The given system can be written in matrix form as:

or A X = B

A = , X = and B =

Now, |A| = 2

= 2(0 – 1) – 6(3 + 2)

= – 2 – 30

= – 32

So, the above system has a unique solution, given by

X = A ^{– 1}B

Cofactors of A are:

C₁₁ = (– 1)^{1 + 1} 0 – 1 = – 1

C₂₁ = (– 1)^{2 + 1}6 + 0 = – 6

C₃₁ = (– 1)^{3 + 1} – 6 = – 6

C₁₂ = (– 1)^{1 + 2} 3 + 2 = 5

C₂₂ = (– 1)^{2 + 1} 2 – 0 = 2

C₃₂ = (– 1)^{3 + 1} – 2 – 0 = 2

C₁₃ = (– 1)^{1 + 2} – 3 – 0 = – 3

C₂₃ = (– 1)^{2 + 1} – 2 – 12 = 14

C₃₃ = (– 1)^{3 + 1} 0 – 18 = – 18

adj A =

=

A ^{– 1} =

Now, X = A ^{– 1}B =

X =

X =

X =

Hence, X = – 2,Y = 1 and Z = 2