Let $$A$$ be a $$3 \times 3$$ matrix such that $$|adj(adj(adj \cdot A))| = 12^4$$. Then $$|A^{-1} adj A|$$ is equal to

Let $$A$$ be a $$3 \times 3$$ matrix such that $$|\text{adj}(\text{adj}(\text{adj}\,A))| = 12^4$$. We need to find $$|A^{-1}\,\text{adj}\,A|$$.

We recall that for an $$n \times n$$ matrix, $$|\text{adj}\,A| = |A|^{n-1}$$, $$\text{adj}\,A = |A|\,A^{-1}$$ when $$A$$ is invertible, and $$|kM| = k^n|M|$$ for a scalar $$k$$. In the $$3\times 3$$ case, $$|\text{adj}\,A| = |A|^2$$.

Setting $$d = |A|$$, it follows that $$|\text{adj}\,A| = d^2$$, $$|\text{adj}(\text{adj}\,A)| = (d^2)^2 = d^4$$, and $$|\text{adj}(\text{adj}(\text{adj}\,A))| = (d^4)^2 = d^8$$. Equating this to $$12^4$$ and taking the positive fourth root gives $$d^8 = 12^4$$, $$d^2 = 12$$, and $$d = \sqrt{12} = 2\sqrt{3}$$.

Since $$\text{adj}\,A = |A|\,A^{-1}$$, we have $$A^{-1}\,\text{adj}\,A = |A|\,A^{-2}$$. Taking determinants yields $$|A^{-1}\,\text{adj}\,A| = \bigl|\,|A|\,A^{-2}\bigr| = |A|^3\,|A^{-2}| = |A|^3\cdot\frac{1}{|A|^2} = |A|\,. $$ Alternatively, $$|A^{-1}\,\text{adj}\,A| = |A^{-1}|\cdot|\text{adj}\,A| = \frac{1}{|A|}\cdot|A|^2 = |A|\,. $$ Hence the required value is $$|A| = 2\sqrt{3}$$.

The answer is Option A: $$2\sqrt{3}$$.