Let P=begin{bmatrix} 3 & -1 & -2 2 & 0 & alp | vedclass

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(A) Given $PQ = kI$,we have $Q = kP^{-1}$. Since $P^{-1} = \frac{1}{\det(P)} 	ext{adj}(P)$,we have $Q = \frac{k}{\det(P)} 	ext{adj}(P)$. First,calcu

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$B, C$

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(A) Given $PQ = kI$,we have $Q = kP^{-1}$. Since $P^{-1} = \frac{1}{\det(P)} 	ext{adj}(P)$,we have $Q = \frac{k}{\det(P)} 	ext{adj}(P)$. First,calculate $\det(P) = 3(0 - (-5\alpha)) - (-1)(0 - 3\alpha) + (-2)(-10 - 0) = 3(5\alpha) + (-3\alpha) + 20 = 15\alpha - 3\alpha + 20 = 12\alpha + 20$. The element $q_{23}$ is the $(2,3)$-th entry of $Q$. $q_{23} = \frac{k}{\det(P)} 	imes (	ext{adj}(P))_{23} = \frac{k}{12\alpha + 20} 	imes (-1)^{2+3} M_{32} = \frac{k}{12\alpha + 20} 	imes (-1) 	imes (3\alpha - (-4)) = \frac{-k(3\alpha + 4)}{12\alpha + 20} = \frac{-k(3\alpha + 4)}{4(3\alpha + 5)}$. Given $q_{23} = -\frac{k}{8}$,we have $\frac{3\alpha + 4}{4(3\alpha + 5)} = \frac{1}{8} \implies 2(3\alpha + 4) = 3\alpha + 5 \implies 6\alpha + 8 = 3\alpha + 5 \implies 3\alpha = -3 \implies \alpha = -1$. Then $\det(P) = 12(-1) + 20 = 8$. Since $Q = kP^{-1}$,$\det(Q) = k^3 \det(P^{-1}) = \frac{k^3}{\det(P)} = \frac{k^3}{8}$. Given $\det(Q) = \frac{k^2}{2}$,we have $\frac{k^3}{8} = \frac{k^2}{2} \implies k = 4$. Check options: $(B)$ $4\alpha - k + 8 = 4(-1) - 4 + 8 = -4 - 4 + 8 = 0$. (Correct) $(C)$ $\det(P 	ext{adj}(Q)) = \det(P) \det(	ext{adj}(Q)) = \det(P) (\det(Q))^2 = 8 	imes (\frac{4^2}{2})^2 = 8 	imes 8^2 = 8^3 = 512 = 2^9$. (Correct) $(D)$ $\det(Q 	ext{adj}(P)) = \det(Q) \det(	ext{adj}(P)) = \det(Q) (\det(P))^2 = \frac{16}{2} 	imes 8^2 = 8 	imes 64 = 512 = 2^9 
eq 2^{13}$. (Incorrect)

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