Proofs of Determinants of Block matrices 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
 A: To prove $(1)$, it suffices to note that
$$
\pmatrix{A &B\\0&D} = \pmatrix{A & 0\\0 & D} \pmatrix{I&A^{-1}B\\0 & I}
$$
From here, it suffices to note that the second matrix is upper-triangular, and to compute the determinant of the first matrix.  It is easy to see that the determinant of the first matrix should be $\det(A)\det(D)$ if we use the Leibniz expansion.
For an example where $(2)$ fails to hold, consider the matrix
$$
\pmatrix{
0&1&0&0\\
0&0&1&0\\
0&0&0&1\\
1&0&0&0
} = 
\pmatrix{B&B^T\\B^T&B}
$$ 
For an example where the diagonal blocks are invertible, add $I$ to the whole matrix.
A: Suppose block $A$ has dimension $r$, block $D$ has dimension $s$. Use the definition of the determinant $\lvert c_{i,j}\rvert,\enspace {1\le i,j\le r+s}$:
$$\begin{vmatrix} A&C\\0& D\end{vmatrix} =\sum_{\sigma \in \mathfrak S_{r+s}}\prod_{1\le j\le r+s}(-1)^{\text{sgn}\, \sigma}c_{\sigma(j),j}.$$
Now the non-zero terms are those such that, if $1\le j\le r$, $\;1\le \sigma(j)\le r$. Similarly, if $r+1\le j\le r+s$, $\;r+1\le \sigma(j)\le r+s$. So the non-zero terms are those for which the permutation $\sigma\in \mathfrak S_{r+s}$ is the concatenation of a permutation of $\mathfrak S_r$ and a permutation in $\mathfrak S_s$, and clearly the signature of $\sigma$ is the product of the signatures of its factors.
The formula for the first formula for the determinant follows by distributivity.
A: 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).
