The given system can be written in matrix form as:

or A X = B

A = , X = and B =

Now, |A| = 2

= 2(– 6 + 1) – 1(– 2 + 3) + 1(1 – 9)

= – 10 – 1 – 8

= – 19

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

X = A ^{– 1}B

Cofactors of A are:

C₁₁ = (– 1)^{1 + 1} – 6 + 1 = – 5

C₂₁ = (– 1)^{2 + 1}(24 + 4) = – 28

C₃₁ = (– 1)^{3 + 1} – 1 – 3 = – 4

C₁₂ = (– 1)^{1 + 2} – 2 + 3 = – 1

C₂₂ = (– 1)^{2 + 1} – 4 – 3 = – 7

C₃₂ = (– 1)^{3 + 1} – 2 – 1 = 3

C₁₃ = (– 1)^{1 + 2}1 – 9 = – 8

C₂₃ = (– 1)^{2 + 1}2 – 3 = – 1

C₃₃ = (– 1)^{3 + 1} 6 – 1 = 5

adj A =

=

A ^{– 1} =

Now, X = A ^{– 1}B =

X =

X =

X =

Hence, X = 1,Y = 1 and Z = – 1