Let P be a square matrix such that P^2 = I - | vedclass

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(C) Given $P^2 = I - P$.We calculate powers of $P$:$P^3 = P(I - P) = P - P^2 = P - (I - P) = 2P - I$.$P^4 = P(2P - I) = 2P^2 - P = 2(I - P) - P = 2I -

Vedclass · Accepted Answer

$24$

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(C) Given $P^2 = I - P$.We calculate powers of $P$:$P^3 = P(I - P) = P - P^2 = P - (I - P) = 2P - I$.$P^4 = P(2P - I) = 2P^2 - P = 2(I - P) - P = 2I - 3P$.$P^5 = P(2I - 3P) = 2P - 3P^2 = 2P - 3(I - P) = 5P - 3I$.$P^6 = P(5P - 3I) = 5P^2 - 3P = 5(I - P) - 3P = 5I - 8P$.$P^7 = P(5I - 8P) = 5P - 8P^2 = 5P - 8(I - P) = 13P - 8I$.$P^8 = P(13P - 8I) = 13P^2 - 8P = 13(I - P) - 8P = 13I - 21P$.Now,$P^8 + P^6 = (13I - 21P) + (5I - 8P) = 18I - 29P$.Comparing with $P^\alpha + P^\beta = \gamma I - 29P$,we get $\alpha = 8, \beta = 6, \gamma = 18$.Also,$P^8 - P^6 = (13I - 21P) - (5I - 8P) = 8I - 13P$.Comparing with $P^\alpha - P^\beta = \delta I - 13P$,we get $\delta = 8$.Thus,$\alpha + \beta + \gamma - \delta = 8 + 6 + 18 - 8 = 24$.

Let $P$ be a square matrix such that $P^2 = I - P$. For $\alpha, \beta, \gamma, \delta \in N$,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 $........$.

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