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anonymous

  • 4 years ago

Can someone help me with finding eigenvectors?

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  1. anonymous
    • 4 years ago
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  2. anonymous
    • 4 years ago
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    I need to use (lambda*I-A)

  3. anonymous
    • 4 years ago
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    I have figured out the eigenvalues but I am having difficulties figuring out th eeigenvectors

  4. anonymous
    • 4 years ago
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    The eigenvalues are Lambda=1,2,3

  5. anonymous
    • 4 years ago
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    I have figured out the eigenvector when lambda=1 but I cant figure out the vector when lambda =2

  6. anonymous
    • 4 years ago
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    |dw:1327811700600:dw|

  7. anonymous
    • 4 years ago
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    I am using the method (2I-A)x=0

  8. anonymous
    • 4 years ago
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    looks good

  9. anonymous
    • 4 years ago
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    did you find the matrix (2I-A) ?

  10. anonymous
    • 4 years ago
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    |dw:1327812063837:dw| This is what i get when i subtract the two matrices

  11. anonymous
    • 4 years ago
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    and it is terrible since there is no x value

  12. anonymous
    • 4 years ago
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    yah, so the first entry of X is unconstrained and can be whatever it likes, tho the other two values have to equal zero ... [ t 0 0 ] <-- column vector

  13. anonymous
    • 4 years ago
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    |dw:1327812148269:dw|

  14. anonymous
    • 4 years ago
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    This is what i get when I reduce the rows

  15. anonymous
    • 4 years ago
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    |dw:1327812171886:dw|

  16. anonymous
    • 4 years ago
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    oh good Thanks BF

  17. anonymous
    • 4 years ago
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    :DDDDDDDDDDDDD

  18. anonymous
    • 4 years ago
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    Hey thanks for coming by :D

  19. anonymous
    • 4 years ago
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    Don't stress urself over it I got the answer already. Thanks

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