If 4 x - a y + 3 z = 0 , x + 2 y + a z = 0 and a x + 2 z = 0 have a non-trivial solution, then the number of real value(s) of a is

Question

If 4 x - a y + 3 z = 0 , x + 2 y + a z = 0 and a x + 2 z = 0 have a non-trivial solution, then the number of real value(s) of a is, 0, 1, 2, 3

MARKS App · Accepted Answer

For non-trivial solution, using Cramer's rule, we get, 4 - a 3 1 2 a a 0 2 = 0 4 4 - 0 + a 2 - a 2 + 3 0 - 2 a = 0 a 3 + 4 a - 16 = 0 ⇒ a - 2 a 2 + 2 a + 8 = 0 a = 2 Hence, only one value of a

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