The given system can be written in matrix form as:

A X = B

Now,

|A| = 1(12 – 5) + 1(9 + 10) + 2(– 3 – 8)

= 7 + 19 – 22

= 4

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

X = A ^{– 1}B

Cofactors of A are:

C₁₁ = (– 1)^{1 + 1} 12 – 5 = 7

C₂₁ = (– 1)^{2 + 1} – 3 + 2 = 1

C₃₁ = (– 1)^{3 + 1} 5 – 8 = – 3

C₁₂ = (– 1)^{1 + 2} 9 + 10 = – 19

C₂₂ = (– 1)^{2 + 1} 3 – 4 = – 1

C₃₂ = (– 1)^{3 + 1} – 5 – 6 = 11

C₁₃ = (– 1)^{1 + 2} – 3 – 8 = – 11

C₂₃ = (– 1)^{2 + 1} – 1 + 2 = – 1

C₃₃ = (– 1)^{3 + 1} 4 + 3 = 7

adj A =

=

A ^{– 1} =

Now, X = A ^{– 1}B =

X =

Hence, X = 2,Y = 1 and Z = 3