This can be written as:

|A| = 1(2) – 1(4) + 1(2)

= 2 – 4 + 2

|A| = 0

So, A is singular. Thus, the given system is either inconsistent or it is consistent with infinitely many solution according to as:

(Adj A)x B≠0 or (Adj A)x B = 0

Cofactors of A are:

C₁₁ = (– 1)^{1 + 1} 14 – 12 = 2

C₂₁ = (– 1)^{2 + 1} 7 – 4 = – 3

C₃₁ = (– 1)^{3 + 1} 3 – 2 = 1

C₁₂ = (– 1)^{1 + 2} 7 – 3 = – 4

C₂₂ = (– 1)^{2 + 1} 7 – 1 = 6

C₃₂ = (– 1)^{3 + 1} 3 – 1 = 2

C₁₃ = (– 1)^{1 + 2} 4 – 2 = 2

C₂₃ = (– 1)^{2 + 1} 4 – 1 = – 3

C₃₃ = (– 1)^{3 + 1} 2 – 1 = 1

adj A =

=

Adj A x B =

Now, AX = B has infinite many solution

Let z = k

Then, x + y = 6 – k

X + 2y = 14 – 3k

This can be written as:

|A| = 1

Adj A =

Now, X = A ^{– 1}B =

=

=

There values of x,y,z satisfy the third equation

Hence, x = k – 2, y = 8 – 2k, z = k