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liliy Group Title

Find all the eigenvalues for the given matrices. if it is a 3x3. and a polynomial. how do i know what to break it down into?

  • 2 years ago
  • 2 years ago

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  1. liliy Group Title
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    [ 4 0 1 -1 -6 -2 5 0 0 ] \[\lambda ^3+2 lambda ^2 - 29 lambda -30]/

    • 2 years ago
  2. UnkleRhaukus Group Title
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    \[\textbf Av=\lambda v\] \[\left(\textbf A-\lambda \textbf I\right)v =0\] \[\left|\textbf A-\lambda \textbf I\right|v =0\] \[\left|\left(\begin{array}\ 4&0&1\\-1&-6&-2\\5&0&0\end{array}\right)-\left(\begin{array}\ \lambda&0&0\\0&\lambda & 0\\0 &0 &\lambda \end{array}\right)\right|=0\] \[\left|\left(\begin{array}\ 4-\lambda&0&1\\-1&-6-\lambda&-2\\5&0&0-\lambda\end{array}\right)\right|=0\] \[(4-\lambda)[-2\times0-(-6-\lambda)(0-\lambda)]-0[-1\times(0-\lambda)]+1[-1\times0-(-2\times5)]=0\]

    • 2 years ago
  3. UnkleRhaukus Group Title
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    *\[(4-\lambda)[-2\times0-(-6-\lambda)(0-\lambda)]\]\[\qquad-0[-1\times(0-\lambda)-(5\times-2)]\]\[\qquad\qquad+1[-1\times0-(-6-\lambda)]=0\]

    • 2 years ago
  4. liliy Group Title
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    right, i know how to do all the calculations but when you get the the last step: when its all combined how do you know what lamda itself is equal to? ( like, you look at the constant and see its factors or something and do something..idk)_)

    • 2 years ago
  5. UnkleRhaukus Group Title
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    well at the last spet you'll have a bunch of factors =0 so to get zero at least one of the factors is zero

    • 2 years ago
  6. UnkleRhaukus Group Title
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    something like this only i have made mistakes \[(\lambda-4)[(6+\lambda)\lambda]+[6+\lambda]=0\]\[(6\lambda+\lambda^2-24-4\lambda)\lambda+6+\lambda=0\]\[\lambda^3+2\lambda^2-24\lambda+\lambda+6=0\]\[\lambda^3+3\lambda^2-24\lambda+6=0\]\[(\lambda-a)(\lambda-b)(\lambda-c)=0\] lambda would be a,b,c

    • 2 years ago
  7. liliy Group Title
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    but how do you know to break it into a b c?

    • 2 years ago
  8. UnkleRhaukus Group Title
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    how to factorize? you could try polynomial long division

    • 2 years ago
  9. liliy Group Title
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    ugh

    • 2 years ago
  10. liliy Group Title
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    \[\lambda^3 +2\lambda^2-29\lambda-30\]

    • 2 years ago
  11. liliy Group Title
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    how would you go about doing this? possibilities: +/- : 1,2,3,5,6,10,30,15

    • 2 years ago
  12. liliy Group Title
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    if you plug in -1, you get =0.then what?

    • 2 years ago
  13. UnkleRhaukus Group Title
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    your expression does factorizes nicely, try dividing by \((\lambda+1)\)

    • 2 years ago
  14. liliy Group Title
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    i dont know how... i knwo that sounds dumb but how do u do that?

    • 2 years ago
  15. UnkleRhaukus Group Title
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    |dw:1341885937030:dw|

    • 2 years ago
  16. UnkleRhaukus Group Title
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    |dw:1341885983310:dw|

    • 2 years ago
  17. UnkleRhaukus Group Title
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    so\[\lambda^3 +2\lambda^2-29\lambda-30=(\lambda+1)(\lambda^2+\lambda-30)\]

    • 2 years ago
  18. UnkleRhaukus Group Title
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    now just factorize the quadratic factor...

    • 2 years ago
  19. liliy Group Title
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    thanks. also: do you know about diagnolization?

    • 2 years ago
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