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MIT 18.06 Linear Algebra, Spring 2010 9 Online
OpenStudy (anonymous):

show that (adj B)^-1 = adj(B^-1)

OpenStudy (anonymous):

It is well known that B^-1=adj(B)/det(B), so adj(B)=det(B).B^-1. Similarly, adj(B^-1)=det(B^-1).B. Hence adj(B^-1).adj(B)=det(B)det(B^-1).B.B^-1=I.

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