Problem: Let $A , B , C , D$ be commuting $n$-square matrices. Consider the $2n$-square block matrix $$M=\begin{pmatrix} A & B \\ C & D\end{pmatrix}$$ Prove that $|M|= |A||D| - |B||C|$, where $|M|$ means the determinant.
I should also state that this from a beginning Linear Algebra book, so I have not studied any fancy determinant formulas yet. My problem here is that everything I can try involves multiplication but there is a minus sign on the right hand side which I cannot presently handle.
Note: (this is not the same question as has been asked before here on this site as the formula here is quite different.)
You cannot prove it because it is not true. Counterexample: $A=B=D=I_2$ and $C=\pmatrix{0&0\\ 0&1}$ over any field. Then $\det M=0$ (the second and the fourth rows are identical to each other), but $\det(A)\det(D)-\det(B)\det(C)=1$.
The correct formula should be $\det M=\det(AD-BC)$. This formula has been discussed many times on this site. See one of my answers for instance.