Math miscellaneousEasy
Question
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
4x + ky + 2z = 0 ; kx + 4y + z = 0 ; 2x + 2y + z = 0 possess a non-zero solution is
Options
A.2
B.1
C.zero
D.3
= 0 ⇒ k2 - 6k + 8 = 0 ⇒ k = 4, 2More Math miscellaneous Questions
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