Let A=(2, -1, -1; 1, 0, -1; 1, -1, 0) and B=A-I. If ω=(√3 i-1)/2, then the number of elements in the set{n ∈{1,2, …, 100}: Aⁿ+(ω B)ⁿ=A+B} is equal to ____________.

Solution

Maths sir · 4 steps

Square A once: A² = A, so every power of A is A and only the B part decides n.

1. Powers of A. Multiplying out gives A² = A, so Aⁿ = A for every n ≥ 1. The condition reduces to (ω B)ⁿ = B.
2. Powers of B. B = A - I has every row equal to (1, -1, -1). Then B² = A² - 2A + I = I - A = -B, so B³ = -B² = B, and in general Bⁿ = (-1)ⁿ⁻¹ B.

   (ω B)ⁿ = ωⁿ (-1)ⁿ⁻¹ B = B ⇔ ωⁿ (-1)ⁿ⁻¹ = 1

3. Solve for n. Here ω is a cube root of unity, so ωⁿ is 1, ω or ω² and is never -1. That rules out even n. For odd n we need ωⁿ = 1, so n must be an odd multiple of 3.
4. So. The values are n = 3, 9, 15, …, 99, which is (99 - 3)/6 + 1 = 17 values.

The common slip: 33. Forgetting the sign (-1)ⁿ⁻¹ from B² = -B and counting every multiple of 3 gives 33, not 17.

The idea to keep. When a matrix power question appears, compute the square first. Idempotent (A² = A) or B² = kB patterns collapse every power into one simple formula.

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