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