anonymous
  • anonymous
find eigenvector for eigenvalue -2...
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
after i plug in the value, i have the matrix |dw:1334524320293:dw|
anonymous
  • anonymous
multiplied 1st row by 2 and add to second row, multiply 1st row by -2 and add to 3rd row... |dw:1334524445998:dw|
anonymous
  • anonymous
multiply second row by -1/2 and add to 3rd row... |dw:1334524531744:dw|

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anonymous
  • anonymous
so now i have these equations v1+v2+2v3=0 -2v2+2v3=0 -6v3=0
anonymous
  • anonymous
i'm not sure what the vector is, if v3=0, then -2v2+2(0)=0 so v2 must be 0 also and then v1+0+0=0 would mean v1 is zero too, right?
anonymous
  • anonymous
but that's what you wanted |A-I(Lambda)| =0
anonymous
  • anonymous
just put any value, for example v3=a and find the other
anonymous
  • anonymous
well for other eigenvalues, i have vectors with nonzero values...ok, if i plug in 3for v3, the last equation wouldn't make sense. and the answer isn't 0 vector. the vectors i have for my other values are correct
phi
  • phi
you should get a singular matrix the last row should be 0
anonymous
  • anonymous
-2v2+2a=0 v2=a v1+v2+2v3=0 v1+a+2a=0 v1 = -3a eigenvector (1,1,3)
anonymous
  • anonymous
it doesn't have to be all 0
anonymous
  • anonymous
@phi is right the matrix should be singular. So you made a mistake somewhere
anonymous
  • anonymous
|A-I(Lambda)| =0
phi
  • phi
remember you are solving det(A- lambda*I)= 0
anonymous
  • anonymous
i was taught the bottom left triangle has to be 0s.
anonymous
  • anonymous
so it would be the bottom left corner, and the values right above and to the right
anonymous
  • anonymous
the determinant has to be 0
anonymous
  • anonymous
i've checked and checked to see if i made a mistake...i will look again
phi
  • phi
what is the original matrix?
anonymous
  • anonymous
i think your eigenvalue is wrong
anonymous
  • anonymous
ugh idiot i did do it wrong. value is right, i will rework this
anonymous
  • anonymous
just trying to help

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