You could start by trying induction on k. If A and B are 1x1 matrices then det(M) is clearly 0 - AB = (-1)^{1} det(A)det(B)
assume its correct for k=K and show it still holds for k=K+1
Let A and B be k x k matrices and let:
Show that
How would I go about this? For example, are the rules for finding the determinants of block matrices similar to finding the determinants of normal matrices (ie multiplying diagonally and summing the result)?
Any help welcome.
The rules are the same. If you start with a matrix of all block matrices you can use the cofactor algorithm using those matrices as elements. Everything just becomes matrix algebra rather than real number algebra.
There's a clever way to do this. It might involve the inverse matrix of M which is easy to compute. If I come up with something I'll post it.
If M is k x k, then A is m x m, and B is (k-m) x (k-m)
Let L = max(m, k-m)
if L is even it takes L/2 column swaps to go from {{0,B},{A,0}} to {{B,0},{0,A}}
if L is odd it takes (L+1)/2 swaps
each column swap adds a factor of -1 to the original determinant.
the determinant of {{B,0},{0,A}} is clearly det(A B) = det(A) det(B)
I'm just having trouble showing that (-1)^{k} = (-1)^(L even ? : L/2, (L+1)/2))
Hi,
Previously, you essentially asked about the determinant of a "diagonal" block matrix. There you found the determinant is the product of the determinants of the diagonal matrices. Now, you know that interchanging 2 columns of a matrix changes the determinant by a factor of -1. So swap the first and (k+1)st columns, the 2nd and (k+2)nd columns, etc. for a total of k swaps. So the determinant is (-1)^{k}det(B)det(A).