Consider a matrix A = 0 1 2 0 - 3 0 1 1 1 . If 6 A - 1 = a A 2 + b A + c I , where a , b , c ∈ Z and I is an identity matrix, then a + 2 b + 3 c is equal to

Step-by-step solution

A = 6 ≠ 0 ( A is non sigular) Given, 6 A - 1 = a A 2 + b A + c I ⇒ 6 I = a A 3 + b A 2 + c A A 2 = 0 1 2 0 - 3 0 1 1 1 0 1 2 0 - 3 0 1 1 1 = 2 - 1 2 0 9 0 1 - 1 3 A 3 = A 2 A = 2 - 1 2 0 9 0 1 - 1 3 0 1 2 0 - 3 0 1 1 1 = 2 7 6 0 - 27 0 3 7 5 Now, a A 3 + b A 2 + c A = 2 a + 2 b 7 a - b + c 6 a + 2 b + 2 c 0 - 27 a + 9 b - 3 c 0 3 a + b + c 7 a - b + c 5 a + 3 b + c 6 I = 6 0 0 0 6 0 0 0 6 ∵ a A 3 + b A 2 + c A = 6 I ⇒ a + b = 3 , 7 a - b + c = 0 , 3 a + b + c = 0 , - 9 a + 3 b - c = 2 , 5 a