Find the inverse of each of the following matrices.

|A| = ![]()
= 2(3 – 0) – 0 – 1(5)
= 6 – 5
= 1
Hence, A – 1 exists
Cofactors of A are:
C11 = 3 C21 = – 1 C31 = 1
C12 = – 15 C22 = 6 C32 = – 5
C13 = – 5 C23 = – 2 C33 = 2
adj A = 
= 
So, adj A = 
Now, A – 1 =
.adj A
So, A – 1 =
.
Hence, A – 1 = 
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