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Determinant proof

  1. Jul 12, 2009 #1
    1. The problem statement, all variables and given/known data
    If A and C are nxn matricies, with C invertible, prove that det(A)=det((C^-1)AC).


    2. Relevant equations



    3. The attempt at a solution
    I think the way to go is to show that if A=(C^-1)AC, then det(A)=det((C^-1)AC), but I'm not sure how to show A=(C^-1)AC. I know CC^-1=(C^-1)C=I, but I just can't see how to put this all together.
     
  2. jcsd
  3. Jul 12, 2009 #2
    Do you know some formula for determinant of a product?
     
  4. Jul 12, 2009 #3
    det((C^-1)AC)=det(C^-1)*det(A)*det(C)=(1/det(C))*det(C)*det(A)=(det(C)/det(C))*det(A)=det(A).

    Is that it?
     
  5. Jul 12, 2009 #4

    HallsofIvy

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    Well, his question was "Do you know some formula for determinant of a product?".

    I assume your answer is "Yes"!:biggrin:

    In that case, so is mine.:tongue:
     
  6. Jul 12, 2009 #5

    epenguin

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    I don't think so - you seem to be confusing the inverse C-1 of a matrix C with the reciprocal 1/det(C) of the number det(C) which does not come into this at all.

    g_edgar asked do you know any formula for a determinant of a product of matrices, e.g. of the determinant of MN where M, N are both n X n matrices.

    You can treat the right hand side using a couple of such formulae.

    Your proposed matrix equation A = C-1AC (equivalent to CA = AC ) is not true in general.
     
  7. Jul 12, 2009 #6

    Office_Shredder

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    det(Identity matrix) = 1 = det(CC-1 = det(C)*det(C-1) and hence det(C-1) = 1/det(C)
     
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