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(B) We are given that $P = 	ext{adj}(A)$ and $|A| = 4$. We know the property of the adjoint of a matrix: $|	ext{adj}(A)| = |A|^{n-1}$,where $n$ is t

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$11$

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(B) We are given that $P = 	ext{adj}(A)$ and $|A| = 4$. We know the property of the adjoint of a matrix: $|	ext{adj}(A)| = |A|^{n-1}$,where $n$ is the order of the matrix. Here,$n = 3$,so $|P| = |	ext{adj}(A)| = |A|^{3-1} = |A|^2$. Given $|A| = 4$,we have $|P| = 4^2 = 16$. Now,calculate the determinant of $P$: $|P| = \begin{vmatrix} 1 & \alpha & 3 \ 1 & 3 & 3 \ 2 & 4 & 4 \end{vmatrix}$ $= 1(3 	imes 4 - 3 	imes 4) - \alpha(1 	imes 4 - 3 	imes 2) + 3(1 	imes 4 - 3 	imes 2)$ $= 1(12 - 12) - \alpha(4 - 6) + 3(4 - 6)$ $= 0 - \alpha(-2) + 3(-2)$ $= 2\alpha - 6$. Equating the two values of $|P|$: $2\alpha - 6 = 16$ $2\alpha = 22$ $\alpha = 11$.

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

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