Let A = a i j 3 × 3 be a square matrix such that A A T = 4 I ,   A < 0 . If a 11 + 4 a 12 a 13 a 21 a 22 + 4 a 23 a 31 a 32 a 33 + 4 = 5 λ A + I . Then, λ is equal to

Step-by-step solution

A A T = 4 I ⇒ A A T = 4 I ⇒ A 2 = 64 ⇒ A = - 8 or 8 As, A < 0 ⇒ A = - 8 a 11 + 4 a 12 a 13 a 21 a 22 + 4 a 23 a 31 a 32 a 33 + 4 = A + 4 I = A + A A T = A I + A T ⇒ - 8 A + I = 5 λ A + I ⇒ λ = - 8 5