The given system can be written in matrix form as:

AX = B

Now,

|A| = 2(75) – 3(– 110) + 10(72)

= 150 + 330 + 720

= 1200

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

X = A ^{– 1}B

Cofactors of A are:

C₁₁ = (– 1)^{1 + 1} 120 – 45 = 75

C₂₁ = (– 1)^{2 + 1} – 60 – 90 = 150

C₃₁ = (– 1)^{3 + 1} 15 + 60 = 75

C₁₂ = (– 1)^{1 + 2} – 80 – 30 = 110

C₂₂ = (– 1)^{2 + 1} – 40 – 60 = – 100

C₃₂ = (– 1)^{3 + 1} 10 – 40 = 30

C₁₃ = (– 1)^{1 + 2} 36 + 36 = 72

C₂₃ = (– 1)^{2 + 1} 18 – 18 = 0

C₃₃ = (– 1)^{3 + 1} – 12 – 12 = – 24

adj A =

=

A ^{– 1} =

Now, X = A ^{– 1}B =

X =

Hence, X = 2,Y = 3 and Z = 5