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

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

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