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Q.

A square matrix P satisfies $P^{2} = I - 2P$ where $I$ is identity matrix. If $P^{2} + P^{3} + P^{4} = a^{2} I - b^{2} P,$ then

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a

$a^{2} + b^{2}$ is prime

b

$a^{2} + b^{4}$ is a perfect square

c

$a^{4} + b^{2}$ is a perfect square

d

$a^{4} + b^{4}$ is prime

answer is B, C, D.

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Detailed Solution

$P^{3} = (1 - 2 P) P = P - 2 (I - 2 P) = 5 P - 2 I, P^{4} = (5 P - 2 I) P = 5 (I - 2 P) - 2 P = 5 I - 12 P$

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