Answer :

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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