The given system can be written in matrix form as:

AX = B

Now, |A| = 8

= 8(– 1) – 4(1) + 3(3)

= – 8 – 4 + 9

= – 3

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

X = A ^{– 1}B

Cofactors of A are:

C₁₁ = (– 1)^{1 + 1} 1 – 2 = – 1

C₂₁ = (– 1)^{2 + 1} 4 – 6 = 2

C₃₁ = (– 1)^{3 + 1} 4 – 3 = 1

C₁₂ = (– 1)^{1 + 2} 2 – 1 = – 1

C₂₂ = (– 1)^{2 + 1} 8 – 3 = 5

C₃₂ = (– 1)^{3 + 1} 8 – 6 = – 2

C₁₃ = (– 1)^{1 + 2} 4 – 1 = 3

C₂₃ = (– 1)^{2 + 1} 16 – 4 = – 12

C₃₃ = (– 1)^{3 + 1} 8 – 8 = 0

adj A =

=

A ^{– 1} =

Now, X = A ^{– 1}B =

X =

X =

X =

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