Let A be a square matrix of order 3 , A T be the transpose matrix of matrix A and A A T = 4 I . If d = 2 A T + A A T + a d j A 2 , then the value of 12 d is equal to A < 0

Step-by-step solution

A A T = 4 I ⇒ A T = 4 A - 1 = 4 × a d j A A ⇒ a d j A = A A T 4 … i ∵ A A T = 4 I ⇒ A A T = 4 I ⇒ A 2 = 64 ⇒ A = - 8   A s   A < 0 From i a d j A = - 2 A T ⇒ 2 A T + a d j A = O d = 4 I 2 = 2 I = 8 ⇒ 12 d = 96