Let P be a square matrix such that P²=I-P. For α, β, γ, δ ∈ ℕ, if P^α+P^β=γ I-29 P and P^α-P^β=δ I-13 P, then α+β+γ-δ is equal to :

Solution

Maths sir · 4 steps

Every power of P collapses to aI + bP, so build a small table of powers and match the P coefficients.

1. Reduce any power. Since P² = I - P, every power can be written as Pⁿ = aₙ I + bₙ P. Multiply by P once more and replace P² to get the next power.

   Pⁿ⁺¹ = aₙ P + bₙ (I - P) = bₙ I + (aₙ - bₙ) P

   Power	aₙ (coefficient of I)	bₙ (coefficient of P)
   P²	1	-1
   P³	-1	2
   P⁴	2	-3
   P⁵	-3	5
   P⁶	5	-8
   P⁷	-8	13
   P⁸	13	-21

2. Match P coefficients. From the two equations, b_α + b_β = -29 and b_α - b_β = -13. Adding gives b_α = -21 and b_β = -8, so α = 8 and β = 6.
3. Coefficients of I. The I coefficients give γ = a₈ + a₆ = 13 + 5 = 18 and δ = a₈ - a₆ = 13 - 5 = 8. Both are natural numbers, as required.
4. So. α + β + γ - δ = 8 + 6 + 18 - 8 = 24, which is option (c).

The idea to keep. When a matrix satisfies a quadratic like P² = I - P, every power is just aI + bP. The coefficients follow a simple recurrence (here Fibonacci numbers with alternating signs), so tabulate them.

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