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3x3 matrix

  • Thread starter rugbygirl2
  • Start date
  • #1

Homework Statement



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




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!
 

Answers and Replies

  • #2
gabbagabbahey
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6
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:
  • #3
jbunniii
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Homework Statement



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




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!
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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