Let A = a i j be a square matrix of order 3 and B = b i j be a matrix such that b i j = 2 i - j a i j for 1 ≤ i , j ≤ 3 , ∀ i , j ∈ N . If the determinant of A is same as its order, then the value of B T - 1 is

Step-by-step solution

∵ A = 3 B = a 11 1 2 a 12 1 4 a 13 2 a 21 a 22 1 2 a 23 4 a 31 2 a 32 a 33 B = 8 a 11 1 2 a 12 1 4 a 13 a 21 1 2 a 22 1 4 a 23 a 31 1 2 a 32 1 4 a 33 = a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 = A = 3 B T - 1 = 1 B T = 1 B = 1 3