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(B) We know that for a $3 	imes 3$ matrix $A$, the determinant of its adjoint matrix is given by $\det(	ext{adj}(A)) = (\det(A))^{n-1}$, where $n$ i

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$11$

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(B) We know that for a $3 	imes 3$ matrix $A$, the determinant of its adjoint matrix is given by $\det(	ext{adj}(A)) = (\det(A))^{n-1}$, where $n$ is the order of the matrix.Here, $n = 3$, so $\det(P) = (\det(A))^{3-1} = (\det(A))^2$.Given $\det(A) = 4$, we have $\det(P) = 4^2 = 16$.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) = 2\alpha - 6$.Equating the two values: $2\alpha - 6 = 16$.$2\alpha = 22 \Rightarrow \alpha = 11$.

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

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