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