Solve the following systems of homogeneous linear equations by matrix method:

x + y – 6z = 0


x – y + 2z = 0


– 3x + y + 2z = 0

The system can be written as



A X = 0


Now, |A| = 1(– 2 – 2) – 1(2 + 6) – 6(1 – 3)


|A| = – 4 – 8 + 12


|A| = 0


Hence, the system has infinite solutions


Let z = k


X + y = 6k


x – y = – 2k



A X = B


|A| = – 1 – 1 = – 2 ≠0 So, A – 1 exist


Now adj A = =


X = A – 1 B =


X =


X =


Hence, x = 2k, y = 4k and z = k


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