Consider the system of equations x+k y=0, y+k z=0 and z+k x=0. The set of all real values of k for which the system has a unique solution, is

Question

Consider the system of equations x+k y=0, y+k z=0 and z+k x=0. The set of all real values of k for which the system has a unique solution, is (A) R - -1  (B) R - 1  (C)  -1  (D)  -1,1

Tardigrade Admin · Accepted Answer

Given, x + ky + 0.z = 0 0 + y + kz = 0 kx + 0y + z = 0 For unique solution, Δ ≠ 0 ⇒ |1 k 0 0 1 k k 0 1| ≠ 0 ⇒ 1(1)- k (- k 2) ≠ 0 ⇒ k 3+1 ≠ 0 ⇒ k ∈ R - -1

Q. Consider the system of equations $x + k y = 0, y + k z = 0$ and $z + k x = 0$ . The set of all real values of $k$ for which the system has a unique solution, is

$R - {- 1}$

$R - {1}$

${- 1}$

${- 1, 1}$

Given,
$x + k y + 0. z = 0$
$0 + y + k z = 0$
$k x + 0 y + z = 0$
For unique solution, $Δ \neq = 0 \Rightarrow ∣ ∣ 10 k k 10 0 k 1 ∣ ∣ \neq = 0 \Rightarrow 1 (1) - k (- k^{2}) \neq = 0$
$\Rightarrow k^{3} + 1 \neq = 0 \Rightarrow k \in R - {- 1}$

Q. Consider the system of equations x+ky=0,y+kz=0 and z+kx=0. The set of all real values of k for which the system has a unique solution, is

R−{−1}

R−{1}

{−1}

{−1,1}