The determinant of a block matrix with commuting lower blocks

Idea. For invertible $D$, block elimination reduces the $2n\times2n$ determinant to an $n\times n$ one. Commutation of $C$ and $D$ lets us simplify the result. The general case follows by density.

Step 1: $D$ invertible. Write the block matrix as a product:
$$\begin{pmatrix}A&B\\C&D\end{pmatrix}=\begin{pmatrix}I&BD^{-1}\\0&I\end{pmatrix}\begin{pmatrix}A-BD^{-1}C&0\\C&D\end{pmatrix}.$$
(Check: the top-left block of the product is $A-BD^{-1}C+BD^{-1}C=A$, the top-right is $BD^{-1}D=B$.) The first factor has determinant $1$, and the second is block lower triangular, so
$$\det\begin{pmatrix}A&B\\C&D\end{pmatrix}=\det(A-BD^{-1}C)\cdot\det D=\det\bigl((A-BD^{-1}C)D\bigr)=\det\bigl(AD-BD^{-1}CD\bigr).$$

Step 2: use $CD=DC$. Multiplying $CD=DC$ on the left by $D^{-1}$ gives $D^{-1}CD=C$. Therefore $BD^{-1}CD=BC$ and the determinant equals $\det(AD-BC)$.

Step 3: general $D$. For $\varepsilon\in\mathbb{R}$, the matrix $D_\varepsilon=D+\varepsilon I$ still commutes with $C$, and it is invertible except for finitely many values of $\varepsilon$ (the eigenvalues of $-D$ that are real). By Step 2,
$$\det\begin{pmatrix}A&B\\C&D_\varepsilon\end{pmatrix}=\det(AD_\varepsilon-BC)$$
for all small $\varepsilon\ne0$. Both sides are polynomials in $\varepsilon$, so they are continuous, and letting $\varepsilon\to0$ gives the identity for $D$.

Remark. The hypothesis is essential: for $A=B=I$, $D=0$... one checks that without $CD=DC$ the formula can fail (take $n=2$ with $C,D$ not commuting). The result is the matrix analogue of $\det\begin{pmatrix}a&b\\c&d\end{pmatrix}=ad-bc$.

Lesson. Block elimination plus density: prove under an invertibility hypothesis, then perturb by $\varepsilon I$ while preserving the algebraic hypothesis (here commutation).