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

Question

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

Tardigrade Admin · 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 - a2)+3(0 - 2 a)=0 a3+4a-16=0⇒ (a - 2)(a2 + 2 a + 8)=0 a=2 Hence, only one value of a

Q. 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

$0$

$1$

$2$

$3$

For non-trivial solution, using Cramer's rule, we get,
$∣ ∣ 41 a - a 20 3 a 2 ∣ ∣ = 0$
$4 (4 - 0) + a (2 - a^{2}) + 3 (0 - 2 a) = 0$
$a^{3} + 4 a - 16 = 0 \Rightarrow (a - 2) (a^{2} + 2 a + 8) = 0$
$a = 2$
Hence, only one value of $a$

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

0

1

2

3

For non-trivial solution, using Cramer's rule, we get, ∣∣​41a​−a20​3a2​∣∣​=0 4(4−0)+a(2−a2)+3(0−2a)=0 a3+4a−16=0⇒(a−2)(a2+2a+8)=0 a=2 Hence, only one value of a

Q. 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

$0$

$1$

$2$

$3$

For non-trivial solution, using Cramer's rule, we get,
$∣ ∣ 41 a - a 20 3 a 2 ∣ ∣ = 0$
$4 (4 - 0) + a (2 - a^{2}) + 3 (0 - 2 a) = 0$
$a^{3} + 4 a - 16 = 0 \Rightarrow (a - 2) (a^{2} + 2 a + 8) = 0$
$a = 2$
Hence, only one value of $a$