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Math Help - Determinants, Adjoint Matrix

  1. #1
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    Determinants, Adjoint Matrix

    I need to prove that |adj(A)|=|A|^(n-1) for every non singular A in F^nxn, while n>2.

    The solution says:


    I can understand the first equality - |adj(A)|=||A|*A^(-1)|,
    but I can't understand the rest.

    Can you please help me?

    Thank you!
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  2. #2
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    I'm sorry, there was a little mistake - only the first row is relevant to my question - I don't understand it...
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  3. #3
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    You need to use two facts to prove this:
    1. If det(A) \neq 0 then det(A^{-1}) = \frac{1}{det(A)}
    2. If c is a scalar in F, det(cA) = c^ndet(A)

    Can you follow from here?
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  4. #4
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    Well.. I got to this line:
    |adj(A)|=||A|*A^(-1)|

    but how can I continue from here?

    I just can't find a way to expand this phrase...
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  5. #5
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    Quote Originally Posted by adam63 View Post
    Well.. I got to this line:
    |adj(A)|=||A|*A^(-1)|

    but how can I continue from here?

    I just can't find a way to expand this phrase...
    det(adj(A)) = det(det(A) * det(A^{-1})) = (det(A))^n * det(A^{-1}) = (det(A))^n * \frac{1}{det(A)}

    det(adj(A)) = \frac{(det(A))^n}{det(A)} = (det(A))^{n-1}
    Last edited by Defunkt; August 29th 2009 at 05:44 AM.
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  6. #6
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    Why:

    det(det(A) * det(A^{-1})) = (det(A))^n * det(A^{-1})

    ?
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  7. #7
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    Quote Originally Posted by adam63 View Post
    Why:

    det(det(A) * det(A^{-1})) = (det(A))^n * det(A^{-1})

    ?
    Using det(\alpha A) = \alpha^n det(A) , we get:

    det(det(A) * det(A^{-1})) = det(A)^n * det(A^{-1})

    Since det(A) is a scalar in F. Also, det(A^{-1}) = \frac{1}{det(A)} \Rightarrow det(det(A) * det(A^{-1})) = \frac{det(A)^n}{det(A)} = det(A)^{n-1}
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  8. #8
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    Okay, I think I can understand the main idea... it's a little hard for me to 'digest' it right now (because intuitively it looks like this 'n' came out of nowhere), but I'll go over it until I'll understand it...

    Thank you !
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