The characteristic equation is as shown:
Expanding determinant:
(-2 – λ) [(1-λ) (5-λ)-2x2] + 4[(-2) x (5-λ) -4x2] + 2[(-2) x 2-4(1-λ)] =0
- λ3 + 4λ2 + 27λ – 90 = 0,
λ3 - 4 λ2 -27 λ + 90 = 0
Here we have an algebraic equation built from the eigenvectors.
By hit and trial:
33 – 4 x 32 - 27 x 3 +90 = 0
Hence, (λ - 3) is a factor:
λ3 - 4 λ2 - 27 λ +90 = (λ – 3) (λ2 – λ – 30)
Eigenvalues are 3,-5,6:
(λ – 3) (λ2 – λ – 30) = (λ – 3) (λ+5) (λ-6),
Calculate eigenvector for λ = 3
For X = 1,
-5 - 4Y + 2Z =0,
-2 - 2Y + 2Z =0
Subtracting the two equations:
3 + 2Y = 0,
Subtracting back into second equation:
Y = -(3/2)
Z = -(1/2)
Similarly, we can calculate the eigenvectors for -5 and 6.
