The number of values of k for which the linear equations 4x + ky + 2z = 0 , kx + 4y + z = 0 and 2x + 2y + z = 0 possess a non-zero solution is

Question

The number of values of k for which the linear equations 4x + ky + 2z = 0 , kx + 4y + z = 0 and 2x + 2y + z = 0 possess a non-zero solution is (A) 2 (B) 1 (C) 5 (D) 3

Tardigrade Admin · Accepted Answer

Δ = 0 ⇒ |4&k&2 k&4&1 2&2&1| = 0 ⇒ 4(4 - 2) - k(k - 2) + 2(2k - 8) = 0 ⇒ 8 - k2 + 2k + 4k - 16 = 0 k2 - 6k + 8 = 0 ⇒ (k - 4 )(k-2) = 0 ⇒ k = 4 , 2

Q. The number of values of k for which the linear equations $4 x + k y + 2 z = 0$ , $k x + 4 y + z = 0$ and $2 x + 2 y + z = 0$ possess a non-zero solution is

2

1

5

3

$Δ = 0$
$\Rightarrow ∣ ∣ 4 k 2 k 42 211 ∣ ∣ = 0$
$\Rightarrow 4 (4 - 2) - k (k - 2) + 2 (2 k - 8) = 0$
$\Rightarrow 8 - k^{2} + 2 k + 4 k - 16 = 0$
$k^{2} - 6 k + 8 = 0$
$\Rightarrow (k - 4) (k - 2) = 0 \Rightarrow k = 4, 2$

Q. The number of values of k for which the linear equations 4x+ky+2z=0 , kx+4y+z=0 and 2x+2y+z=0 possess a non-zero solution is

2

1

5

3

Δ=0 ⇒∣∣​4k2​k42​211​∣∣​=0 ⇒ 4(4−2)−k(k−2)+2(2k−8)=0 ⇒ 8−k2+2k+4k−16=0 k2−6k+8=0 ⇒ (k−4)(k−2)=0 ⇒ k=4,2

Q. The number of values of k for which the linear equations $4 x + k y + 2 z = 0$ , $k x + 4 y + z = 0$ and $2 x + 2 y + z = 0$ possess a non-zero solution is

$Δ = 0$
$\Rightarrow ∣ ∣ 4 k 2 k 42 211 ∣ ∣ = 0$
$\Rightarrow 4 (4 - 2) - k (k - 2) + 2 (2 k - 8) = 0$
$\Rightarrow 8 - k^{2} + 2 k + 4 k - 16 = 0$
$k^{2} - 6 k + 8 = 0$
$\Rightarrow (k - 4) (k - 2) = 0 \Rightarrow k = 4, 2$