The given system can be written in matrix form as:

or A X = B

A = , X = and B =

Now, |A| = 1

= (– 20) – 1(– 17) – 1(11)

= – 20 + 17 + 11 = 8

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

X = A ^{– 1}B

Cofactors of A are:

C₁₁ = (– 1)^{1 + 1} – 21 + 1 = – 20

C₂₁ = (– 1)^{2 + 1} – 7 – 1 = 8

C₃₁ = (– 1)^{3 + 1} 1 + 3 = 4

C₁₂ = (– 1)^{1 + 2} – 14 – 3 = 17

C₂₂ = (– 1)^{2 + 1} – 7 + 3 = – 4

C₃₂ = (– 1)^{3 + 1} 1 + 2 = – 3

C₁₃ = (– 1)^{1 + 2} – 2 – 9 = – 11

C₂₃ = (– 1)^{2 + 1} – 1 – 3 = 4

C₃₃ = (– 1)^{3 + 1} 3 – 2 = 1

adj A =

=

Now, X = A ^{– 1}B =

X =

X =

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