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

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

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$1$

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(A) We know that for a square matrix $A$ of order $n$,$|\operatorname{adj} A| = |A|^{n-1}$.Given that $B = \operatorname{adj} A$ and $n = 3$,we have $|B| = |A|^{3-1} = |A|^2$.Given $|A| = 4$,so $|B| = 4^2 = 16$.Now,calculate the determinant of $B$:$|B| = \begin{vmatrix} 3 & \alpha & -1 \ 1 & 3 & 1 \ -1 & 1 & 3 \end{vmatrix} = 3(9-1) - \alpha(3+1) - 1(1+3) = 3(8) - 4\alpha - 4 = 24 - 4\alpha - 4 = 20 - 4\alpha$.Equating the two values of $|B|$:$20 - 4\alpha = 16$$4\alpha = 4$$\alpha = 1$.

If $B = \begin{bmatrix} 3 & \alpha & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3 \end{bmatrix}$ is the adjoint of a $3 \times 3$ matrix $A$ and $|A| = 4$,then $\alpha$ is equal to:

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