Math, asked by Avinag, 9 months ago

The Number of linearly independent solutions of the system

2x + 3ky + (3k + 4)z = 0, x + (k + 4) + (4k +2)z = 0, x+2(k+1) + (3k + 4)z = 0

for k = 2 is…….

Answers

Answered by hariniakshu

1

1)2x + 3ky + (3k+4)z = 0

2x + 3(2)y + (3(2) + 4)z = 0

2x + 6y + 10z

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