Let $$A$$ and $$B$$ be two $$3 \times 3$$ matrices such that $$AB = I$$ and $$|A| = \frac{1}{8}$$ then $$|adj(B \cdot adj(2A))|$$ is equal to

We are given $$3 \times 3$$ matrices $$A$$ and $$B$$ with $$AB = I$$ and $$|A| = \frac{1}{8}$$. Since $$AB = I$$, it follows that $$|A|\cdot|B| = 1$$, so $$|B| = \frac{1}{|A|} = 8$$. For an $$n \times n$$ matrix $$M$$, the formula $$|adj(M)| = |M|^{n-1}$$ holds. Hence $$|2A| = 2^3 \cdot |A| = 8 \times \frac{1}{8} = 1$$ and $$|adj(2A)| = |2A|^{3-1} = 1^2 = 1$$. It follows that $$|B \cdot adj(2A)| = |B| \cdot |adj(2A)| = 8 \times 1 = 8$$. Finally, since for a $$3 \times 3$$ matrix $$M$$ we have $$|adj(M)| = |M|^2$$, we get $$|adj(B \cdot adj(2A))| = |B \cdot adj(2A)|^2 = 8^2 = 64$$.

The correct answer is Option C: $$64$$.