This can be written as:

|A| = 5(260 – 4) – 3(30 – 14) + 7(6 – 182)

= 5(256) – 3(16) + 7(176)

|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} 260 – 4 = 256

C₂₁ = (– 1)^{2 + 1} 30 – 14 = – 16

C₃₁ = (– 1)^{3 + 1} 6 – 182 = – 176

C₁₂ = (– 1)^{1 + 2} 30 – 14 = – 16

C₂₂ = (– 1)^{2 + 1} 50 – 49 = 1

C₃₂ = (– 1)^{3 + 1} 10 – 21 = 11

C₁₃ = (– 1)^{1 + 2} 6 – 182 = – 176

C₂₃ = (– 1)^{2 + 1} 10 – 21 = 11

C₃₃ = (– 1)^{3 + 1} 130 – 9 = 121

adj A =

=

Adj A x B =

Now, AX = B has infinite many solution

Let z = k

Then, 5x + 3y = 4 – 7k

3x + 26y = 9 – 2k

This can be written as:

|A| = 121

Adj A =

Now, X = A ^{– 1}B =

=

=

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