Given $A = \begin{bmatrix} 1 & 2 \\ 2 & -1 \end{bmatrix}$ The characteristic equation of matrix A is

$$|A - \lambda I| = 0 \\ \phi(\lambda) = \lambda^2 – 5 = 0 \\ \therefore \lambda = \pm \sqrt{5}$$

To show that A satisfies this equation i.e. to show that

$$\phi(A) = A^2 – 5I = 0 \\ A^2 = A \cdot A \\ = \begin{bmatrix} 1 & 2 \\ 2 & -1 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 2 & -1 \end{bmatrix} \\ = \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix} \\ = 5 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \\ A^2 = 5I \\ A^2 – 5I = 0$$

Hence Cayley – Hamilton theorem is verified

$$A^4 = A^2 \cdot A^2 = 25I \\ A^8 = A^4 \cdot A^4 \\ = 25 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \cdot 25 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \\ = 625 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \\ A^8 = 625I$$

Hence $A^8 = 6251$