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2 Answers

Best answer
3 votes
3 votes

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

selected by
0 votes
0 votes
Option A is correct ,$adj.adjA = |A|^{^{n-2}}$

 

So option A is perfectly Valid.

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