Consider the matrix A = 3 - 2 0 1 and B be a square matrix of order 2 such that B B T = B T B = I . Let C = B A B T and D = d i j 2 × 2 = B T C 6 B , then d 11 is equal to

Step-by-step solution

D = B T B A B T B A B T B A B T B A B T B A B T B A B T B D = A 6 Now, A 2 = 3 - 2 0 1 3 - 2 0 1 = 3 - 2 3 - 2 0 1 A 3 = A 2 ⋅ A = 3 - 2 3 - 2 0 1 3 − 2 0 1 = 3 3 − 2 3 − 8 0 1 A 6 = A 3 ⋅ A 3 = 3 3 − 2 3 − 8 0 1 3 3 − 2 3 − 8 0 1 = 27 - 26 - 26 3 0 1 = D ⇒ d 11 = 27