If B = begin{bmatrix} 1 & alpha & 2 1 & 2 & | vedclass

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(B) We know that for a square matrix $A$ of order $n$,$|\operatorname{Adj} A| = |A|^{n-1}$.Given that $B = \operatorname{Adj}(A)$ and $A$ is a $3 	im

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$27$

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(B) We know that for a square matrix $A$ of order $n$,$|\operatorname{Adj} A| = |A|^{n-1}$.Given that $B = \operatorname{Adj}(A)$ and $A$ is a $3 	imes 3$ matrix,we have $n = 3$.Therefore,$|B| = |A|^{3-1} = |A|^2$.Given $|A| = 5$,we have $|B| = 5^2 = 25$.Now,calculate the determinant of $B$:$|B| = \begin{vmatrix} 1 & \alpha & 2 \ 1 & 2 & 2 \ 2 & 3 & 3 \end{vmatrix} = 1(2 	imes 3 - 2 	imes 3) - \alpha(1 	imes 3 - 2 	imes 2) + 2(1 	imes 3 - 2 	imes 2)$.$|B| = 1(6 - 6) - \alpha(3 - 4) + 2(3 - 4)$.$|B| = 0 - \alpha(-1) + 2(-1) = \alpha - 2$.Equating the two values of $|B|$:$\alpha - 2 = 25$.$\alpha = 27$.

If $B = \begin{bmatrix} 1 & \alpha & 2 \\ 1 & 2 & 2 \\ 2 & 3 & 3 \end{bmatrix}$ is the adjoint of a $3 \times 3$ matrix $A$ and $|A| = 5$,then $\alpha$ is equal to

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