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Finding the eigenvalues of a 3x3

  1. Jul 13, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the eigenvalues of:
    |13 -30 0|
    |1 0 0|
    |0 1 0|

    2. Relevant equations

    Equation for the eigenvalues: det(A-λI)=0

    Cofactor expansion = det A = a11(a22a33-a23a32)+a12(a23a31-a21a33)+a13(a21a32-a22a31)

    3. The attempt at a solution

    |13-λ -30 0 |
    | 1 -λ 0 |
    | 0 1 -λ|

    (13 -λ)(λ[itex]^{2}[/itex])-30(-λ)+0 = 0
    13λ[itex]^{2}[/itex]-λ[itex]^{3}[/itex]+30
    -λ[itex]^{3}[/itex]+13λ[itex]^{2}[/itex]+ 30λ
    -λ(λ[itex]^{2}[/itex]-13λ-30)
    (λ+2)(λ-15)

    Eigenvalues = 0, 2, 15

    However, this not the correct answer according the the software I'm using. Can anyone see what I'm doing wrong?
     
  2. jcsd
  3. Jul 13, 2012 #2

    micromass

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    The -λ in bold should be just λ

     
  4. Jul 13, 2012 #3

    I like Serena

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    Hmm, I'd say that the -30(-λ) should really be --30(-λ).
     
  5. Jul 13, 2012 #4

    micromass

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    Yes, I agree that that is the most common way of seeing it. But I guess that the OP wanted to follow his formula:

    In that case, it is [itex]-30\lambda[/itex]. But it is a weird way of doing things, agreed.
     
  6. Jul 13, 2012 #5

    HallsofIvy

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    Expanding on the last column gives [itex]-\lambda(\lambda^2- 13\lambda+ 30)[/itex], as I Like Serena and micromass are saying, not [itex]-\lambda(\lambda^2- 13\lambda- 30)[/itex].

    That factors as [itex]-\lambda(\lambda- 10)(\lambda- 3)= 0[/itex].
     
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