The given system can be written in matrix form as:

or A X = B

A = , X = and B =

Now, |A| = 1

= (3 – 1) – 1(– 6 – 2) + 1(2 + 2)

= 2 + 8 + 4

= 14

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

X = A ^{– 1}B

Cofactors of A are:

C₁₁ = (– 1)^{1 + 1} 3 – 1 = 2

C₂₁ = (– 1)^{2 + 1} – 3 – 1 = 4

C₃₁ = (– 1)^{3 + 1} 1 + 1 = 2

C₁₂ = (– 1)^{1 + 2} – 6 – 2 = 8

C₂₂ = (– 1)^{2 + 1} – 3 – 2 = – 5

C₃₂ = (– 1)^{3 + 1} 1 – 2 = 1

C₁₃ = (– 1)^{1 + 2} 2 + 2 = 4

C₂₃ = (– 1)^{2 + 1} 1 – 2 = 1

C₃₃ = (– 1)^{3 + 1} – 1 – 2 = – 3

adj A =

=

Now, X = A ^{– 1}B =

X =

X =

Hence, X = ,Y = and Z =