The given system can be written in matrix form as:

or A X = B

A = , X = and B =

Now, |A| = 3

= 3(3 – 6) – 4(– 6 – 3) + 7(4 + 1)

= – 9 + 36 + 35

= 62

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

X = A ^{– 1}B

Cofactors of A are:

C₁₁ = (– 1)^{1 + 1} 3 – 6 = – 3

C₂₁ = (– 1)^{2 + 1} – 12 – 14 = 26

C₃₁ = (– 1)^{3 + 1}12 + 7 = 19

C₁₂ = (– 1)^{1 + 2} – 6 – 3 = 9

C₂₂ = (– 1)^{2 + 1} – 3 – 7 = – 10

C₃₂ = (– 1)^{3 + 1} 9 – 14 = 5

C₁₃ = (– 1)^{1 + 2} 4 + 1 = 5

C₂₃ = (– 1)^{2 + 1} 6 – 4 = – 2

C₃₃ = (– 1)^{3 + 1} – 3 – 8 = – 11

adj A =

=

A ^{– 1} =

Now, X = A ^{– 1}B =

X =

X =

X =

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