  We compute the determinant of $\sum_{n=1}^{N} \vec{A}^{(n)} \otimes
\vec{B}^{(n)}$, where $\vec{A}^{(n)}$ is square and
${\vec{B}^{(n)}=\vec{x}^{(n)}{\vec{y}^{(n)}}^T}$ where $\vec{x}^{(n)}$ and
$\vec{y}^{(n)}$ have length $N$.
