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.