# Homework Help: Determinant question

1. Nov 17, 2007

### ehrenfest

1. The problem statement, all variables and given/known data

A is a k by k matrix, B is a k by l matrix, D is an l by l matrix.

Let M be the (k+l) by (k+l) matrix that has A in the upper left, B in the upper right, D in the lower right, and 0 in the lower left.

Show that det(M) = det(A) det(D)

2. Relevant equations

3. The attempt at a solution

2. Nov 17, 2007

### robphy

3. Nov 17, 2007

### ehrenfest

I would have posted my attempt if it had gone anywhere. But anyway, I tried to use the fact the determinant is an alternating multilinear map. I tried to use cofactor expansion and that seemed like it might work but could get really messy and I thought that there is probably a better way to do it. On second thought though, cofactor expansion might be the best approach...

4. Nov 18, 2007

### morphism

Cofactor expansion on the first row is a good idea. Try induction on k.

I've seen a combinatorial proof of this, but unfortunately have long forgotten it.

5. Nov 18, 2007

### ehrenfest

Wait. Do you mean the first column? The first row doesn't (necessarily) have any 0s in it?

Last edited: Nov 18, 2007
6. Nov 18, 2007

### Marco_84

i think tou can diagonalize the bigger M matrix with an SO(k+l) rotation, the determinant should be invariant under this operation.
Now u get a block reduced matrix and you equation hold.

7. Nov 18, 2007

### ehrenfest

Never mind. Induction worked well. Now I want to show that if our matrix is

(A B)
(C D)

where A is k by k, B is k by l, C is l by k, D is l by l

Then its determinant is detA det(C A^(-1) B)

I have seen this before and it involves factoring the matrix into 2 triangular matrices. I forget exactly how it goes. Does anyone remember why you can do this:

(A B) = ( E 0)(A B )
(C D) ( C A^(-1) E)( 0 D - C A^(-1) B )

if that makes sense. E is the identity matrix.

Last edited: Nov 18, 2007
8. Nov 18, 2007

### ehrenfest

Never mind. I just realized you can multiply "block" matrices block by block.