The Sherman-Morrison-Woodbury formula expresses the inverse of an “update” to a matrix $A$ in terms of $A^{-1}$ and the factors of the update:
$$\displaystyle (A + UCV)^{-1} = A^{-1} – A^{-1} U\left( C^{-1} + VA^{-1}U \right)^{-1}VA^{-1} $$
where $A$ and $C$ are invertible square matrices and $U$ and $V$ have dimensions such that $UCV$ has the same dimensions as $A$. Let’s say we don’t actually want the inverse of $A+UCV$, only its determinant.
Let $M$ be the block matrix:
$$\displaystyle M = \begin{bmatrix} I & CV \\ -U & A \end{bmatrix} $$
We can perform block Gaussian elimination, adding $U$ times the first row to the second row, to produce an upper triangular matrix $M_1$:
$$\displaystyle M_1 = \begin{bmatrix} I & CV \\ 0 & A + UCV \end{bmatrix} $$
Or by subtracting $CVA^{-1}$ times the second row from the first to get a lower triangular matrix $M_2$:
$$\displaystyle M_2 = \begin{bmatrix} I + CVA^{-1}U & 0 \\ -U & A \end{bmatrix} $$
Since adding adding linear combinations of rows to other rows of a matrix does not change its determinant, the determinants of these three matrices are equal:
$$\displaystyle \det(M) = \det(M_1) = \det(M_2) $$
Since $M_1$ and $M_2$ are block triangular, their determinants can be read off from the diagonal blocks:
$$\displaystyle \begin{aligned} \det(M_1) &= \det(A + UCV) \\ \det(M_2) &= \det(I + CVA^{-1}U) \det(A) \end{aligned} $$
So we get what we were after:
$$\displaystyle \det(A + UCV) = \det(A) \det(I + CVA^{-1}U) $$
We didn’t really need the $C$ matrix, but retained it in order to match the form of the Sherman-Morrison-Woodbury formula. But note that for our determinant formula we do not need $C$ to be invertible which is necessary for Sherman-Morrison-Woodbury.
We get an interesting special case by letting $A = I$ and $C = I$:
$$\displaystyle \det(I + UV) = \det(I + VU) $$
This is known as the Weinstein-Aronszajn identity.