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(B) The characteristic equation of matrix $A$ is given by $|A - \lambda I| = 0$. $|A - \lambda I| = \begin{vmatrix} 2-\lambda & 1 & 1 \ 0 & 1-\lambda

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$\frac{-1}{3}$

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(B) The characteristic equation of matrix $A$ is given by $|A - \lambda I| = 0$. $|A - \lambda I| = \begin{vmatrix} 2-\lambda & 1 & 1 \ 0 & 1-\lambda & 0 \ 1 & 1 & 2-\lambda \end{vmatrix} = (1-\lambda) \begin{vmatrix} 2-\lambda & 1 \ 1 & 2-\lambda \end{vmatrix} = (1-\lambda)((2-\lambda)^2 - 1) = (1-\lambda)(\lambda^2 - 4\lambda + 3) = (1-\lambda)(\lambda-1)(\lambda-3) = -(\lambda-1)^2(\lambda-3) = -(\lambda^2 - 2\lambda + 1)(\lambda-3) = -(\lambda^3 - 3\lambda^2 - 2\lambda^2 + 6\lambda + \lambda - 3) = -(\lambda^3 - 5\lambda^2 + 7\lambda - 3) = -\lambda^3 + 5\lambda^2 - 7\lambda + 3 = 0$. By Cayley-Hamilton theorem, $A^3 - 5A^2 + 7A - 3I = 0$. Multiplying by $A^{-1}$, we get $A^2 - 5A + 7I - 3A^{-1} = 0$, which implies $3A^{-1} = A^2 - 5A + 7I$, or $A^{-1} = \frac{1}{3}A^2 - \frac{5}{3}A + \frac{7}{3}I$. Comparing this with $A^{-1} = \alpha A^2 + \beta A + \gamma I$, we get $\alpha = \frac{1}{3}$, $\beta = -\frac{5}{3}$, and $\gamma = \frac{7}{3}$. Thus, $17\alpha + 5\beta + \gamma = 17(\frac{1}{3}) + 5(-\frac{5}{3}) + \frac{7}{3} = \frac{17 - 25 + 7}{3} = \frac{-1}{3}$.

Let $A = \begin{bmatrix} 2 & 1 & 1 \\ 0 & 1 & 0 \\ 1 & 1 & 2 \end{bmatrix}$. If $A^{-1} = \alpha A^2 + \beta A + \gamma I$, where $\alpha, \beta, \gamma$ are real numbers and $I$ is a $3 \times 3$ identity matrix, then $17 \alpha + 5 \beta + \gamma =$

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