If P = begin{bmatrix} 1 & alpha & 3 1 & 3 & | vedclass

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(B) We know that for a $3 	imes 3$ matrix $A$,the determinant of its adjoint matrix $P$ is given by $|P| = |A|^{n-1}$,where $n$ is the order of the m

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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 $P$ is given by $|P| = |A|^{n-1}$,where $n$ is the order of the matrix.Here,$n = 3$ and $|A| = 4$,so $|P| = |A|^{3-1} = |A|^2 = 4^2 = 16$.Now,calculate the determinant of $P = \begin{bmatrix} 1 & \alpha & 3 \ 1 & 3 & 3 \ 2 & 4 & 4 \end{bmatrix}$:$|P| = 1(3 	imes 4 - 3 	imes 4) - \alpha(1 	imes 4 - 3 	imes 2) + 3(1 	imes 4 - 3 	imes 2)$$|P| = 1(12 - 12) - \alpha(4 - 6) + 3(4 - 6)$$|P| = 0 - \alpha(-2) + 3(-2)$$|P| = 2\alpha - 6$.Equating the two values of $|P|$:$2\alpha - 6 = 16$$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 $\alpha$ is equal to:

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