1. **Problem statement:** Given matrix $$A = \begin{pmatrix}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{pmatrix}$$, show that $$A^2 = 4A + 5I$$ where $$I$$ is the 3x3 identity matrix. Then use this to find $$A^{-1}$$. 2. **Calculate $$A^2$$:** Multiply $$A$$ by itself: $$ A^2 = A \times A = \begin{pmatrix}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{pmatrix} \times \begin{pmatrix}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{pmatrix} $$ Calculate each element: - First row, first column: $$1\times1 + 2\times2 + 2\times2 = 1 + 4 + 4 = 9$$ - First row, second column: $$1\times2 + 2\times1 + 2\times2 = 2 + 2 + 4 = 8$$ - First row, third column: $$1\times2 + 2\times2 + 2\times1 = 2 + 4 + 2 = 8$$ - Second row, first column: $$2\times1 + 1\times2 + 2\times2 = 2 + 2 + 4 = 8$$ - Second row, second column: $$2\times2 + 1\times1 + 2\times2 = 4 + 1 + 4 = 9$$ - Second row, third column: $$2\times2 + 1\times2 + 2\times1 = 4 + 2 + 2 = 8$$ - Third row, first column: $$2\times1 + 2\times2 + 1\times2 = 2 + 4 + 2 = 8$$ - Third row, second column: $$2\times2 + 2\times1 + 1\times2 = 4 + 2 + 2 = 8$$ - Third row, third column: $$2\times2 + 2\times2 + 1\times1 = 4 + 4 + 1 = 9$$ So, $$ A^2 = \begin{pmatrix}9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9\end{pmatrix} $$ 3. **Calculate $$4A + 5I$$:** $$ 4A = 4 \times \begin{pmatrix}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{pmatrix} = \begin{pmatrix}4 & 8 & 8 \\ 8 & 4 & 8 \\ 8 & 8 & 4\end{pmatrix} $$ $$ 5I = 5 \times \begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix} = \begin{pmatrix}5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5\end{pmatrix} $$ Add them: $$ 4A + 5I = \begin{pmatrix}4+5 & 8+0 & 8+0 \\ 8+0 & 4+5 & 8+0 \\ 8+0 & 8+0 & 4+5\end{pmatrix} = \begin{pmatrix}9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9\end{pmatrix} $$ 4. **Verify the equality:** We see that $$ A^2 = 4A + 5I $$ which confirms the given relation. 5. **Find $$A^{-1}$$:** From the equation, $$ A^2 = 4A + 5I $$ Rewrite as $$ A^2 - 4A - 5I = 0 $$ Factor as $$ A^2 - 4A = 5I $$ Multiply both sides by $$A^{-1}$$ (assuming $$A$$ is invertible): $$ A \cancel{A^{-1}} A - 4 \cancel{A} \cancel{A^{-1}} = 5 A^{-1} $$ which simplifies to $$ A - 4I = 5 A^{-1} $$ Therefore, $$ A^{-1} = \frac{1}{5}(A - 4I) $$ **Final answer:** $$ \boxed{A^{-1} = \frac{1}{5}(A - 4I)} $$