Let B=begin{bmatrix} 1 & 3 & alpha 1 & 2 & 3 | vedclass

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(C) Given $B = 	ext{adj}(A)$. We know that $|B| = |	ext{adj}(A)| = |A|^{n-1}$. Here $n=3$ and $|A|=2$,so $|B| = 2^{3-1} = 2^2 = 4$.Calculating the d

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(C) Given $B = 	ext{adj}(A)$. We know that $|B| = |	ext{adj}(A)| = |A|^{n-1}$. Here $n=3$ and $|A|=2$,so $|B| = 2^{3-1} = 2^2 = 4$.Calculating the determinant of $B$: $|B| = 1(8 - 3\alpha) - 3(4 - 3\alpha) + \alpha(\alpha - 2\alpha) = 4$$8 - 3\alpha - 12 + 9\alpha - \alpha^2 = 4$$-\alpha^2 + 6\alpha - 4 = 4$$\alpha^2 - 6\alpha + 8 = 0$$(\alpha - 2)(\alpha - 4) = 0$Since $\alpha > 2$,we have $\alpha = 4$.Now,substitute $\alpha = 4$ into the expression:$B = \begin{bmatrix} 1 & 3 & 4 \ 1 & 2 & 3 \ 4 & 4 & 4 \end{bmatrix}$The expression is $X^T B X$ where $X = \begin{bmatrix} \alpha \ -2\alpha \ \alpha \end{bmatrix} = \begin{bmatrix} 4 \ -8 \ 4 \end{bmatrix}$.$X^T B X = \begin{bmatrix} 4 & -8 & 4 \end{bmatrix} \begin{bmatrix} 1 & 3 & 4 \ 1 & 2 & 3 \ 4 & 4 & 4 \end{bmatrix} \begin{bmatrix} 4 \ -8 \ 4 \end{bmatrix}$$= \begin{bmatrix} 4(1) - 8(1) + 4(4) & 4(3) - 8(2) + 4(4) & 4(4) - 8(3) + 4(4) \end{bmatrix} \begin{bmatrix} 4 \ -8 \ 4 \end{bmatrix}$$= \begin{bmatrix} 12 & 12 & -8 \end{bmatrix} \begin{bmatrix} 4 \ -8 \ 4 \end{bmatrix}$$= 12(4) + 12(-8) - 8(4) = 48 - 96 - 32 = -80$.

Let $B=\begin{bmatrix} 1 & 3 & \alpha \\ 1 & 2 & 3 \\ \alpha & \alpha & 4 \end{bmatrix}, \alpha > 2$ be the adjoint of a matrix $A$ and $|A|=2$. Then the value of $\begin{bmatrix} \alpha & -2\alpha & \alpha \end{bmatrix} B \begin{bmatrix} \alpha \\ -2\alpha \\ \alpha \end{bmatrix}$ is equal to:

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