Determinant of a $2n$ square block matrix in which all blocks commute

Question

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.)

What book is this taken from? – user1551 Dec 21 '15 at 10:14 — user1551, Dec 21 '15 at 10:14
@ user 1551 :Schaum's outline of Linear Algebra...... – herashefat Dec 21 '15 at 11:41 — herashefat, Dec 21 '15 at 11:41

score 3 · Accepted Answer · edited Apr 13 '17 at 12:21

3

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.

edited Apr 13 '17 at 12:21

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answered Dec 21 '15 at 01:29

user1551

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Determinant of a $2n$ square block matrix in which all blocks commute

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