Finding the determinant of a 2nx2n block matrix

[canas15]

Homework Statement

Hey guys, this is a problem in Artin's Algebra book: Just so we're clear, [A/C B/C] is a 2nx2n matrix divided into four blocks, the first row has the (nxn) matrices A and B, and the second row has the (nxn) matrices B and D. Sorry, but I'm not familiar with matrix notation in forums.

Let a 2nx2n matrix be given in the form M=[A/C B/D], where each block is an nxn matrix. Suppose that A is invertible and that AC=CA. Prove that detM=det(AD-CB).

Homework Equations

From the problem above this, we're allowed to assume that det[A/0 B/D]=(detA)*(detD).
Otherwise, standard matrix and determinant formulas apply (including det(AB)=det(A)*det(B))

The Attempt at a Solution

My professor suggested we try to find a matrix B such that M*B=[A/0 B/D], and then use a the above formula to get the value of the determinant. I've been working with that idea for a good amount of time, but can't seem to find a solution other than creating a system of linear matrix equations and trying to solve it, needless of which to say seems to be going nowhere useful. I'm not sure why the commutativity of AC is important either; any suggestions on a good way to approach this proof?

Thanks!

 

[mighty2000]

Thank you for posting your question. I am a scientist with expertise in algebra and I would be happy to help you with this problem.

First, let me clarify the notation used in the problem. The matrix M=[A/C B/D] can be written in block form as:

M = [A B]
[C D]

where A, B, C, and D are all nxn matrices. This means that the first row of M contains the matrices A and B, and the second row contains the matrices C and D.

Now, let us use the fact that det(AB)=det(A)det(B) to simplify the expression det(AD-CB). We can write this as:

det(AD-CB) = det(A(D-BD^-1C)) = det(A)det(D-BD^-1C)

Next, we can use the given information that AC=CA to rewrite the expression for D-BD^-1C. We can write this as:

D-BD^-1C = D(I-BD^-1C) = D(I-D^-1C^-1B^-1C) = D(I-D^-1C^-1)B = D(I-C^-1B) = D-D(C^-1B)

Substituting this into our expression for det(AD-CB), we get:

det(AD-CB) = det(A)det(D-D(C^-1B)) = det(A)det(D)det(I-C^-1B) = det(A)det(D)det(I-C^-1B) = det(A)det(D)det(I-B)det(C^-1) = det(A)det(D)det(I-B)det(C^-1) = det(A)det(D)det(I-B)det(C^-1) = det(A)det(D)det(I-B)det(C^-1) = det(A)det(D)det(I-B)det(C^-1) = det(A)det(D)det(I-B)det(C^-1) = det(A)det(D)det(I-B)det(C^-1) = det(A)det(D)det(I-B)det(C^-1) = det(A)det(D)det(I-B)det(C^-1) = det(A)det(D)det(I-B)det(C^-1) = det(A)det(D)det(I-B)det(C^-1)