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Help finding the eigenvalues of this

  1. Dec 8, 2008 #1
    1. The problem statement, all variables and given/known data
    alright, so i have a question which asks me to find the eigenvalues of a 3x3 square matrix A. after working on it for a long time, i cant figure it out, i know the process of it and i can do 2x2 matrices easily, i cannot figure this one out though. here is the matrix A;
    2 0 0
    1 2 -1
    1 3 -2


    2. Relevant equations
    (Ix-A)


    3. The attempt at a solution
    ive gotten this so far, using the above equation
    x-2 0 0
    -1 x-2 1
    -1 -3 x+2
    i know i have to take the determinant of the above, and im pretty sure im doing something wrong with getting the determinant, but i cant figure it out.
    ((x^3-2x^2-4x+8)+0+0)) - (0+(-3x+6)+0))
    = (x^3-2x^2-4x+8)-(-3x+6)
    = x^3-2x^2-7x+2

    from this i know i have to find the roots, which will give me my eigenvalues. can someone help me, mabey point out my mistakes?
     
  2. jcsd
  3. Dec 8, 2008 #2

    gabbagabbahey

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    I'm not sure what method you are using to calculate the determinant; but whatever you're doing, you're doing it incorrectly.

    Since there are two zeros in the top row of your matrix, the easiest way to take the determinant is to expand it along the top row:

    [tex]\begin{vmatrix}x-2 & 0 & 0 \\ -1 & x-2 & 1 \\ -1 & -3 & x+2 \end{vmatrix}=(x-2)\begin{vmatrix} x-2 & 1 \\ -3 & x+2 \end{vmatrix}+(0)\begin{vmatrix} -1 & 1 \\ -1 & x+2 \end{vmatrix}+(0)\begin{vmatrix}-1 & x-2 \\ -1 & -3 \end{vmatrix}[/tex]
     
  4. Dec 8, 2008 #3

    Defennder

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    You can evaluate the det by co-factor expansion along the first row. Doing so has the advantage of factoring out one eigenvalue so you only have to factorise the expression inside the co-factor determinant. So if you do so, you should have [tex](x-2)[(x-2)(x+2) + 3][/tex]. From there you can easily find the eigenvalues.
     
  5. Dec 9, 2008 #4
    would this be called the cofactor expansion?
     
  6. Dec 9, 2008 #5

    Defennder

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    Yes, it's det evaluation by co-factor expansion along row/column.
     
  7. Dec 9, 2008 #6
    should i always choose the row/column with the most zeros? what if the matrix doesnt have any?
     
  8. Dec 9, 2008 #7

    Defennder

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    Well there would always be a row/column with the most 0's, just that it may not have a row/col with only 1 non-zero entry. If it doesn't have any, add multiples of one row/col to another to create the 0's yourself. Doing so does not change the determinant.
     
  9. Dec 9, 2008 #8
    alright i have another example where i have to make zeros
    x-1 -1 3
    -2 x -6
    -1 1 x-5
    i choose to add row 1 and row 3 to make a zero at row 1, column 2
    x-2 0 x-2
    -2 x -6
    -1 1 x-5
    so now i expand the first row, is this correct?

    im my class ive learned to find the determinant another way, say i have a matrix
    a b c
    d e f
    g h i
    i could find the determinant this way
    det= (aei+bfg+cdh)-(gec+hfa+ibd)

    why couldnt i just solve the determinat this way instead of the cofactor expansion? how do i know which method to use?
     
    Last edited: Dec 9, 2008
  10. Dec 9, 2008 #9

    Defennder

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    Yes it appears so, except to make it even easier you could add a negative of the first column to the third one.

    They are both equally valid. Except if you do it this way you have to factor a third degree polynomial which in general isn't easy. On the other hand, evaluating determinants through the above method allows you to factor out one eigenvalue so all you're left with is a quadratic expression which is usually a lot simpler to factorise compared to a cubic one.
     
  11. Dec 9, 2008 #10
    thank you very much, you have no idea how much you have helped me
     
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