Let P be a square matrix such that P 2 = I - P . For , if P + P = I - 29 P and P - P = I - 13 P , then is equal to [2023]

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Solution

Q.
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 [2023]

1 40

2 22

3 18

4 24

Ans.
(4)

We have,

$p^{α} + p^{β} = γ I - 29 p ...(i)$

$p^{α} - p^{β} = δ I - 13 p ...(ii)$

Now, $p^{2} = I - p$

$p^{3} = p - p^{2} = p - I + p = 2 p - I$

$p^{4} = 2 p^{2} - p = 2 (I - p) - p = 2 I - 3 p$

$p^{5} = 2 p - 3 p^{2} = 2 p - 3 (I - p) = 5 p - 3 I$

$p^{6} = 5 p^{2} - 3 p = 5 (I - p) - 3 p = 5 I - 8 p$

$p^{7} = 5 p - 8 p^{2} = 5 p - 8 (I - p) = 13 p - 8 I$

$p^{8} = 13 p^{2} - 8 p = 13 (I - p) - 8 p = 13 I - 21 p$

$p^{8} + p^{6} = 18 I - 29 p ...(iv)$

$p^{8} - p^{6} = 8 I - 13 p ...(v)$

After comparing (iv) with (i) and (v) with (ii), we get:

$α = 8, β = 6, γ = 18, δ = 8$

$∴ α + β + γ - δ = 8 + 6 + 18 - 8 = 24$

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