Find the inverse of each of the following matrices.
|A| =
= 1(1 + 3) – 2( – 1 + 2) + 5(3 + 2)
= 4 – 2 + 25
= 27
Hence, A – 1 exists
Cofactors of A are:
C11 = 4 C21 = 17 C31 = 3
C12 = – 1 C22 = – 11 C32 = 6
C13 = 5 C23 = 1 C33 = – 3
adj A =
=
So, adj A =
Now, A – 1 = .adj A
So, A – 1 = .
Hence, A – 1 =