The correct option is D (adj A)−1=adj(A−1) for all invertible matrix A.
A is symmetric and B is skew-symmetric.
Let P=AB
Now, PT=(AB)T=BTAT=−BA
⇒PT≠−P
∴AB is not skew symmetric.
We know that
A(adj A)=|A|I3
⇒(A(adj A))T=(|A|I3)T
⇒(adj A)TAT=|A|I3
⇒(adj A)T=|AT|(AT)−1
∴(adj A)T=adj(AT) for all invertible matrix A.
Let P=AB+BA
∴PT=(AB+BA)T=(AB)T+(BA)T
=BTAT+ATBT=BA+AB
=AB+BA=P
∴AB+BA is symmetric for all symmetric matrices A and B.
adj(A−1)=A|A| ⋯(1)
Also, A(adj A)=|A|I3
⇒(adj A)=|A|A−1
⇒(adj A)−1=A|A| ⋯(2)
From (1) and (2), we get
(adj A)−1=adj(A−1)