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Determinant Proof Help Please

  1. Nov 18, 2012 #1
    1. The problem statement, all variables and given/known data

    A square (nn) matrix is called skew-symmetric (or antisymmetric) if AT =
    -A. Prove that if A is skew-symmetric and n is odd, then detA = 0. Is this true
    for even n?

    2. Relevant equations
    Det(A) = Det(AT) where AT= the transpose of matrix A


    3. The attempt at a solution
    I started to try and say that since AT=-A then Det(AT) = Det(-A) so Det(A) = Det(-A) b/c the law that Det(A)=Det(AT) but I didnt know where to go from here.. specifically what impact does the odd number of n have to do with anything.. Any help
     
  2. jcsd
  3. Nov 18, 2012 #2
    and I assume that I need to say that Det(-A) = -Det(A) for the odd number on n so then I could conclude 2Det(A)=0 but i don't see why: Det(-A) = -Det(A) is true for odd number n matrices
     
  4. Nov 18, 2012 #3

    micromass

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    You can easily calculate det(-I), no?? (where I is the identity matrix)
    Then you can write det(-A)= det((-I)A).
     
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