Let S={m ∈ Z: A^(m²)+Aᵐ=3 I-A⁻⁶}, where A=[2, -1; 1, 0]. Then n(S) is equal to __________.

Solution

Maths sir · 4 steps

Write A as I + N with N² = O, and every power of A becomes a straight line in N.

1. Split off I. Let N = A - I = (1, -1; 1, -1). Multiplying gives N² = O, so N is nilpotent.
2. Powers of A. Expand (I+N)ᵏ by the binomial theorem. Every term from N² onwards vanishes, so Aᵏ = I + kN. This also holds for negative k, because (I+N)(I-N) = I - N² = I, which gives A⁻¹ = I - N.

   A^(m²) + Aᵐ = 2I + (m²+m)N, 3I - A⁻⁶ = 3I - (I - 6N) = 2I + 6N

3. Compare coefficients. N is not the zero matrix, so the two sides are equal only when m² + m = 6. That is (m-2)(m+3) = 0, so m = 2 or m = -3.
4. So. S = {2, -3} and n(S) = 2.

The common slip: 1. Keeping only the positive root m = 2 and dropping m = -3 gives 1. S is a set of integers, so negative powers count, and A⁻³ exists because det A = 1.

The idea to keep. When A - I squares to zero, Aᵏ = I + k(A - I) for every integer k. Turn a matrix equation into one equation in k and solve that.

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