Let P=begin{bmatrix} -30 & 20 & 56 90 & 140 | vedclass

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(C) Let $M = (P^{-1}AP - I)^{2}$.We know that $\det(P^{-1}AP - I) = \det(P^{-1}(A - I)P) = \det(P^{-1}) \det(A - I) \det(P) = \det(A - I)$.Therefore,$

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$36$

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(C) Let $M = (P^{-1}AP - I)^{2}$.We know that $\det(P^{-1}AP - I) = \det(P^{-1}(A - I)P) = \det(P^{-1}) \det(A - I) \det(P) = \det(A - I)$.Therefore,$\det(M) = (\det(A - I))^{2}$.Now,calculate $A - I$:$A - I = \begin{bmatrix} 2-1 & 7 & \omega^{2} \ -1 & -\omega-1 & 1 \ 0 & -\omega & -\omega+1-1 \end{bmatrix} = \begin{bmatrix} 1 & 7 & \omega^{2} \ -1 & -\omega-1 & 1 \ 0 & -\omega & -\omega \end{bmatrix}$.Using the property $1 + \omega + \omega^{2} = 0$,we have $-\omega - 1 = \omega^{2}$.$A - I = \begin{bmatrix} 1 & 7 & \omega^{2} \ -1 & \omega^{2} & 1 \ 0 & -\omega & -\omega \end{bmatrix}$.Calculating the determinant:$\det(A - I) = 1(\omega^{2}(-\omega) - (1)(-\omega)) - 7((-1)(-\omega) - (1)(0)) + \omega^{2}((-1)(-\omega) - (\omega^{2})(0))$$= 1(-\omega^{3} + \omega) - 7(\omega) + \omega^{2}(\omega)$$= -1 + \omega - 7\omega + \omega^{3} = -1 - 6\omega + 1 = -6\omega$.Thus,$\det(M) = (-6\omega)^{2} = 36\omega^{2}$.Given $\det(M) = \alpha \omega^{2}$,we get $\alpha = 36$.

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