1. **Problem statement:** Given the matrix $$J = \begin{pmatrix}4 & 1 & 1 \\ 1 & 4 & 1 \\ 1 & 1 & 4\end{pmatrix}$$, find the matrix $$J^{-n}$$ for a positive integer $$n$$. 2. **Key idea:** To find $$J^{-n}$$, we first need to find the eigenvalues and eigenvectors of $$J$$, then express $$J$$ in its diagonalized form $$J = PDP^{-1}$$ where $$D$$ is diagonal with eigenvalues, and $$P$$ contains eigenvectors. Then $$J^{-n} = PD^{-n}P^{-1}$$ where $$D^{-n}$$ is the diagonal matrix with eigenvalues raised to the power $$-n$$. 3. **Step 1: Find eigenvalues of $$J$$.** Note that $$J = 4I + A$$ where $$A$$ is the matrix with zeros on the diagonal and ones off-diagonal: $$A = \begin{pmatrix}0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0\end{pmatrix}$$ The eigenvalues of $$A$$ are known: one eigenvalue is $$2$$ (since each row sums to 2), and the other two eigenvalues are $$-1$$ (with multiplicity 2). Therefore, eigenvalues of $$J$$ are $$4 + 2 = 6$$ and $$4 - 1 = 3$$ (with multiplicity 2). 4. **Step 2: Write eigenvalues:** $$\lambda_1 = 6, \quad \lambda_2 = 3, \quad \lambda_3 = 3$$ 5. **Step 3: Diagonalize $$J$$:** $$J = PDP^{-1}$$ where $$D = \mathrm{diag}(6,3,3)$$. 6. **Step 4: Compute $$J^{-n}$$:** $$J^{-n} = PD^{-n}P^{-1} = P \begin{pmatrix}6^{-n} & 0 & 0 \\ 0 & 3^{-n} & 0 \\ 0 & 0 & 3^{-n} \end{pmatrix} P^{-1}$$ 7. **Summary:** The matrix $$J^{-n}$$ is obtained by diagonalizing $$J$$, raising its eigenvalues to the power $$-n$$, and transforming back. This method works because $$J$$ is symmetric and diagonalizable. **Final answer:** $$J^{-n} = P \begin{pmatrix}6^{-n} & 0 & 0 \\ 0 & 3^{-n} & 0 \\ 0 & 0 & 3^{-n} \end{pmatrix} P^{-1}$$ where $$P$$ diagonalizes $$J$$.