Prove $\det \left[\begin{smallmatrix} A&B\\\\C&D\\ \end{smallmatrix}\right] =\det(AD-BC)$ for $A,B,C,D$ upper triangular complex matrices

Question

Let $A,B,C$ and $D$ be upper-triangular $n \times n$ complex matrices. Let

$$E=\begin{bmatrix} A&B\\C&D\\ \end{bmatrix}$$

Prove $\det(E)=\det(AD-BC)$.

I did this problem in the case that the matrices commute but I cannot figure out this case.

Answer 1

It suffices to prove the identity when the upper triangular parts of $A,B,C,D$ are independent indeterminates (alternatively, prove the identity for invertible $D$ first, then pass $D$ to the limit). Try to justify the first equality below using properties of Schur complement (see also this Wikipedia entry) and the second equality below using the condition that $A,B,C,D$ are triangular: $$ \det(E) =\det(A-BD^{-1}C)\det(D) =\det(A-BCD^{-1})\det(D) =\det(AD-BC). $$ (Edit. The following part is wrong. See darij grinberg's comment.) You may also try to prove that in Leibniz formula for determinant of $E$, only two generalised diagonals of $E$ are non-vanishing, one given by the main diagonal and the other formed by the diagonals of $B$ and $D$. But I find this argument harder to be worded clearly.