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Eigenvalues of a 3x3 matrix

  1. Jul 19, 2011 #1
    Ive been trying for 3 hours now and cant seem to find the eigenvalues, the long polynomials are getting me confused, the matrix is [2 2 1:1 3 1:1 2 2]




    So far i did [2-L 2 1:1 3-L 1:1 2 2-L] then I do the normal way to find the determinant but after that I get a horrible polynomial. Please help anyone!

    Thanks
     
  2. jcsd
  3. Jul 19, 2011 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    What horrible polynomial did you get? Multiply it out. It will be a cubic, but you can factor it. There's no way anyone can help until you show more of your work.
     
  4. Jul 19, 2011 #3

    Mark44

    Staff: Mentor

    After taking the determinant, I get
    [itex](2 - \lambda)(6 - 5\lambda + \lambda^2 - 2) - (4 - 2\lambda - 2) + 2 - (3 - \lambda)[/itex]

    [itex]= (2 - \lambda)(\lambda^2 - 5\lambda + 4) + 3\lambda - 3 [/itex]

    Instead of multiplying all that stuff out, factor the quadratic and the last two terms and notice that there is a common factor.

    I get [itex]\lambda = 1~and~\lambda = 5[/itex].
     
  5. Jul 19, 2011 #4
    I get (2-L)((3-L)(2-L)-2) - 2((2-L)-1) + 2-(3-L)

    2-L(4 - 3L - 2L - L^2) + 3L - 3 I tried doing all sorts of stuff to this, just cant get it.
     
    Last edited: Jul 19, 2011
  6. Jul 19, 2011 #5
    Ahh I see now, thanks alot.
     
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