adj(A) = |A| . A-1
$\therefore$ adj(adjA)
= |adj A| . (adj A)-1
= |A|n-1 . (adj A)-1 .................(1)
adj(A) = |A| . A-1
Premultiply both sides by (adj A)-1
$\therefore$ I = (adj A)-1 .|A| . A-1
$\therefore$ $\frac{A}{|A|}$ = (adj A)-1 .........(2)
Substituting (2) in (1),
we get adj(adj A) = |A|n-2 . A = 3. A