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Math Help - Linear algebra - Eigenvalues can't get the answer

  1. #1
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    I need to solve for the eigenvalues of the matrix 1 -1 -1 1 3 2 -1 -1 0 So I subtract lambda I from the matrix and find the determinant I get: (1-x)(3-x)(-x)-(-1)(2)(-1)-(-1)(1)(-1)+(-1)(3-x)(-1)+(-1)(1)(-x)+(1-x)(2)(-1) = -x^3 + 4x^2 - x - 2 So (x-1) is a factor and then I get this remaining -(x^2-3x-2) which can't be factored..

    Did I calculate the determinant wrongly?
    Last edited by mr fantastic; February 15th 2010 at 09:18 PM. Reason: Merged posts
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  2. #2
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    Quote Originally Posted by zeion View Post
    I need to solve for the eigenvalues of the matrix 1 -1 -1 1 3 2 -1 -1 0 So I subtract lambda I from the matrix and find the determinant I get: (1-x)(3-x)(-x)-(-1)(2)(-1)-(-1)(1)(-1)+(-1)(3-x)(-1)+(-1)(1)(-x)+(1-x)(2)(-1) = -x^3 + 4x^2 - x - 2 So (x-1) is a factor and then I get this remaining -(x^2-3x-2) which can't be factored..

    Did I calculate the determinant wrongly?
    I haven't checked your determinant (I can't follow your calculation) but I will say that if you're studying eigenvalues then you ought to know how to factorise a quadratic function using surds (that the quadratic can be factorised using real numbers should be obvious from the fact that the discriminant is greater than zero).
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  3. #3
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    In other words, any quadratic equation can be solved by completing the square or using the quadratic formula. You don't need to be able to "factor" it.
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