If P is a non null matrix of order 3×3 with real entries such that P3=O then (consider O as 3×3 null matrix and I as 3×3 identity matrix)

if matrix P has all integral entries then absolute value of determinant of I − 4 P 2 is 1 , I − 2 P is an invertible matrix , if matrix P has all integral entries then determinant of I − 4 P 2 is zero

If P is a non null matrix of order 3×3 with real entries such that P3=O then (consider O as 3×3 null matrix and I as 3×3 identity matrix)

8 P 3 + I = ( 2 P + I ) ( 4 P 2 − 2 P + I ) = I Taking determinant | 2 P + I | ≠ 0 , | 4 P 2 − 2 P + I | ≠ 0 Again I − 8 P 3 = ( I − 2 P ) ( 4 P 2 + 2 P + I ) = I Taking determinant | I − 2 P | ≠ 0 So | I + 2 P | | I − 2 P | ≠ 0 ⇒ | I − 4 P 2 | ≠ 0 If P has integral entries | I − 2 P | and | I + 2 P | can be 1 or – 1. So | | I − 4 P 2 | | = 1.

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If $P$ is a non null matrix of order $3 \times 3$ with real entries such that $P^{3} = O$ then (consider $O$ as $3 \times 3$ null matrix and $I$ as $3 \times 3$ identity matrix)

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if matrix $P$ has all integral entries then absolute value of determinant of $I - 4 P^{2}$ is 1

$I - 2 P$ is an invertible matrix

determinant of $4 P^{2} - 2 P + I$ is non zero

if matrix $P$ has all integral entries then determinant of $I - 4 P^{2}$ is zero

answer is A, B, D.

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Detailed Solution

$8 P^{3} + I = (2 P + I) (4 P^{2} - 2 P + I) = I$

Taking determinant $| 2 P + I | \neq 0, | 4 P^{2} - 2 P + I | \neq 0$
Again $I - 8 P^{3} = (I - 2 P) (4 P^{2} + 2 P + I) = I$
Taking determinant $| I - 2 P | \neq 0$
So $| I + 2 P | | I - 2 P | \neq 0$
$\Rightarrow | I - 4 P^{2} | \neq 0$
If $P$ has integral entries $| I - 2 P |$ and $| I + 2 P |$ can be 1 or – 1. So $| | I - 4 P^{2} | | = 1.$