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(C) Given $A = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 1 \ 0 & -2 & 4 \end{bmatrix}$.First,we find the characteristic equation of $A$. The characteristi

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$(-6, 11)$

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(C) Given $A = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 1 \ 0 & -2 & 4 \end{bmatrix}$.First,we find the characteristic equation of $A$. The characteristic polynomial is $|A - \lambda I| = 0$.$|A - \lambda I| = \begin{vmatrix} 1-\lambda & 0 & 0 \ 0 & 1-\lambda & 1 \ 0 & -2 & 4-\lambda \end{vmatrix} = (1-\lambda) [(1-\lambda)(4-\lambda) + 2] = (1-\lambda) [\lambda^2 - 5\lambda + 6] = 0$.So,$\lambda^3 - 6\lambda^2 + 11\lambda - 6 = 0$.By Cayley-Hamilton theorem,$A^3 - 6A^2 + 11A - 6I = 0$.Multiplying by $A^{-1}$,we get $A^2 - 6A + 11I - 6A^{-1} = 0$.$6A^{-1} = A^2 - 6A + 11I$.$A^{-1} = \frac{1}{6}[A^2 - 6A + 11I]$.Comparing this with $A^{-1} = \frac{1}{6}[A^2 + cA + dI]$,we get $c = -6$ and $d = 11$.Thus,the pair $(c, d) = (-6, 11)$.

If $A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & -2 & 4 \end{bmatrix}$ and $I = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$,and $A^{-1} = \frac{1}{6}[A^2 + cA + dI]$ where $c, d \in R$,then the pair of values $(c, d)$ is:

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