Let A be a square matrix of order 3 such that A + A T = 10 4 6 a 21 + a 12 6 a 23 + a 32 a 31 + a 13 8 4 , where a 12 , a 23 , a 31 are positive roots of the equation x 3 - 6 x 2 + p x - 8 = 0 , ∀ p ∈ R , then the absolute value of A is equal to

Question

Let A be a square matrix of order 3 such that A + A T = 10 4 6 a 21 + a 12 6 a 23 + a 32 a 31 + a 13 8 4 , where a 12 , a 23 , a 31 are positive roots of the equation x 3 - 6 x 2 + p x - 8 = 0 , ∀ p ∈ R , then the absolute value of A is equal to,

MARKS App · Accepted Answer

Clearly, A + A T is a symmetric matrix Also, a 12 + a 23 + a 31 = 6 , a 12 a 23 a 31 = 8 A M = G M ⇒ a 12 = a 23 = a 31 = 2 a 21 + a 12 = 4 ⇒ a 21 = a 12 = 2 a 31 + a 13 = 6 ⇒ a 31 = a 13 = 3 a 23 + a 32 = 8 ⇒ a 23 = a 32 = 4 For diagonal elements 2 a 11 = 10 ⇒ a 11 = 5 2 a 22 = 6 ⇒ a 22 = 3 2 a 33 = 4 ⇒ a 33 = 2 A = 5 2 3 2 3 4 3 4 2 A = 5 6 - 16 - 2 4 - 12 + 3 8 - 9 = - 50 + 16 - 3 = - 3 7

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