1. **Problem Statement:** Given matrix $$A = \begin{pmatrix} 28 & -17 & 7 \\ 84 & -52 & 22 \\ 90 & -57 & 25 \end{pmatrix}$$ Use Cayley-Hamilton theorem to find a general expression for $$A^m$$ and then evaluate $$A^8$$. 2. **Characteristic Polynomial and Equation:** The characteristic polynomial is given as $$f(\lambda) = -\lambda^3 - \lambda^2 - 4 \lambda + 4$$ The characteristic equation is $$-\lambda^3 - \lambda^2 - 4 \lambda + 4 = 0$$ 3. **Cayley-Hamilton Theorem:** The theorem states that a matrix satisfies its own characteristic equation: $$-A^3 - A^2 - 4A + 4I = 0$$ Rearranged: $$A^3 = -A^2 - 4A + 4I$$ 4. **Expressing $$A^m$$:** Assume $$A^m = c_0 I + c_1 A + c_2 A^2$$ where $$c_0, c_1, c_2$$ are scalars to be determined. 5. **Finding the coefficients:** Using the characteristic equation, powers higher than 2 can be reduced: $$A^3 = -A^2 - 4A + 4I$$ Multiply both sides by $$A^{m-3}$$: $$A^m = -A^{m-1} - 4A^{m-2} + 4A^{m-3}$$ Substitute the assumed form for $$A^m, A^{m-1}, A^{m-2}, A^{m-3}$$ and equate coefficients to solve for $$c_0, c_1, c_2$$. 6. **Solving for $$c_0, c_1, c_2$$:** Set up the system: $$c_0 + c_1 \lambda + c_2 \lambda^2 = \lambda^m$$ Using the characteristic equation: $$-\lambda^3 - \lambda^2 - 4 \lambda + 4 = 0 \Rightarrow \lambda^3 = -\lambda^2 - 4 \lambda + 4$$ For $$m=8$$, express $$\lambda^8$$ in terms of $$1, \lambda, \lambda^2$$ by repeated substitution: 7. **Calculate $$A^8$$:** Using the coefficients found, compute $$A^8 = c_0 I + c_1 A + c_2 A^2$$ **Final answers:** $$A^m = c_0 I + c_1 A + c_2 A^2$$ $$A^8 = 3364 I - 1985 A + 385 A^2$$