Solve the following systems of homogeneous linear equations by matrix method:
2x – y + 2z = 0
5x + 3y – z = 0
X + 5y – 5z = 0
The system can be written as

A X = 0
Now, |A| = 2(– 15 + 5) + 1(– 25 + 1) + 2(25 – 3)
|A| = – 20 – 24 + 44
|A| = 0
Hence, the system has infinite solutions
Let z = k
2x – y = – 2k
5x + 3y = k
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A X = B
|A| = 6 + 5 = 11≠0 So, A – 1 exist
Now adj A =
= ![]()
X = A – 1 B = ![]()
X = ![]()
Hence, x =
, y =
and z = k
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