JEE Main 2025 · 7 April · morning · Single correct · +4 / −1
Let A be a 3 × 3 matrix such that |adj(adj(adj A))|=81.
If S={n ∈ ℤ:(|adj(adj A)|)^((n-1)²/2)=|A|^((3 n²-5 n-4))}, then Σn ∈ S|A^((n²+n))| is equal to :
- (a) 820
- (b) 866
- (c) 750
- (d) 732
Answer and solution
Answer (d) 732
Solution
Maths sir · 5 steps
Turn every adjoint into a power of |A|, then compare exponents and the sets drop out neatly.
- Adjoint as powers. For a 3 × 3 matrix, |adj A| = |A|². Each extra adj squares the determinant again, so |adj(adj A)| = |A|⁴ and |adj(adj(adj A))| = |A|⁸.
- Find the determinant. |A|⁸ = 81 = 3⁴, so |A|² = 3. Then |A| = ±√3, which is not 0 or ± 1, so two equal powers of |A| must have equal exponents.
(|A|⁴)^((n-1)²/2) = |A|^(3n²-5n-4) ⇒ 2(n-1)² = 3n²-5n-4
- Solve for n. Expanding gives 2n²-4n+2 = 3n²-5n-4, that is n²-n-6 = 0, so (n-3)(n+2)=0. Hence S = {3, -2}.
- Add the terms. Use |Aᵏ| = |A|ᵏ. For n = 3 the exponent is 12, giving |A|¹² = 3⁶ = 729. For n = -2 the exponent is 2, giving |A|² = 3. Both exponents are even, so the sign of |A| does not matter.
- So. The sum is 729 + 3 = 732, option (d).
The idea to keep. For an n × n matrix, |adj A| = |A|ⁿ⁻¹, and k adjoints give |A|^((n-1)ᵏ). Reduce everything to powers of |A| and match exponents.
Answer from the NTA final answer key. Solution by PrepWiser.