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nubeer

  • 2 years ago

Anyone good at Eigen Vectors.

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  1. UnkleRhaukus
    • 2 years ago
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    im good at eigen values

  2. nubeer
    • 2 years ago
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    ohh finally.. ok i have a question.. i know how to solve.. just for oncae type of case i dont know how to solve.. i will post up the question [ 6 5 2 2 0 -8 5 4 0 ] this is the matrix.

  3. zzr0ck3r
    • 2 years ago
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    do you have your eigen values ?

  4. nubeer
    • 2 years ago
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    yes i have .. its 2

  5. waterineyes
    • 2 years ago
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    If I am not getting it wrong, then I think we will do like this to find eigen values : A - IX = 0 Something like that... No knowledge... @UnkleRhaukus please help your friend (me)...

  6. waterineyes
    • 2 years ago
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    Where is lambda?? Oh my God...

  7. UnkleRhaukus
    • 2 years ago
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    \[\textbf Ax=\lambda x\]\[(\textbf A-\lambda \textbf I)x=0\]

  8. waterineyes
    • 2 years ago
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    Oh yeah.... I forgot all the things... One day, I found a page thrown by someone on the road, I read that.. There on the page, it was shown how to find Eigen values but the question was not complete.. And I have forgot that too...

  9. nubeer
    • 2 years ago
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    i have found lammida.. and its 2

  10. UnkleRhaukus
    • 2 years ago
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    there is repeated roots?

  11. nubeer
    • 2 years ago
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    yes all 3 values of lamida are 2

  12. waterineyes
    • 2 years ago
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    I try to find eigen value now... You can carry on @UnkleRhaukus

  13. UnkleRhaukus
    • 2 years ago
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    \[\textbf A=\left[\begin{array}{ccc}6&5&2\\2&0&-8\\5&4&0\end{array}\right]\]\[x=\left[\begin{array}{ccc} x_1\\x_2\\x_3\end{array}\right]\]\[\lambda=2\] \[\left[\begin{array}{ccc}6&5&2\\2&0&-8\\5&4&0\end{array}\right]\left[\begin{array}{ccc} x_1\\x_2\\x_3\end{array}\right]=2\left[\begin{array}{ccc} x_1\\x_2\\x_3\end{array}\right]\]

  14. UnkleRhaukus
    • 2 years ago
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    \[\left[\begin{array}{ccc}6-2&5&2\\2&0-2&-8\\5&4&0-2\end{array}\right]\left[\begin{array}{ccc} x_1\\x_2\\x_3\end{array}\right]=\left[\begin{array}{ccc} 0\\0\\0\end{array}\right]\]

  15. nubeer
    • 2 years ago
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    ok.. got this far...

  16. UnkleRhaukus
    • 2 years ago
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    you have three simultaneous equations

  17. nubeer
    • 2 years ago
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    ok so i have to just solve simultaneously to get first eigen vector?

  18. UnkleRhaukus
    • 2 years ago
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    something like that, i always go confused when it comes to eigenvectors

  19. nubeer
    • 2 years ago
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    i think first eigen vector is x1= = 2 x2 = -2 x3 = 1

  20. waterineyes
    • 2 years ago
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    \[4x_1 + 5x_2 + 2x_3 = 0\] \[2x_1 -2x_2 - 8x_3 = 0\] \[5x_1 + 4x_2 - 2x_3 = 0\]

  21. UnkleRhaukus
    • 2 years ago
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    @nubeer , that certainly solves the three equations, how did you get there?

  22. nubeer
    • 2 years ago
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    i have found x1 and x2 in terms of x3 while using subsitution method.

  23. nubeer
    • 2 years ago
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    ok thanks guys.. i think i have done it... thank you very much espacially @UnkleRhaukus :)

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