Let P=[√3/2, 1/2; -1/2, √3/2], A=[1, 1; 0, 1] and Q=P A Pᵀ. If Pᵀ Q²⁰⁰⁷ P=[a, b; c, d], then 2 a+b-3 c-4 d equal to :

Solution

Maths sir · 4 steps

P is orthogonal, so every inner Pᵀ P collapses to I and Q²⁰⁰⁷ unwraps to A²⁰⁰⁷.

1. P is orthogonal. The rows of P are (√3/2, 1/2) and (-1/2, √3/2). Each has length 1 and their dot product is 0, so PPᵀ = PᵀP = I.
2. Powers of Q. Q² = PAPᵀ PAPᵀ = PA(PᵀP)APᵀ = PA²Pᵀ. Repeating this gives Q²⁰⁰⁷ = PA²⁰⁰⁷Pᵀ.

   PᵀQ²⁰⁰⁷P = (PᵀP) A²⁰⁰⁷ (PᵀP) = A²⁰⁰⁷

3. Power of A. A² = [1, 2; 0, 1], and in general Aⁿ = [1, n; 0, 1]. So a = 1, b = 2007, c = 0, d = 1.
4. So. 2a + b - 3c - 4d = 2 + 2007 - 0 - 4 = 2005, option (d).

The idea to keep. When you see PAPᵀ with P orthogonal, powers pass straight through: (PAPᵀ)ⁿ = PAⁿPᵀ. Check orthogonality first, then only Aⁿ needs work.

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