Singularity forced by an odd-dimensional transpose-inverse identity ($(CB{)}^{\mathsf{T}}+B^{-1}C=0$)

Let $B$ be an invertible matrix and $C$ a real square matrix of the same odd order. If the displayed identity holds, multiply it on the left by $B$ to obtain $B{B}^{\mathsf{T}}{C}^{\mathsf{T}}=-C$. Taking determinants givesThus $((\det B{)}^2+1)\det C=0$. Since $\det B$ is real, the first factor is positive, forcing $C$ to be a singular matrix. Both odd dimension and reality matter: an invertible real antisymmetric matrix exists in even dimension, while over the complex numbers a squared determinant can equal $-1$.