Use the cayley hamilton theorum to work out high powers of matrices

Question

Let Matrix

$$A= \left( \begin{array}{ccc} 1 & 2& 3 \\ 0 & 1 & 0 \\ 0 & 5 & -1 \end{array} \right) $$

Compute $A^{25}$ using the cayley hamilton theorum

I know i use $-A^3+A^2+A-1=0$ but how do i make $A^{25}$ simper using this?

thanks for any help, much appreciated

Answer 1

$$A^{25}=A(A^3)^8=A(A^2+A-I)^8=\cdots$$

Answer 2

Solve the characteristic polynomial equation for $A^3$. Now you have $A^3$ in terms of lower powers of $A$. Square the expression for $A^3$ and replace all $A^3$ with the expression in terms of lower powers (note $A^4=AA^3$). This gives you $A^6$. Square $A^6$ to get $A^{12}$, simplify. Square one more time to get $A^{24}$, simplify then multiply by $A$ and simplify to get the answer.