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Math Help - Calculation of a Determinant?

  1. #1
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    Calculation of a Determinant?

    So our professor at university gave us this assignment for home, we need to calculate the determinant of the following square(N*N) matrix



    I have already found that |AN|=a|AN-1| - |AN-2| , but from that point I can't go any further, heard that I need to use difference equations, but unfortunately I am not familiar with them. Thank you in advance, I would be really grateful for any help given
    Last edited by mr fantastic; December 13th 2011 at 12:33 PM. Reason: Title.
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  2. #2
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    Re: Can anyone help me with the Calculation of a Determinant???

    You're on the right track trying to find a recursive formula. I would say solve the first few determinants and see if a pattern begins to form, it usually will.

    Also, take a look at this, it should help.

    Recurrence relation - Wikipedia, the free encyclopedia
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  3. #3
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    Re: Can anyone help me with the Calculation of a Determinant???

    Thanks for your answer, I already solved the first few determinants but I am unable to find the recursive formula, the link you provided was very helpful I will look into it. Thanks again
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  4. #4
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    Re: Can anyone help me with the Calculation of a Determinant???

    No problem, I'm actually just in linear now so I'm no expert but if I finish my take home test and you haven't solved I'll look into it more.
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  5. #5
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    Re: Can anyone help me with the Calculation of a Determinant???

    Just a few hints about solving recurrence relations, which is actually not too difficult and is explained very well in the Wikipedia article that Johngalt13 linked to. To solve the equation |A_N| - a|A_{N-1}|+|A_{N-2}| = 0, you first need to solve the auxiliary equation \lambda^2-a\lambda+1=0. Its roots \lambda_1 and \lambda_2 are \tfrac12(a\pm r), where r=\sqrt{a^2-4}. The solution to the difference equation is |A_N| = C\lambda_1^N + D\lambda_2^N, where C and D are constants that you have to determine by ensuring that the formula works for N = 1 and 2.

    The solution works well if a>2. If a<2 then the roots of the auxiliary equation are complex numbers, but the formula for |A_N| still works.

    If a=2 then r=0 and you will find that the formula breaks down completely. But in fact the formula for |A_N| becomes much simpler when a=2. You can guess what it is by evaluating |A_N| directly from the determinant for small values of N, and then verifying your guess using a proof by induction.
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