Explanation:
(D) : We have,
\(A=\frac{1}{5 ! 6 ! 7 !}\left[\begin{array}{lll}
5 ! & 6 ! & 7 ! \\ 6 ! & 7 ! & 8 ! \\ 7 ! & 8 ! & 9 ! \end{array}\right]\)
Now, \(\quad|A|=\frac{1}{5 ! 6 ! 7 !} \left\lvert\, \begin{array}{lll}6 ! & 7 ! & 8 ! \\ 7 ! & 8 ! & 9 !\end{array}\right.\)
Taking out 5!, 6!, 7! Common from first, second and third row respectively.
\(=\frac{5 ! 6 ! 7 !}{5 ! 6 ! 7 !}\left|\begin{array}{lll}
1 & 6 & 42 \\ 1 & 7 & 56 \\ 1 & 8 & 72 \end{array}\right|=\left|\begin{array}{lll}
1 & 6 & 42 \\ 1 & 7 & 56 \\ 1 & 8 & 72 \end{array}\right|\)
Operating \(R_{3} \rightarrow R_{3}-R_{2}\)
\(=\left|\begin{array}{lll}1 & 6 & 42 \\ 1 & 7 & 56 \\ 0 & 1 & 16\end{array}\right|\)
Operating \(\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1}\)
\(=\left|\begin{array}{lll}
1 & 6 & 42 \\ 0 & 1 & 14 \\ 0 & 1 & 16 \end{array}\right|\)
On expanding along Row 1 , we get
\(=16-14=2\)
Thus, \(|\mathrm{A}|=2\)
Now, \(|\operatorname{adj}(\operatorname{adj} 2 \mathrm{~A})|\)
\(=|(2 \mathrm{~A})|^{(3-1)^{2}}=|(2 \mathrm{~A})|^{(2)^{2}}=|(2 \mathrm{~A})|^{4}\)
\(=\left[2^{3}|\mathrm{~A}|\right]^{4}=\left[2^{3} \times 2\right]^{4}=\left[2^{4}\right]^{4}=2^{16}\)
Hence, option (d) is correct.