1. **Problem Statement:** Find $P(C^{-1})$ where $P(x) = 2x^2 - 4x - 2I$ and $$C = \begin{bmatrix}1 & 3 & 2 \\ 2 & 0 & 3 \\ 1 & -1 & 1\end{bmatrix}$$ 2. **Step 1: Find $C^{-1}$** Calculate the inverse of matrix $C$ using the formula $C^{-1} = \frac{1}{\det(C)} \mathrm{adj}(C)$. - Compute $\det(C)$: $$\det(C) = 1 \times \begin{vmatrix}0 & 3 \\ -1 & 1\end{vmatrix} - 3 \times \begin{vmatrix}2 & 3 \\ 1 & 1\end{vmatrix} + 2 \times \begin{vmatrix}2 & 0 \\ 1 & -1\end{vmatrix}$$ $$= 1(0 \times 1 - (-1) \times 3) - 3(2 \times 1 - 3 \times 1) + 2(2 \times (-1) - 0 \times 1)$$ $$= 1(3) - 3(2 - 3) + 2(-2) = 3 - 3(-1) - 4 = 3 + 3 - 4 = 2$$ - Find the adjugate matrix $\mathrm{adj}(C)$ by calculating cofactors and transposing: $$\mathrm{adj}(C) = \begin{bmatrix}3 & -1 & -3 \\ -1 & -1 & 5 \\ 6 & -1 & -6\end{bmatrix}$$ - Therefore, $$C^{-1} = \frac{1}{2} \begin{bmatrix}3 & -1 & -3 \\ -1 & -1 & 5 \\ 6 & -1 & -6\end{bmatrix} = \begin{bmatrix}1.5 & -0.5 & -1.5 \\ -0.5 & -0.5 & 2.5 \\ 3 & -0.5 & -3\end{bmatrix}$$ 3. **Step 2: Compute $P(C^{-1}) = 2(C^{-1})^2 - 4C^{-1} - 2I$** - Calculate $(C^{-1})^2 = C^{-1} \times C^{-1}$: $$ (C^{-1})^2 = \begin{bmatrix}1.5 & -0.5 & -1.5 \\ -0.5 & -0.5 & 2.5 \\ 3 & -0.5 & -3\end{bmatrix} \times \begin{bmatrix}1.5 & -0.5 & -1.5 \\ -0.5 & -0.5 & 2.5 \\ 3 & -0.5 & -3\end{bmatrix} = \begin{bmatrix}3.75 & -0.25 & -6.75 \\ -7.5 & 1.5 & 9.5 \\ 7.5 & -1.5 & -12\end{bmatrix}$$ - Multiply by 2: $$2(C^{-1})^2 = \begin{bmatrix}7.5 & -0.5 & -13.5 \\ -15 & 3 & 19 \\ 15 & -3 & -24\end{bmatrix}$$ - Multiply $C^{-1}$ by 4: $$4C^{-1} = \begin{bmatrix}6 & -2 & -6 \\ -2 & -2 & 10 \\ 12 & -2 & -12\end{bmatrix}$$ - Multiply identity matrix $I$ by 2: $$2I = \begin{bmatrix}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{bmatrix}$$ - Now compute: $$P(C^{-1}) = 2(C^{-1})^2 - 4C^{-1} - 2I = \begin{bmatrix}7.5 & -0.5 & -13.5 \\ -15 & 3 & 19 \\ 15 & -3 & -24\end{bmatrix} - \begin{bmatrix}6 & -2 & -6 \\ -2 & -2 & 10 \\ 12 & -2 & -12\end{bmatrix} - \begin{bmatrix}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{bmatrix}$$ - Perform subtraction element-wise: $$= \begin{bmatrix}7.5 - 6 - 2 & -0.5 + 2 - 0 & -13.5 + 6 - 0 \\ -15 + 2 - 0 & 3 + 2 - 2 & 19 - 10 - 0 \\ 15 - 12 - 0 & -3 + 2 - 0 & -24 + 12 - 2\end{bmatrix} = \begin{bmatrix}-0.5 & 1.5 & -7.5 \\ -13 & 3 & 9 \\ 3 & -1 & -14\end{bmatrix}$$ **Final answer:** $$P(C^{-1}) = \begin{bmatrix}-0.5 & 1.5 & -7.5 \\ -13 & 3 & 9 \\ 3 & -1 & -14\end{bmatrix}$$