# Solve the followi

2x – y + z = 0

3x + 2y – z = 0

X + 4y + 3z = 0

The system can be written as

A X = 0

Now, |A| = 2(6 + 4) + 1(9 + 1) + 1(12 – 2)

|A| = 2(10) + 10 + 10

|A| = 40 ≠ 0

Since, |A|≠ 0, hence x = y = z = 0 is the only solution of this homogeneous equation.

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