For square matrices A and B of the same order, we have adj(AB)=?

Question

Philoid · Accepted Answer

We know that (AB)^-1 = adj(AB)/

adj (AB)= (AB)^-1

We also know that (AB)^-1 = B^-1. A^-1

Putting them in 1

Adj (AB) = B^-1. A^-1.

= (A^-1.) (B^-1)

= adj(A) adj(B)

Since, adj (A)= (A)^-1

adj (B)= (B)^-1