Computing the determinant of a block matrix

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I am having trouble using a well-known formula for computing the determinant of a block matrix. That is, if
in which A and D are square matrices and exists then
.
The code snippet below should achieve this but it returns two different results. I'm sure it's a very trivial mistake I've made but I've probably been staring at it too long now to find it.
As a follow up, does anyone know if there is a suitably adjusted version of this formula for the case when it is B and C which are the square matrices?
function [detM,detBlock] = BlockDeterminant()
M = magic(4);
detM = det(M);
A = M(1:3,1:3);
B = M(1:3,4);
C = M(4,1:3);
D = M(4,4);
detBlock = det(A)*det(D-C*inv(A)*B);
end

采纳的回答

KSSV
KSSV 2022-8-1
A = rand(2) ;
B = rand(2) ;
C = rand(2) ;
D = rand(2) ;
M = [A B; C D] ;
det(M)
ans = -0.0221
det(A)*det(D-C*inv(A)*B)
ans = -0.0221
In your case, D is a scalar.
  14 个评论
RickyBoy
RickyBoy 2022-8-1
编辑:RickyBoy 2022-8-1
Aha!
OK, so for a bit of enlightenment, @Bruno Luong, let me demonstrate how this is so helpful...
In my setup (and it happens to be that . Therefore using detBlockAlt, we can achieve
.
Which is a much simpler form of the determinant than a general formulation will result in.
Thank you so much for your engagement!
Bruno Luong
Bruno Luong 2022-8-1
Good. I don't get why contribution is so helpful, but if you say so...

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更多回答(1 个)

Walter Roberson
Walter Roberson 2022-8-1
Just round-off error. You are calculating two different order of operations and that affects the results .
If you use M = sym(magic(4)) you will get exact zero for each

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