a matrix by minus one changes its determinant's sign. HenceThe assumption $\det (B^{\mathsf{T}})\ne 0$ is the same as $\det B\ne 0$, so multiplication by $\det B$ givesSince the matrices are real, the first factor is strictly positive. Thus $\boxed{\det C=0}$, proving that $C$ is a singular matrix. This is singularity forced by an odd-dimensional transpose-inverse identity; no orthogonality of $B$ was used.