Determinant of large block matrix (3x3)

[grif...@tamu.edu]

Greetings all,

I am familiar with the determinant relation for a 2x2 structured block
matrix, e.g. det([A B;C D]) = det(A)*det(D - C*inv(A)*B). However, I
am looking for a similar relation, an extension, for a larger block
structure, say 3x3.

det([A B C;D E F;G H J]) = ?

I envision a solution of the form:

det([A B C;D E F;G H J]) = det(A)*det(schur_complement(A)) -
det(B)*det(schur_complement(B)) + .....

I'm not sure if this form is in the ballbark. Even if so I'm not sure
about how to compute the Schur complement.

I'd imagine there must be some good references which detail the
solution to this problem, and the computation of the Schur complent.
However, I am not aware of them.

Any help on this problem would be greatly appreciated.

Regards,

Todd Griffith
Dept. of Aerospace Engineering
Texas A&M University
grif...@tamu.edu

[Robert Israel]

In article <951a5c7e.03022...@posting.google.com>,

<grif...@tamu.edu> wrote:
>Greetings all,
>
>I am familiar with the determinant relation for a 2x2 structured block
>matrix, e.g. det([A B;C D]) = det(A)*det(D - C*inv(A)*B). However, I
>am looking for a similar relation, an extension, for a larger block
>structure, say 3x3.
>
>det([A B C;D E F;G H J]) = ?

Well, you can apply the 2x2 block result to your matrix M, thinking of it as

[ A | B C ]
[ --+---- ]
[ D | E F ]
[ G | H J ]

so det(M) = det(A) det([E F; H J] - [D;G] A^(-1) [B C])
= det(A) det(R)

where
[ E - DA^(-1)B F - DA^(-1)C ]
R = [ H - GA^(-1)B J - GA^(-1)C ]

and the 2 x 2 block result can again be applied to this, obtaining

det(M) = det(A) det(E - DA^(-1)B)
det(J - GA^(-1)C - (H - GA^(-1)B)(E-DA^(-1)B)^(-1)(F - DA^(-1)C))

Robert Israel isr...@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2