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3x3 matrix

  1. May 12, 2009 #1
    1. The problem statement, all variables and given/known data

    I am having trouble figuring out lambda for the 3x3, I have only done 2x2.
    My matrix:
    4 1 4
    1 7 1
    4 1 4




    3. The attempt at a solution
    I know that when I add lambda (L) in it looks like
    4-L 1 4
    1 7-L 1
    4 1 4-L

    When I do a 2x2, I would just cross multiply as my next step then solve for L values, but I don't know how to do that here. Any help would be appreciated!
     
  2. jcsd
  3. May 12, 2009 #2

    gabbagabbahey

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    You can expand a determinant along any row or column. For example, expanding along the top row gives:

    [tex]\begin{vmatrix} 4-\lambda & 1 & 4 \\ 1 & 7-\lambda & 1 \\ 4 & 1 & 4-\lambda \end{vmatrix} = (4-\lambda) \begin{vmatrix}7-\lambda& 1 \\ 1& 4-\lambda\end{vmatrix}-(1)\begin{vmatrix}1& 1 \\ 4& 4-\lambda \end{vmatrix}+(4)\begin{vmatrix} 1& 7-\lambda \\ 4 & 1 \end{vmatrix}[/tex]

    And each of the 2x2 determinants can be found by "cross-multiplication" as usual.
     
    Last edited: May 12, 2009
  4. May 13, 2009 #3

    jbunniii

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    You can tell immediately just by looking at the matrix that 0 is one of the eigenvalues, with

    [tex]\left[\begin{array}{c}
    1 \\
    0 \\
    -1 \end{array}\right][/tex]

    a corresponding eigenvector. Can you see why?
     
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