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The inverse appearing in the assumption requires $\det C\ne 0$. Taking determinants of $(BC{)}^{\mathsf{T}}=-C^{-1}B$ giveswhere the minus sign comes from the odd matrix order. Multiplying by $\det C$ gives $\det B((\det C{)}^2+1)=0$. Since $C$ is real, the second factor is positive. Thus $\boxed{\det B=0}$ and $B$ is a singular matrix. This is the singularity forced by an odd-dimensional transpose-inverse identity, with the roles of the two matrix letters exchanged.

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.