A = -1 2 -1 0 1 1 0 0 2

Solution:

p() = (-1)³ (+1) (-1) (-2)    (upper triangular matrix => eigenvalues lie on the main diagonal)
= - (³ - 2 ² - + 2 )
= - ³ + 2 ² + - 2

-A matrix satisfies its characteristic equation:
=> -A³ + 2A² + A - 2I = 0 <=> A³ = 2A² + A - 2I   | ·A
=> A⁴ = 2A³ + A² - 2A | Substitute A³ = 2A² + A - 2I
= 2(2A² + A - 2I) + A² - 2A = 5A² - 4I
=> A⁵ = 5A³ - 4A | Substitute A³ = 2A² + A - 2I
= 5(2A² + A - 2I) - 4A = 10A² +A - 10I

= 10 -1 2 -1 0 1 1 0 0 2

-1 2 -1 0 1 1 0 0 2

+ -1 2 -1 0 1 1 0 0 2

- 10 1 0 0 0 1 0 0 0 1

= 10 0 10 0 10 30 0 0 40

+ -1 2 -1 0 1 1 0 0 2

- 10 0 0 0 10 0 0 0 10

= -1 2 9 0 1 31 0 0 32