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 $$|adj(adj A^2)|$$ is equal to

Given $$A = \begin{bmatrix} 1 & \log_x y & \log_x z \\ \log_y x & 2 & \log_y z \\ \log_z x & \log_z y & 3 \end{bmatrix}$$ with $$x, y, z > 1$$.

Finding $$|A|$$:

Let $$p = \ln x, q = \ln y, r = \ln z$$. Multiply rows 1, 2, 3 by $$p, q, r$$ respectively:

$$M = \begin{bmatrix} p & q & r \\ p & 2q & r \\ p & q & 3r \end{bmatrix}$$

$$\det(M) = pqr \cdot \det(A)$$

Applying $$R_2 - R_1$$ and $$R_3 - R_1$$:

$$\det(M) = \begin{vmatrix} p & q & r \\ 0 & q & 0 \\ 0 & 0 & 2r \end{vmatrix} = 2pqr$$

Therefore $$\det(A) = 2$$.

Verification: Setting $$x = y = z$$: $$A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 3 \end{bmatrix}$$, $$\det = 5 - 2 - 1 = 2$$ ✓

Computing $$|\text{adj}(\text{adj}(A^2))|$$:

For an $$n \times n$$ matrix $$M$$: $$|\text{adj}(M)| = |M|^{n-1}$$

$$|\text{adj}(\text{adj}(M))| = |M|^{(n-1)^2}$$

With $$n = 3$$ and $$M = A^2$$:

$$|A^2| = |A|^2 = 4$$

$$|\text{adj}(\text{adj}(A^2))| = |A^2|^{(3-1)^2} = 4^4 = 256 = 2^8$$

The answer is Option B: $$2^8$$.