TITLE: Proofs of Determinants of Block matrices
QUESTION [11 upvotes]: I know that there are three important results when taking the Determinants of Block matrices
$$\begin{align}\det \begin{bmatrix}
A & B \\
0 & D 
\end{bmatrix} &= \det(A) \cdot \det(D) \ \ \ \ & (1) \\ \\
\det \begin{bmatrix}
A & B \\
C & D 
\end{bmatrix} &\neq AD - CB & (2) \\ \\
\det \begin{bmatrix}
A & B \\
C & D 
\end{bmatrix} &= \det \begin{bmatrix}
A & B \\
0 & D - CA^{-1}B
\end{bmatrix} \\ \\
&= \underbrace{\det(A)\cdot \det\left(D-CA^{-1}B\right)}_\text{if $A^{-1}$ exists} \\ \\
&= \underbrace{\det\left(AD-CB\right)}_\text{if $AC=CA$} & (3)
\end{align}$$
Now I understand in result $(3)$, that all that row operations are being performed to bring it into the form we see in $(1)$, but I can't seem to convince myself that result $(1)$ is true in the first place.
Furthermore in result $(3)$, I understand that,  $\det(A)\cdot \det\left(D-CA^{-1}B\right) = \det\left(A(D-CA^{-1}B)\right)= \det(AD-CB)$, via the product rule for determinants I also understand that we need $A^{-1}$ to exist, for the initial row operation to reduce the matrix into an upper triangular form $U$, and I understand that we require $AC = CA$, to allow commutativity when we multiply $ACA^{-1}B$ to reduce it to $CB$.
Can someone provide proofs for results $(1)$ and $(2)$, as I can't seem to find proofs for them in any of the textbooks I have at my disposal

REPLY [9 votes]: Proof of the 3rd identity.
It is a consequence of the following "block diagonalization" identity:
$$\pmatrix{
A&B\\
C&D
}=\pmatrix{
I&0\\
CA^{-1}&I
}\pmatrix{
A&0\\
0&S
}\pmatrix{
I&A^{-1}B\\
0&I
} \ \ \text{with} \ \ S:=D-CA^{-1}B$$
(S = "Schur's complement" (https://en.wikipedia.org/wiki/Schur_complement)),
Then, it suffices to take determinants on both sides.
Remark : For many matrix formulas, take a look at the amazing compendium : "Matrix Mathematics: Theory, Facts, and Formulas" Second Edition by Dennis S. Bernstein (Princeton University Press, 2009).