(a) Since A and B commute: AB = BA
Pre- and post-multiplying both sides by A-1, we get
A-1(AB)A-1 = A-1(BA)A-1
⇒ (A-1A)(BA-1) = A-1B(AA-1) (by associativity)
⇒ I(BA-1) = (A-1B)I
⇒ BA-1 = A-1B
Now, (A-1B)′ = (BA-1′ = (A-1)′B′ (by reversal law)
= A-1B [as B′ = B (symmetric) and (A-1)′ = (A′) -1 = A-1]
Hence, A-1B is symmetric.
(b) Pre-and post-multiplying by B-1, we get
B-1(AB)B-1 = B-1(BA)B-1
⇒ (B-1A)BB-1 = B-1B(AB-1)
⇒ B-1A = AB-1
Now, (AB-1)′= (B-1A)′ = (A′B-1)′
= AB-1 [as A = A′ (symmetric) and (B-1)′ = (B′)-1 = B-1]
Hence, AB-1 is symmetric.
(c) Since A and B are symmetric, we have
AB = BA ⇒ (BA) -1 = (AB) -1
⇒ A-1B-1 = B-1A-1
⇒ (A-1B-1)′ = (B-1A-1)′ = (A-1)′ ⋅ (B-1)′ = A-1B-1
[as (A-1)′ = A-1 and (B-1)′ = B-1]
Hence, A-1B-1 is symmetric.