if A, B are square matrices of the same order, then prove that adj (

1.	if A, B are square matrices of the same order, then prove that adj (AB) = (adj B) (adj A).
Answer» We know that (AB) adj(AB) = \|AB\| = adj(AB)(AB) ..(i) (AB)(adj B adj A) = A.B adj B.adj A = A(B adj B) adj A =A(\|B\|)adj A [ :. B adj B = \|B\|I] = \|B\|(A.adj A) = \|B\| \|A\| I [:. A adj A = \|A\|I] = \|A\| \|B\| I = \|AB\|I ....(ii) From (i) and (ii), we get AB(adj AB) = AB(adj B.adj A) Pre-multiplying both sides by(AB)^-1,we get (AB)^-1[(AB)adj AB] = (AB)^-1 [(AB)adj B.adj A] or adj AB = adj B.adj A

if A, B are square matrices of the same order, then prove that adj (AB) = (adj B) (adj A).

Answer»

We know that

(AB) adj(AB) = |AB| = adj(AB)(AB) ..(i)

(AB)(adj B adj A)

= A.B adj B.adj A = A(B adj B) adj A

=A(|B|)adj A [ :. B adj B = |B|I]

= |B|(A.adj A)

= |B| |A| I [:. A adj A = |A|I]

= |A| |B| I

= |AB|I ....(ii)

From (i) and (ii), we get

AB(adj AB) = AB(adj B.adj A)

Pre-multiplying both sides by(AB)^-1,we get (AB)^-1[(AB)adj AB]

= (AB)^-1 [(AB)adj B.adj A]

or adj AB = adj B.adj A