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

Question

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

Tardigrade Admin · Accepted Answer

For non-trivial solution of given system of linear equations |4&k&2 k&4&1 2&2&1|=0 ⇒ 8+k(2-k)+2(2 k-8)=0 ⇒ -k2+6 k-8=0 ⇒ k2-6 k+8=0 ⇒ k=2,4 Clearlv there exists two values of k

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

Zero

3

2

1

For non-trivial solution of given system of linear equations
$∣ ∣ 4 k 2 k 42 211 ∣ ∣ = 0$
$\Rightarrow 8 + k (2 - k) + 2 (2 k - 8) = 0$
$\Rightarrow - k^{2} + 6 k - 8 = 0$
$\Rightarrow k^{2} - 6 k + 8 = 0$
$\Rightarrow k = 2, 4$
Clearlv there exists two values of $k$

Q. 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

Zero

3

2

1

For non-trivial solution of given system of linear equations ∣∣​4k2​k42​211​∣∣​=0 ⇒8+k(2−k)+2(2k−8)=0 ⇒−k2+6k−8=0 ⇒k2−6k+8=0 ⇒k=2,4 Clearlv there exists two values of k

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

For non-trivial solution of given system of linear equations
$∣ ∣ 4 k 2 k 42 211 ∣ ∣ = 0$
$\Rightarrow 8 + k (2 - k) + 2 (2 k - 8) = 0$
$\Rightarrow - k^{2} + 6 k - 8 = 0$
$\Rightarrow k^{2} - 6 k + 8 = 0$
$\Rightarrow k = 2, 4$
Clearlv there exists two values of $k$