Answer :

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


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