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Determinant question

  1. Nov 17, 2007 #1
    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. jcsd
  3. Nov 17, 2007 #2


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    your attempt?
  4. Nov 17, 2007 #3
    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...
  5. Nov 18, 2007 #4


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    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.
  6. Nov 18, 2007 #5
    Wait. Do you mean the first column? The first row doesn't (necessarily) have any 0s in it?
    Last edited: Nov 18, 2007
  7. Nov 18, 2007 #6
    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.
  8. Nov 18, 2007 #7
    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
  9. Nov 18, 2007 #8
    Never mind. I just realized you can multiply "block" matrices block by block.
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