Find high powers of a matrix with the Cayley Hamilton theorem

Question

Let A = \begin{bmatrix} 0 & 1 & 0\\ 0 & 0 & 1\\ -1 & -1 &-1\\ \end{bmatrix}

Compute $A^{10000} + A^{9998}$

I know this should be done by the Cayley-Hamilton theorem. I get as characteristic polynomial $-A^3 - A^2 - A - I = 0$ but I don't see how to calculate $A^{10000} + A^{9998}$ from there. I hope someone can help me out!

Answer 1

Via the Cayley Hamilton Theorem: $$ A^4 = I + (A-I)(A^3 + A^2+A+I)=I;\\ A^{10000} + A^{9998} = I^{2500} + I^{2499}A^2 = I + A^2. $$