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Proving an exhausting relation in Algebra.

  1. Aug 29, 2011 #1
    1. The problem statement, all variables and given/known data

    Hello, thanks for reading.

    So, here's the problem. I'm given n2 functions fij(x), and the matrix A = (fij)
    I get the definition of a new function: F(x) = detA

    I need to prove, that F'(x) = [itex]\sum^{n}_{k=1}det(A_{k})[/itex]
    Where Ak is a matrix identical to A, with only the k'th row being replaces with the row of the derivatives of the functions on the k'th row of A (bothersome!)

    2. Relevant equations

    Other than the expansion formula for a determinate I can't think of any.

    3. The attempt at a solution

    Seeing no other choice, I've tried proving this with induction. The problem is, I get lost in a maze of indexes and cannot seem to prove it.

    for n=2 it's easy to show.
    I assume it holds for n-1 and want to prove it for n.

    That's as far as I got:

    Here A[itex]^{ij}[/itex] is the minor of A at ij.

    F'(x) = (detA)' = [itex](\sum_{k=1}^{n}f_{k1}detA^{k1})' = \sum_{k=1}^{n}[f'_{k1}detA^{k1} + f_{k1}(detA^{k1})'][/itex]

    Now if we use the notation Fij(x) = det(A[itex]^{ij})[/itex], we get according to the induction assumtion:

    = [itex]\sum_{k=1}^{n}[f'_{k1}detA^{k1} + f_{k1}\sum_{k=1}^{n}detA_{k}^{k1}][/itex]

    And that's where I've pretty much given up. I doubt I'm on the right track.

    Does anyone see a way out?

    Thanks a lot! Tomer
     
  2. jcsd
  3. Aug 29, 2011 #2

    micromass

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    What about proving it with the definition of the determinant:

    [tex]\det(A)=\sum_{\sigma\in S_n}{sgn(\sigma)f_{1,\sigma(1)}f_{2,\sigma(2)}...f_{n,\sigma(n)}}[/tex]

    This should be a lot easier...
     
  4. Aug 29, 2011 #3
    I now realize I forgot the "sign" element in the expansion of the determinate above (how terribly bothersome!).

    What is this "sigma" notation?
     
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