Let A = 1 a a 0 1 b 0 0 1 , a , b ∈ ℝ . If for some n ∈ N , A n = 1 48 2160 0 1 96 0 0 1 then n + a + b is equal to _______.

Step-by-step solution

Given, A = 1 a a 0 1 b 0 0 1 Now on rearranging we get, A = 1 0 0 0 1 0 0 0 1 + 0 a a 0 0 b 0 0 0 = I + B Now finding B 2 = 0 a a 0 0 b 0 0 0 0 a a 0 0 b 0 0 0 = 0 0 a b 0 0 0 0 0 0 And B 3 = 0 Now by using binomial expansion we get, A n = 1 + B n = C 0 n I + C 1 n B + C 2 n B 2 + C 3 n B 3 + . . . = 1 0 0 0 1 0 0 0 1 + 0 n a n a 0 0 n b 0 0 0 + 0 0 n n - 1 a b 2 0 0 0 0 0 0 as   B 3   and higher power will become zero = 1 n a n a + n n - 1 2 a b 0 1 n b 0 0 1 = 1 48 2160 0 1 96 0