Let P be a square matrix such that $P^{2}=I-P$. For $\alpha, \beta, \gamma, \delta \in \mathbb{N}$, if $P^{\alpha}+P^{\beta}=\gamma I-29 P$ and $P^{\alpha}-P^{\beta}= \delta I-13 P$, then $\alpha+\beta+\gamma-\delta$ is equal to

JEE Main · 2023 · 06 Apr 2023 · Shift 2 · Matrices

Let P be a square matrix such that

P^{2} = I - P

. For

α, β, γ, δ \in 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

Answer: (C)

From

P^{2} = I - P

, we compute higher powers:

P^{4} = (P^{2})^{2} = (I - P)^{2} = I - 2 P + P^{2} = I - 2 P + (I - P) = 2 I - 3 P

. Similarly,

P^{6} = P^{4} \cdot P^{2} = (2 I - 3 P) (I - P) = 2 I - 2 P - 3 P + 3 P^{2} = 2 I - 5 P + 3 (I - P) = 5 I - 8 P

. And

P^{8} = (P^{4})^{2} = (2 I - 3 P)^{2} = 4 I - 12 P + 9 P^{2} = 4 I - 12 P + 9 (I - P) = 13 I - 21 P

. Now,

P^{8} + P^{6} = (13 I - 21 P) + (5 I - 8 P) = 18 I - 29 P

, and

P^{8} - P^{6} = (13 I - 21 P) - (5 I - 8 P) = 8 I - 13 P

. Comparing with the given equations,

α = 8

β = 6

γ = 18

δ = 8

. Thus

α + β + γ - δ = 8 + 6 + 18 - 8 = 24

JEE Main · 2023 · 06 Apr 2023 · Shift 2 · Single correct option