If $$A = \frac{1}{5!6!7!} \begin{vmatrix} 5! & 6! & 7! \\ 6! & 7! & 8! \\ 7! & 8! & 9! \end{vmatrix}$$, then adj $$2A$$ is equal to

We wish to find $$\text{adj}(2A)$$ where $$A = \dfrac{1}{5!\,6!\,7!} \begin{vmatrix} 5! & 6! & 7! \\ 6! & 7! & 8! \\ 7! & 8! & 9! \end{vmatrix}$$.

Let $$M = \begin{pmatrix} 5! & 6! & 7! \\ 6! & 7! & 8! \\ 7! & 8! & 9! \end{pmatrix}$$. By factoring $$5!$$ from the first row, $$6!$$ from the second row, and $$7!$$ from the third row, we obtain

$$M = \begin{pmatrix} 5! & 0 & 0 \\ 0 & 6! & 0 \\ 0 & 0 & 7! \end{pmatrix} \begin{pmatrix} 1 & 6 & 42 \\ 1 & 7 & 56 \\ 1 & 8 & 72 \end{pmatrix},$$

since $$\dfrac{6!}{5!} = 6$$, $$\dfrac{7!}{5!} = 42$$, $$\dfrac{7!}{6!} = 7$$, $$\dfrac{8!}{6!} = 56$$, $$\dfrac{8!}{7!} = 8$$, and $$\dfrac{9!}{7!} = 72$$.

Substituting this into the definition of $$A$$ yields

$$A = \dfrac{1}{5!\,6!\,7!}\cdot M = \dfrac{5!\,6!\,7!}{5!\,6!\,7!}\begin{pmatrix} 1 & 6 & 42 \\ 1 & 7 & 56 \\ 1 & 8 & 72 \end{pmatrix} = \begin{pmatrix} 1 & 6 & 42 \\ 1 & 7 & 56 \\ 1 & 8 & 72 \end{pmatrix}.$$

To compute $$\det(A)$$, perform the row operations $$R_2\to R_2-R_1$$ and $$R_3\to R_3-R_1$$, which transform $$A$$ into

$$\begin{pmatrix} 1 & 6 & 42 \\ 0 & 1 & 14 \\ 0 & 2 & 30 \end{pmatrix}$$

and hence

$$\det(A) = 1\cdot(1\times30 - 14\times2) = 30 - 28 = 2.$$

For any $$n\times n$$ matrix $$M$$ there is the identity $$|\text{adj}(M)| = |M|^{n-1}$$. Here $$n=3$$, so first

$$|2A| = 2^3\cdot|A| = 8\times2 = 16,$$

and therefore

$$|\text{adj}(2A)| = |2A|^{2} = 16^2 = 256 = 2^8.$$

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