Find the inverse of the matrix and show that aA – 1 = (a2 + bc + 1) I – aA.

A =


Now, |A| = bc =


Hence, A – 1 exists.


Cofactors of A are


C11 = C12 = – c


C21 = – b C22 = a


Since, adj A =


(adj A) =


=


Now, A – 1 = .adj A


A – 1 = .


A – 1 =


To show. aA – 1 = (a2 + bc + 1) I – aA.


LHS aA – 1 = a


=


RHS (a2 + bc + 1) I – aA =


=


Hence, LHS = RHS


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