The system can be written as

A X = 0

Now, |A| = 1(4) – 1(– 8) – 1(12) = 0

|A| = 4 + 8 – 12

|A| = 0

Hence, the system has infinite solutions

Let z = k

X + y = k

x – 2y = – k

A X = B

|A| = – 2 – 1 = – 3 ≠0 So, A ^{– 1} exist

Now adj A = =

X = A ^{– 1} B =

X =

X =

Hence, x = , y = and z = k