How to prove $\det \begin{bmatrix}A & B \\ C & D \end{bmatrix} =\det(AD-BC)$, where $A, B, C,$ and $D$ are $n\times n$ and commuting.

Question

I can solve for the cases in which $C=0$ or $B=0$. But I cannot find any idea for the case in which both $B$ and $C$ are nonzero matrices.

Answer 1

I assume that $A$ is an $n\times n$ matrix.

First suppose that $A$ is invertible. By Gauss elimination, $$\begin{pmatrix}I & 0 \\ -CA^{-1} & I\end{pmatrix}\begin{pmatrix}A & B \\ C & D\end{pmatrix}=\begin{pmatrix}A & B \\ 0 & D-CA^{-1}B\end{pmatrix}.$$ Hence, by what you already solved, the determinant is $\det(A)\det(D-CA^{-1}B)=\det(AD-ACA^{-1}B)=\det(AD-CB)$, since $AC=CA$.

For $A$ not invertible, use approximation.