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(B) Given $A = \begin{bmatrix} 1 & \log_x y & \log_x z \ \log_y x & 2 & \log_y z \ \log_z x & \log_z y & 3 \end{bmatrix}$.Using the property $\log_a

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$2^8$

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(B) Given $A = \begin{bmatrix} 1 & \log_x y & \log_x z \ \log_y x & 2 & \log_y z \ \log_z x & \log_z y & 3 \end{bmatrix}$.Using the property $\log_a b = \frac{\ln b}{\ln a}$,we can write $A = \begin{bmatrix} 1 & \frac{\ln y}{\ln x} & \frac{\ln z}{\ln x} \ \frac{\ln x}{\ln y} & 2 & \frac{\ln z}{\ln y} \ \frac{\ln x}{\ln z} & \frac{\ln y}{\ln z} & 3 \end{bmatrix}$.Multiply $R_1$ by $\ln x$,$R_2$ by $\ln y$,and $R_3$ by $\ln z$:$|A| = \frac{1}{\ln x \ln y \ln z} \begin{vmatrix} \ln x & \ln y & \ln z \ \ln x & 2 \ln y & \ln z \ \ln x & \ln y & 3 \ln z \end{vmatrix}$.Factoring out $\ln x, \ln y, \ln z$ from the columns:$|A| = \frac{\ln x \ln y \ln z}{\ln x \ln y \ln z} \begin{vmatrix} 1 & 1 & 1 \ 1 & 2 & 1 \ 1 & 1 & 3 \end{vmatrix} = 1(6-1) - 1(3-1) + 1(1-2) = 5 - 2 - 1 = 2$.We know that $|\operatorname{adj}(\operatorname{adj} M)| = |M|^{(n-1)^2}$,where $n$ is the order of the matrix.Here $n=3$,so $|\operatorname{adj}(\operatorname{adj} A^2)| = |A^2|^{(3-1)^2} = |A^2|^4 = (|A|^2)^4 = |A|^8$.Since $|A| = 2$,$|A|^8 = 2^8$.

Let $x, y, z > 1$ and $A = \begin{bmatrix} 1 & \log_x y & \log_x z \\ \log_y x & 2 & \log_y z \\ \log_z x & \log_z y & 3 \end{bmatrix}$. Then $|\operatorname{adj}(\operatorname{adj} A^2)|$ is equal to

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