Let S = n : 1 ⩽ n ⩽ 50 and n is odd . Let a ∈ S and A = 1 0 a - 1 1 0 - a 0 1 . If Σ a ∈ S det adj A = 100 λ , then λ is equal to

Step-by-step solution

We know that A d j   A = A n - 1 Here n is 3 × 3 matrix so A d j   A = A 3 - 1 = A 2 Now find A = 1 0 a - 1 1 0 - a 0 1 = 1 + a 2 Now A d j   A = A 2 = 1 + a 2 2 Now Σ a ∈ S det adj A = Σ 1 + a 2 2 = 100 λ ⇒ 2 2 + 4 2 . . . + 48 2 + 50 2 = 100 λ ⇒ 4 1 2 + 2 2 . . . + 25 2 = 100 λ ⇒ 4 × 25 × 26 × 51 6 = 100 λ ⇒ 221 = λ