Let for A = 1 2 3 α 3 1 1 1 2 , A = 2 . If | 2 adj ( 2 adj ( 2 A ) ) | = 32 n , then 3 n + α is equal to

Step-by-step solution

Given that A = 1 2 3 α 3 1 1 1 2 ⇒ | A | = 2 ⇒ 1 6 - 1 - α - 7 = 2 ⇒ - α - 2 = 2 ⇒ α = - 4 ⇒ | 2 adj ( 2 adj ( 2 A ) ) | = 32 n We know that k A = k n A , adj ( adj ( A ) ) = A n - 1 2 and adj k A = k n - 1 adj A . = 2 3 | adj ( 2 adj ( 2 A ) ) | = 2 3 | 2 3 - 1 adj ( adj ( 2 A ) ) | = 2 3 · ( 4 ) 3 | adj ( adj ( 2 A ) ) | = 2 9 | 2 A | ( 2 ) 2 = 2 9 · | 2 A | 4 = 2 9 · 2 12 | A | 4 = 2 21 · 2 4 = 2 25 = ( 32 ) n = ( 2 ) 5 n ∴ n = 5 ⇒