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) = 100λ , then λ is equal to :

Solution

Maths sir · 4 steps

Find det A first, then use det(adj A) = (det A)² for a 3 × 3 matrix.

1. Find det A. Expand along the second column. Its only non-zero entry is the 1 in the middle, so det A = 1 × |1, a; -a, 1| = 1 + a².
2. Adjoint rule. For an n × n matrix, det(adj A) = (det A)ⁿ⁻¹. Here n = 3, so det(adj A) = (1 + a²)². Since a = √n with n odd, a² runs through 1, 3, 5, …, 49.

   Σ_(a ∈ S) (1 + a²)² = 2² + 4² + … + 50² = 4Σₖ₌₁²⁵ k² = 4 × (25 × 26 × 51)/6

3. Add them up. (25 × 26 × 51)/6 = 5525, so the sum is 4 × 5525 = 22100.
4. So. 100λ = 22100, which gives λ = 221, option (b).

The idea to keep. For an n × n matrix, det(adj A) = (det A)ⁿ⁻¹. Always reduce to det A first, then look for a standard sum like Σ k².

Answer from the NTA final answer key. Solution by PrepWiser.