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

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(C) Given that $P = 	ext{adj}(A)$ and $|A| = 4$. For a $3 	imes 3$ matrix $A$,the property of the adjoint matrix is $|	ext{adj}(A)| = |A|^{n-1}$,wh

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$11$

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(C) Given that $P = 	ext{adj}(A)$ and $|A| = 4$. For a $3 	imes 3$ matrix $A$,the property of the adjoint matrix is $|	ext{adj}(A)| = |A|^{n-1}$,where $n$ is the order of the matrix. Here,$n = 3$,so $|	ext{adj}(A)| = |A|^{3-1} = |A|^2$. Given $|A| = 4$,we have $|	ext{adj}(A)| = 4^2 = 16$. Since $P = 	ext{adj}(A)$,we have $|P| = 16$. Calculating the determinant of $P$: $|P| = \begin{vmatrix} 1 & \alpha & 3 \ 1 & 3 & 3 \ 2 & 4 & 4 \end{vmatrix} = 1(12 - 12) - \alpha(4 - 6) + 3(4 - 6) = 16$. $0 - \alpha(-2) + 3(-2) = 16$. $2\alpha - 6 = 16$. $2\alpha = 22$. $\alpha = 11$.

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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