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(A) Given $A = \begin{bmatrix} \alpha & \beta & \gamma \ \alpha^{2} & \beta^{2} & \gamma^{2} \ \beta+\gamma & \gamma+\alpha & \alpha+\beta \end{bmat

Vedclass · Accepted Answer

$42$

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(A) Given $A = \begin{bmatrix} \alpha & \beta & \gamma \ \alpha^{2} & \beta^{2} & \gamma^{2} \ \beta+\gamma & \gamma+\alpha & \alpha+\beta \end{bmatrix}$.Applying $R_{3} ightarrow R_{3} + R_{1}$,we get $|A| = \begin{vmatrix} \alpha & \beta & \gamma \ \alpha^{2} & \beta^{2} & \gamma^{2} \ \alpha+\beta+\gamma & \alpha+\beta+\gamma & \alpha+\beta+\gamma \end{vmatrix}$.Taking $(\alpha+\beta+\gamma)$ common from $R_{3}$,we have $|A| = (\alpha+\beta+\gamma) \begin{vmatrix} \alpha & \beta & \gamma \ \alpha^{2} & \beta^{2} & \gamma^{2} \ 1 & 1 & 1 \end{vmatrix}$.Using the determinant of a Vandermonde matrix,$|A| = -(\alpha+\beta+\gamma)(\alpha-\beta)(\beta-\gamma)(\gamma-\alpha)$.Since $|\operatorname{adj}(\operatorname{adj}(\operatorname{adj}(\operatorname{adj} A))))| = |A|^{(n-1)^4} = |A|^{2^4} = |A|^{16}$,where $n=3$.Substituting this into the given equation: $\frac{|A|^{16}}{(\alpha-\beta)^{16}(\beta-\gamma)^{16}(\gamma-\alpha)^{16}} = 2^{32} 	imes 3^{16}$.This simplifies to $(\alpha+\beta+\gamma)^{16} = 2^{32} 	imes 3^{16} = (2^2 	imes 3)^{16} = 12^{16}$.Thus,$\alpha+\beta+\gamma = 12$.Since $\alpha, \beta, \gamma \in \mathbb{N}$,the number of positive integer solutions is $\binom{12-1}{3-1} = \binom{11}{2} = 55$.We must exclude cases where $\alpha, \beta, \gamma$ are not distinct. If $\alpha=\beta=\gamma$,then $3\alpha=12 \Rightarrow \alpha=4$,which is $1$ case $(4,4,4)$.If two are equal,say $\alpha=\beta$,then $2\alpha+\gamma=12$. Possible values for $\alpha$ are $1, 2, 3, 5$ (since $\alpha=4$ gives $\gamma=4$). There are $4$ such pairs for each permutation of $(\alpha, \beta, \gamma)$,totaling $4 	imes 3 = 12$ cases.Total distinct tuples = $55 - 1 - 12 = 42$.

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