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The number of values of k for which the linear equations 4x ky 2z 0 ++= ; kx 4y z 0 + + = ; 2x 2y z 0 + += possess a non-zero solution is

The number of values of k for which the linear equations 4x ky 2z 0 ++= ; kx 4y z 0 + + = ; 2x 2y z 0 + += possess a non-zero solution is

Grade:12

1 Answers

Latika Leekha
askIITians Faculty 165 Points
7 years ago
The signs have not been placed at the correct place. The correct equations would be 4x + ky+ 2z = 0;
kx + 4y + z = 0;
2x + 2y + z = 0.
Now, for the system of equations to possess a non -zero solution, the value of the determinant should be zero.
Hence, D = 0 gives, [4(4-2) – k(k-2) + 2(2k-8)]
Hence, we have (16 – 8 – k2 + 2k + 4k – 16) = 0
This gives -k2 + 6k -8 = 0
Hence, k = 4, 2.
The given system has a non-zero solution for two values of k.
Thanks & Regards
Latika Leekha
askIITians Faculty

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