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Proving det(AB)=det(A)det(B)

  1. Feb 17, 2013 #1
    1. The problem statement, all variables and given/known data

    Prove the relation det(AB)=det(A)det(B)


    2. Relevant equations

    det A = [itex]\sum_{k}A_{kq} C_{kq}[/itex]

    3. The attempt at a solution

    Here is what I have done:

    (detA)(detB)= [itex]\sum_{k}A_{kq} C_{kq}[/itex] [itex]\sum_{k}B_{kq} C_{kq}[/itex]
    = [itex]\sum_{k}(A_{kq}B_{kq}) C_{kq}[/itex]
    = [itex]\sum_{k}(AB)_{kq} C_{kq}[/itex]
    =det(A+B)

    My question is: is it mathematically proper to factor out the Ckq in the summation, or am I breaking some matrix/summation rule here?

    Thanks!
     
    Last edited: Feb 17, 2013
  2. jcsd
  3. Feb 17, 2013 #2

    Ray Vickson

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    Homework Helper

    This is riddled with serious errors. First: the cofactors ##C_{ij}## depend on the matrix, so you have one cofactor ##C_{ij}(A)## for matrix ##A## and a different one ##C_{ij}(B)## for matrix ##B##. Second, you cannot use the same summation index in both factors, so you need to write something like
    [tex] \det(A) \det(B) = \sum_{k} a_{kp}C_{kp}(A) \sum_{m} b_{mq} C_{mq}(B),[/tex]
    and this does not really lead anywhere.
     
  4. Feb 17, 2013 #3
    Ok, I will try to prove it without summation notation. Thank you
     
  5. Feb 17, 2013 #4

    Zondrina

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    Why don't you try considering two cases, one where A is singular and another where A is nonsingular?
     
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