Let P and Q be `3xx3` matrices with `P!=Q` . If `P^3=""Q^3a n d""P^2Q""=""Q^2P` , then determinant of `(P^2+""Q^2)` is equal to (1) `2` (2) 1 (3) 0 (4

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Let P and Q be `3xx3` matrices with `P!=Q` . If `P^3=""Q^3a n d""P^2Q""=""Q^2P` , then determinant of `(P^2+""Q^2)` is equal to (1) `2` (2) 1 (3) 0 (4

Let P and Q be `3xx3` matrices with `P!=Q` . If `P^3=""Q^3a n d""P^2Q""=""Q^2P` , then determinant of `(P^2+""Q^2)` is equal to (1) `2` (2) 1 (3) 0 (4) `1`

Monjur Alahi

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2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Correct Answer - A
Given , `P^(3) = Q^(3)" "...(i) `
` and P^(2) Q = Q^(2) P " "...(ii)`
Subtracting Eq. (i) and (ii), we get
`P^(3) - P^(2) Q = Q^(3) - Q^(2) P `
` P^(2) (P-Q) = - Q^(2) (P-Q)`
`rArr (P^(2) +Q^(2)) (P-Q) = O`
`rArr abs((P^(2) +Q^(2)) (P-Q)) =abs(O)`
`rArr abs((P^(2) +Q^(2)))abs( (P-Q)) =0`
`rArr abs((P^(2) +Q^(2)))=0 " " [ because P ne Q]`