= 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. Hence
$$
\det C\,\det B=-\frac{\det C}{\det B}.
$$
The assumption $\det(B^{\mathsf T})\ne0$ is the same as $\det B\ne0$, so multiplication by $\det B$ gives
$$
\bigl((\det B)^2+1\bigr)\det C=0.
$$
Since 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.
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