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The number of values of k for which the linear equations4x + ky + 2z = 0kx + 4y + z = 02x + 2y + z = 0possess a non-zero solution is

The number of values of k for which the linear equations4x + ky + 2z = 0kx + 4y + z = 02x + 2y + z = 0possess a non-zero solution is

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1 Answers

Latika Leekha
askIITians Faculty 165 Points
6 years ago
The given system is
4x + ky + 2z = 0
kx + 4y + z = 0
2x + 2y + z = 0
For a non-zero solution, the determinant of the system must be zero.
Hence, 4(4-2) – k(k-2) + 2(2k-8) = 0
This yields (16 – 8 – k2 + 2k + 4k -16) = 0.
Hence, -k2 + 6k – 8 = 0
This gives k = 4, 2.
Hence, the given system has a non-zero solution for two values of k.
Thanks & Regards
Latika Leekha
askIITians Faculty

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