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anonymous

  • 4 years ago

Can someone help me figure out the eigenvector?

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  1. anonymous
    • 4 years ago
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    |dw:1327807423151:dw|

  2. anonymous
    • 4 years ago
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    the eigenvalue is 2 but I cant figure out the eigenvector

  3. anonymous
    • 4 years ago
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    After subtracting the two matrices this is what I got |dw:1327807555690:dw|

  4. anonymous
    • 4 years ago
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    What wld the answer be?

  5. anonymous
    • 4 years ago
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    preetha i thought u were going to chat with my older sis I hid her laptop so she wouldn't keep emailing you

  6. anonymous
    • 4 years ago
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    ok so the eigenvalue is 2. The eigenvector associated with 2 we will call X and we need: \[AX=2X\] 0 0 -1 x 2x 0 -1 -4 y = 2y 0 0 1 z 2z So use matrix multiplication: -z = 2x, -y -4z = 2y, z=2z the last one implies z=0, and that implies that x=0. We have then -y=2y which means y=0 too, so the 0 vector is the eigenvector associated with 2...

  7. anonymous
    • 4 years ago
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    maybe someone can verify this is right...I need to sleep!

  8. anonymous
    • 4 years ago
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    ok I am gonna look this over

  9. anonymous
    • 4 years ago
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    Thanks :D

  10. anonymous
    • 4 years ago
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    Obviously here X = (x,y,z)^t and you do this for each eigenvalue...

  11. anonymous
    • 4 years ago
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    ummm i think u used the wrong matrice

  12. anonymous
    • 4 years ago
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    cuzz I was using a diff method so that is y i showed u that matrice

  13. anonymous
    • 4 years ago
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    ohh, you were using the method: \[(A-2I)X = 0\] ?

  14. anonymous
    • 4 years ago
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    yes

  15. anonymous
    • 4 years ago
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    well use the correct matrix and just use my steps above as a blueprint, you can't go far wrong...

  16. anonymous
    • 4 years ago
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    idk okk whtvr

  17. anonymous
    • 4 years ago
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    Thanks I really appreciate ur help :DDDDDDDDDDD

  18. anonymous
    • 4 years ago
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    :D

  19. anonymous
    • 4 years ago
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    NO I may have sounded rude but seriously I realy appreciate ur help :DDD

  20. mathmate
    • 4 years ago
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    |dw:1327836820216:dw| For \( \lambda = 2 \), since the first column is all zeroes, this would be your free variable and the eigenvector is (1,0,0). Have you found the eigenvector corresponding to the other eigenvalues?

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