B’AB is a symmetric matrix.
Solution:
Given A is symmetric matrix
⇒ A’=A ..(1)
Now in B’AB,
Let AB=C ..(2)
⇒ B’AB=B’C
Now Using Property (AB)’=B’A’
⇒ (B’C)’=C’(B’)’ (As (B’)’=B)
⇒ C’(B’)’=C’B
⇒ C’B=(AB)’B (Using Property (AB)’=B’A’)
⇒ (AB)’B=B’A’B (Using (1))
⇒ B’A’B= B’AB
⇒ Hence (B’AB)’= B’AB