|A| = 2(14) – 2(14) – 2(0)

|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}8 + 6 = 14

C₂₁ = (– 1)^{2 + 1} 4 + 12 = – 16

C₃₁ = (– 1)^{3 + 1} – 2 + 8 = 6

C₁₂ = (– 1)^{1 + 2} 8 + 6 = – 14

C₂₂ = (– 1)^{2 + 1} 4 + 12 = 16

C₃₂ = (– 1)^{3 + 1} – 2 + 8 = – 6

C₁₃ = (– 1)^{1 + 2} 24 – 24 = 0

C₂₃ = (– 1)^{2 + 1} 12 – 12 = 0

C₃₃ = (– 1)^{3 + 1} 8 – 8 = 0

adj A =

=

Adj A x B =

Now, AX = B has infinite many solution

Let z = k

Then, 2x + 2y = 1 + 2k

4x + 4y = 2 + k

This can be written as:

Hence, |A| = 0 z = 0

Hence, The given equation doesn’t satisfy .