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(C) We know that for any square matrix $A$ of order $n$,the property $\det(	ext{adj}(A)) = (\det(A))^{n-1}$ holds true.Here,the order of matrix $A$ i

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$11$

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(C) We know that for any square matrix $A$ of order $n$,the property $\det(	ext{adj}(A)) = (\det(A))^{n-1}$ holds true.Here,the order of matrix $A$ is $n = 3$ and $\det(A) = 4$.Therefore,$\det(P) = \det(	ext{adj}(A)) = (\det(A))^{3-1} = (4)^2 = 16$.Now,calculate the determinant of matrix $P = \begin{bmatrix} 1 & \alpha & 3 \ 1 & 3 & 3 \ 2 & 4 & 4 \end{bmatrix}$:$\det(P) = 1(3 	imes 4 - 3 	imes 4) - \alpha(1 	imes 4 - 3 	imes 2) + 3(1 	imes 4 - 3 	imes 2)$$\det(P) = 1(12 - 12) - \alpha(4 - 6) + 3(4 - 6)$$\det(P) = 0 - \alpha(-2) + 3(-2)$$\det(P) = 2\alpha - 6$.Equating the two values of $\det(P)$:$2\alpha - 6 = 16$$2\alpha = 22$$\alpha = 11$.Thus,the value of $\alpha$ is $11$.

If $P = \begin{bmatrix} 1 & \alpha & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4 \end{bmatrix}$ is the adjoint of a matrix $A$ and $\det(A) = 4$,then the value of $\alpha$ is

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