Given,
A=5!6!7!1[5!6!7!6!7!8!7!8!9!]
⇒∣A∣=5!6!7!1∣5!6!7!6!7!8!7!8!9!∣
⇒∣A∣=5!6!7!15!×6!×7!∣111678425672∣
⇒∣A∣=2
Now using the property of adjoint of matrix we get,
|adj(adj(A))|={|A|}^{(n-1{)}^{2}}&|kA|={k}^{n}|A|, where n is order of square matrix,
Now using the above two formula we get,
∣adj(adj(2A))∣=∣2A∣4=212∣A∣4
⇒∣adj(adj(2A))∣=212⋅24
⇒∣adj(adj(2A))∣=216