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(B) Given $A = \begin{bmatrix} 1 & 0 & 0 \ 0 & \alpha & \beta \ 0 & \beta & \alpha \end{bmatrix}$.The determinant of $A$ is $|A| = 1(\alpha^2 - \bet

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$5$

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(B) Given $A = \begin{bmatrix} 1 & 0 & 0 \ 0 & \alpha & \beta \ 0 & \beta & \alpha \end{bmatrix}$.The determinant of $A$ is $|A| = 1(\alpha^2 - \beta^2) - 0 + 0 = \alpha^2 - \beta^2$.Using the property $|kA| = k^n|A|$ for a matrix of order $n 	imes n$,here $n=3$,so $|2A| = 2^3|A| = 8|A|$.Given $|2A|^3 = 2^{21}$,we have $(8|A|)^3 = 2^{21}$.Taking the cube root on both sides,$8|A| = (2^{21})^{1/3} = 2^7 = 128$.Thus,$|A| = \frac{128}{8} = 16$.Substituting $|A| = \alpha^2 - \beta^2$,we get $\alpha^2 - \beta^2 = 16$,which can be written as $(\alpha - \beta)(\alpha + \beta) = 16$.Since $\alpha, \beta \in \mathbb{Z}$,we look for integer factors of $16$. Possible pairs $(\alpha-\beta, \alpha+\beta)$ are $(2, 8), (4, 4), (8, 2), (-2, -8), (-4, -4), (-8, -2)$.For $(\alpha-\beta, \alpha+\beta) = (2, 8)$,adding gives $2\alpha = 10 \Rightarrow \alpha = 5$.For $(\alpha-\beta, \alpha+\beta) = (4, 4)$,adding gives $2\alpha = 8 \Rightarrow \alpha = 4$.Comparing with the given options,$5$ is a valid value for $\alpha$.

Let $A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \alpha & \beta \\ 0 & \beta & \alpha \end{bmatrix}$ and $|2A|^3 = 2^{21}$,where $\alpha, \beta \in \mathbb{Z}$. Then a value of $\alpha$ is:

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