1. The problem is to find the expression for $J^{-n}$ where $J$ is the matrix $$J = \begin{pmatrix}4 & 1 & 1 \\ 1 & 4 & 1 \\ 1 & 1 & 4\end{pmatrix}$$ 2. To find $J^{-n}$, we first need to diagonalize $J$ or find its eigenvalues and eigenvectors because powers of matrices are easier to compute in diagonal form. 3. The matrix $J$ is symmetric and has the form $J = 4I + A$ where $$A = \begin{pmatrix}0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0\end{pmatrix}$$ 4. The eigenvalues of $A$ are known: one eigenvalue is $2$ (corresponding to the vector $(1,1,1)$) and the other two are $-1$ (with multiplicity 2). 5. Therefore, the eigenvalues of $J$ are $4 + 2 = 6$ and $4 - 1 = 3$ (with multiplicity 2). 6. The diagonalization of $J$ is $$J = PDP^{-1}$$ where $D = \text{diag}(6,3,3)$. 7. Then, $$J^{-n} = PD^{-n}P^{-1} = P \begin{pmatrix}6^{-n} & 0 & 0 \\ 0 & 3^{-n} & 0 \\ 0 & 0 & 3^{-n}\end{pmatrix} P^{-1}$$ 8. Using the spectral decomposition, the matrix $J$ can be expressed as $$J = 6P_1 + 3P_2$$ where $P_1$ is the projection onto the eigenspace of eigenvalue 6 and $P_2$ is the projection onto the eigenspace of eigenvalue 3. 9. The projection $P_1$ corresponds to the vector $\frac{1}{\sqrt{3}}(1,1,1)$: $$P_1 = \frac{1}{3} \begin{pmatrix}1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1\end{pmatrix}$$ 10. The projection $P_2 = I - P_1$. 11. Therefore, $$J^{-n} = 6^{-n} P_1 + 3^{-n} P_2 = 6^{-n} \frac{1}{3} \begin{pmatrix}1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1\end{pmatrix} + 3^{-n} \left(I - \frac{1}{3} \begin{pmatrix}1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1\end{pmatrix}\right)$$ 12. Simplifying, $$J^{-n} = \frac{6^{-n}}{3} \begin{pmatrix}1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1\end{pmatrix} + 3^{-n} \begin{pmatrix}1 - \frac{1}{3} & -\frac{1}{3} & -\frac{1}{3} \\ -\frac{1}{3} & 1 - \frac{1}{3} & -\frac{1}{3} \\ -\frac{1}{3} & -\frac{1}{3} & 1 - \frac{1}{3}\end{pmatrix} = \frac{6^{-n}}{3} J_1 + 3^{-n} J_2$$ where $$J_1 = \begin{pmatrix}1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1\end{pmatrix}, \quad J_2 = \begin{pmatrix}\frac{2}{3} & -\frac{1}{3} & -\frac{1}{3} \\ -\frac{1}{3} & \frac{2}{3} & -\frac{1}{3} \\ -\frac{1}{3} & -\frac{1}{3} & \frac{2}{3}\end{pmatrix}$$ Final answer: $$\boxed{J^{-n} = \frac{6^{-n}}{3} \begin{pmatrix}1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1\end{pmatrix} + 3^{-n} \begin{pmatrix}\frac{2}{3} & -\frac{1}{3} & -\frac{1}{3} \\ -\frac{1}{3} & \frac{2}{3} & -\frac{1}{3} \\ -\frac{1}{3} & -\frac{1}{3} & \frac{2}{3}\end{pmatrix}}$$