|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} =