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Homework Help: Det(A) = det(AT) = det(−A) = (−1)^n*det(A). Hence det(A) = 0 when n is odd Why?

  1. Feb 28, 2010 #1
    1. The problem statement, all variables and given/known data
    det(A) = det(AT) = det(−A) = (−1)^n*det(A). Hence det(A) = 0 when n is odd
    I understand all of this except for the last statement: =(-1)^n*det(A).
    Why is this? Does it have to do with the cofactor expansion, and the "checkerboard" (as my professor puts it) of signs? I'm just not quite sure why that has to be. A quick explanation would be more than adequate. Thank you very much for your time. I didn't think that my post fit the rest of the template so I deleted it.
  2. jcsd
  3. Feb 28, 2010 #2
    It follows from the fact that: if B is formed from multiplying a single row of A by scalar x, then det(B) = x(det(A)). This can be shown easily using the definition of the determinant - you know there is some bij = xaij (where i is the row you multiplied to), so you can factor out the x. The desired result comes from multiplying all rows of A by x.

    Also, note that det(A) = det(AT) is true but det(A) = det(-A) only if the size of n is even, and your statement: "det(A) = 0 if n when n is odd" doesn't make sense. det(A)=0 tells you about the invertibility of A.
  4. Feb 28, 2010 #3
    Wow I'm an idiot. I forgot to mention that I was talking about a skew symmetric matrix. So I guess it does make sense. Thank you for your help.
  5. Feb 28, 2010 #4
    Okay, I thought you were talking about something specific but thought I'd throw in that last bit anyway to be safe. Cheers.
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