Let A be a 3 × 3 matrix and det(A)=2. If n=det(adj(adj(… . .(adj A))₂₀₂₄₋ₜᵢₘₑₛ)), then the remainder when n is divided by 9 is equal to __________.

Solution

Maths sir · 5 steps

Each adjoint squares the determinant, so the answer is a tower of powers of 2. Reduce it with the cycle mod 9.

1. One adjoint. For an n × n matrix, det(adj A) = (det A)ⁿ⁻¹. Here n = 3, so each adjoint squares the determinant.
2. Repeat 2024 times. After one adjoint the determinant is 2², after two it is 2⁴, and after k it is 2^(2ᵏ). So n = 2^(2²⁰²⁴).
3. Cycle mod 9. Powers of 2 mod 9 run 2, 4, 8, 7, 5, 1 and then repeat, because 2⁶ = 64 = 63 + 1. So only the remainder of the exponent 2²⁰²⁴ on division by 6 matters.
4. Reduce the exponent. 2²⁰²⁴ - 4 = 4(4¹⁰¹¹ - 1), and 4¹⁰¹¹ - 1 is divisible by 4 - 1 = 3. So 2²⁰²⁴ = 6m + 4 for some whole number m.

   n = 2⁶ᵐ⁺⁴ = 16 × (64)ᵐ ≡ 16 × 1 ≡ 7 (mod 9)

5. So. The remainder when n is divided by 9 is 7.

The idea to keep. Use |adj A| = |A|ⁿ⁻¹. Repeated k times it gives |A|^((n-1)ᵏ). For a huge power mod m, find the cycle length first, then reduce only the exponent by that length.

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