Let $$A$$ be a $$3 \times 3$$ invertible matrix. If $$|\text{adj}(24A)| = |\text{adj}(3 \text{ adj}(2A))|$$, then $$|A|^2$$ is equal to

We need to find $$|A|^2$$ given that $$|\text{adj}(24A)| = |\text{adj}(3\,\text{adj}(2A))|$$ for a $$3 \times 3$$ invertible matrix $$A$$.

Using the key properties for an $$n \times n$$ matrix, namely $$|\text{adj}(M)| = |M|^{n-1}$$ and $$|kM| = k^n|M|$$, we first simplify the left side. Since for a 3×3 matrix $$|\text{adj}(24A)| = |24A|^{3-1} = |24A|^2$$ and $$|24A| = 24^3 |A|$$, we have $$|\text{adj}(24A)| = (24^3)^2 |A|^2 = 24^6 |A|^2$$.

Next, the right side involves a nested adjugate. First, $$|2A| = 2^3 |A| = 8|A|$$, so $$|\text{adj}(2A)| = |2A|^2 = 64|A|^2$$. Then, $$|3\,\text{adj}(2A)| = 3^3 \cdot |\text{adj}(2A)| = 27 \times 64|A|^2 = 1728|A|^2$$. Finally, $$|\text{adj}(3\,\text{adj}(2A))| = |3\,\text{adj}(2A)|^2 = (1728)^2 |A|^4$$.

Equating the two expressions gives $$24^6 |A|^2 = 1728^2 |A|^4$$, which implies $$|A|^2 = \frac{24^6}{1728^2}$$. Since $$24 = 2 \times 12$$ and $$1728 = 12^3$$, this becomes $$\frac{24^6}{1728^2} = \frac{(2 \times 12)^6}{(12^3)^2} = \frac{2^6 \times 12^6}{12^6} = 2^6 = 64$$.

Therefore, $$|A|^2 = 2^6$$.

The correct answer is Option A.