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- The number of values of k for which thelinear equations 4x + ky +2 z=0,kx + 4y + z=0 and 2 x +2y+z=0 posses anon-zero solution, is[AIEEE 2011](a) 2 (b) 1 (0) 0 (d) 3 |
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Answer» 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) = 0This gives -k2 + 6k -8 = 0Hence, k = 4, 2.The given system has a non-zero solution for two values of k. |
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