If A is an invertible matrix of order 3 and B is another matrix of the same order as of A , such that B = 2 ,   A T A B = A B B T . If A B - 1 a d j A T B - 1 = K , then the value of 16 K is equal to

Step-by-step solution

∵ A T A B = A B B T Taking determinant on both sides, we get, A T A B = A B B T ⇒ A = 2 Now, A B - 1 a d j A T B - 1 | A | × 1 | B | × 1 | a d j ( A T B ) | A × 1 B × 1 A T B 2 = A B 3 A 2 = 1 16 i.e. 4 K = 1 4 = 0 . 25