Question
The number of values of k for which the linear equations
4x + ky + 2z = 0
kx + 4y + z = 0
2x + 2y + z = 0
posses a non-zero solution is:
-
3
-
2
-
1
-
0
Solution
B.
2
⇒ 8 - k(k -2) - 2(2k - 8) = 0
⇒ 8 - k2 + 2k - 4k + 16 = 0
⇒ -k2 - 2k + 24 = 0
⇒ k2 + 2k - 24 = 0
⇒ (k + 6)(k - 4) = 0
⇒ k = - 6, 4
Number of values of k is 2
