1. **State the problem:** Find the inverse of matrix $A$ using the Cayley–Hamilton theorem, where $$A=\begin{bmatrix}2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2\end{bmatrix}$$ 2. **Recall Cayley–Hamilton theorem:** Every square matrix satisfies its own characteristic equation. If the characteristic polynomial of $A$ is $p(\lambda) = \det(\lambda I - A)$, then $p(A) = 0$. 3. **Find the characteristic polynomial:** $$p(\lambda) = \det(\lambda I - A) = \det\begin{bmatrix}\lambda - 2 & -1 & -1 \\ -1 & \lambda - 2 & -1 \\ -1 & -1 & \lambda - 2\end{bmatrix}$$ 4. Calculate the determinant: $$p(\lambda) = (\lambda - 2)^3 - 3(\lambda - 2) + 2$$ Expanding: $$p(\lambda) = (\lambda - 2)^3 - 3(\lambda - 2) + 2$$ 5. Expand $(\lambda - 2)^3$: $$ (\lambda - 2)^3 = \lambda^3 - 6\lambda^2 + 12\lambda - 8 $$ 6. Substitute back: $$p(\lambda) = \lambda^3 - 6\lambda^2 + 12\lambda - 8 - 3\lambda + 6 + 2 = \lambda^3 - 6\lambda^2 + 9\lambda = 0$$ Simplify constants: $$-8 + 6 + 2 = 0$$ So, $$p(\lambda) = \lambda^3 - 6\lambda^2 + 9\lambda = 0$$ 7. Factor out $\lambda$: $$p(\lambda) = \lambda(\lambda^2 - 6\lambda + 9) = 0$$ 8. The quadratic factors as: $$\lambda^2 - 6\lambda + 9 = (\lambda - 3)^2$$ So, $$p(\lambda) = \lambda(\lambda - 3)^2 = 0$$ 9. By Cayley–Hamilton theorem: $$p(A) = A(A - 3I)^2 = 0$$ 10. Rearrange to express $A^{-1}$: Since $A$ is invertible (eigenvalues are 0 and 3, but 0 eigenvalue means not invertible; however, check carefully), note that eigenvalue 0 implies $A$ is singular, so $A^{-1}$ does not exist. 11. But since $p(\lambda)$ has a root at 0, $A$ is singular and does not have an inverse. **Final answer:** Matrix $A$ is singular and does not have an inverse.