Proofing that det$\begin{pmatrix}A&B\\ C&D\end{pmatrix}$ = $\det(DA -BC)$ When B Invert And $BD=DB$

Question

I have the matrices $A,B,C,D \in\mathcal{M}_{n\times n}(\mathbb{R})$ lets $X = \begin{pmatrix}A&B\\ C&D\end{pmatrix}$ and I know that $BD=DB$ and B is invertible.

How can I prove that $\det(X)=\det(DA -BC)$?

Answer 1

Notice first that

$$X=\begin{bmatrix}A & B \\ C & D \end{bmatrix}= \begin{bmatrix}0 & B \\ I & D \end{bmatrix}\begin{bmatrix}C - D B^{-1} A & 0\\ B^{-1} A& I \end{bmatrix}$$

Also

$$\det \begin{bmatrix}0 & B \\ I & D \end{bmatrix}=(-1)^n \det \begin{bmatrix}B& 0 \\ D & I \end{bmatrix} = (-1)^n \det(B) = \det(-B)$$

Then $$\det(X) = \det(-B) \det(C-DB^{-1}A)=\det(B (DB^{-1}A-C))=\det(DA-BC)$$

because $BD=DB$.