The given system can be written in matrix form as:

or A X = B

A = , X = and B =

Now, |A| = 5

= 5(4 – 6) – 3(8 – 3) + 1(4 – 2)

= – 10 – 15 + 3

= – 22

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

X = A ^{– 1}B

Cofactors of A are:

C₁₁ = (– 1)^{1 + 1} (4 – 6) = – 2

C₂₁ = (– 1)^{2 + 1}(12 – 2) = – 10

C₃₁ = (– 1)^{3 + 1}(9 – 1) = 8

C₁₂ = (– 1)^{1 + 2} (8 – 3) = – 5

C₂₂ = (– 1)^{2 + 1} 20 – 1 = 19

C₃₂ = (– 1)^{3 + 1} 15 – 2 = – 13

C₁₃ = (– 1)^{1 + 2} (4 – 2) = 2

C₂₃ = (– 1)^{2 + 1} 10 – 3 = – 7

C₃₃ = (– 1)^{3 + 1} 5 – 6 = – 1

adj A =

=

A ^{– 1} =

Now, X = A ^{– 1}B =

X =

X =

X =

Hence, X = 1,Y = 2 and Z = 5