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any good mathematicians here?

I am lost and need guidance :/
Original post by Chinese Noodles
I am lost and need guidance :/


To find an eigenvector corresponding to an eigenvalue λ \lambda you need to find a vector (x,y,z)T (x, y, z)^T satisfying:

(0λ1152λ1805λ)(xyz)=(000) \begin{pmatrix} {0-\lambda} & {-1} & {1} \\5 & {-2-\lambda} & {-1} \\{-8} & 0 & {5 - \lambda} \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end {pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}

If you don't know how to do this, multiply out the LHS to get three simultaneous equations and solve them in terms of a parameter.

EDIT: Corrected my matrix.
(edited 8 years ago)
Original post by 16Characters....
To find an eigenvector corresponding to an eigenvalue λ \lambda you need to find a vector (x,y,z)T (x, y, z)^T satisfying:

(0λ1152λ1805λ)(xyz)=(000) \begin{pmatrix} {0-\lambda} & {-1} & {1} \\5 & {-2-\lambda} & {-1} \\{-8} & 0 & {5 - \lambda} \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end {pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}

If you don't know how to do this, multiply out the LHS to get three simultaneous equations and solve them in terms of a parameter.

EDIT: Corrected my matrix.


i got the cubic equation where x = lamnda


x^3 + 2x^2 -18x -1 = 0



doesnt give integers :/ one of the soloution is 1
Original post by Chinese Noodles
i got the cubic equation where x = lamnda


x^3 + 2x^2 -18x -1 = 0



doesnt give integers :/ one of the soloution is 1


How did you get that equation?
Original post by 16Characters....
How did you get that equation?


ok i got x^3 + 3x^2 + 3x -1 = 0 now




still no luck :frown:



i just used the determinant method
Original post by Chinese Noodles
ok i got x^3 + 3x^2 + 3x -1 = 0 now




still no luck :frown:



i just used the determinant method


So you are trying to find the eigenvalues? OK I thought you were looking for the eigenvector.

If so you are one sign wrong, the characteristic equation should be λ33λ2+3λ1=0 \lambda^3 - 3\lambda^2 + 3 \lambda - 1 = 0 .

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