# Find a power of matrix by Cayley-Hamilton theorem

Let $$A= \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 1 & 0 & 0 \\ \end{pmatrix}$$ And I should calculate $A^2$ and $A^{12}$ by Cayley Hamilton theorem.

I found that the characteristic polynomial is $f_A(x)=x^3-x-1$ and thus by Cayley Hamilton: $A^3-A=I_{3}$ .

I tried to multiply by $A^9$ but it didn't lead to something simple to express $A^2$ and $A^{12}$ by.

Any suggestions?

First calculate $A^2$, then calculate $A^6=(A+I_3)^2=A^2 + 2A + I_3$ and finally $A^{12}=(A^6)^2$
• You mean that I need to calculate clearly $A^2$ and than use it for the others? Thank You – A-H May 22 '16 at 18:48