MathematicsMedium244×since 2002Q4545If A=15!6!7![5!6!7!6!7!8!7!8!9!]\mathrm{A}=\frac{1}{5 ! 6 ! 7 !}\left[\begin{array}{ccc}5 ! & 6 ! & 7 ! \\ 6 ! & 7 ! & 8 ! \\ 7 ! & 8 ! & 9 !\end{array}\right]A=5!6!7!15!6!7!6!7!8!7!8!9!, then ∣adj(adj(2 A))∣|\operatorname{adj}(\operatorname{adj}(2 \mathrm{~A}))|∣adj(adj(2 A))∣ is equal to :A2122^{12}212B2202^{20}220C282^{8}28D2162^{16}216Check answerSkip