If the system of equations 2x + 3y+5 = 0, x + ky + 5 = 0, kx - 12y - 14 = 0 has non-trivial solution, then the value of k is

Question

If the system of equations 2x + 3y+5 = 0, x + ky + 5 = 0, kx - 12y - 14 = 0 has non-trivial solution, then the value of k is (A) -2, (12/5) (B) -1, (1/5) (C) -6, (17/5) (D) -6, (12/5)

Tardigrade Admin · Accepted Answer

The homogeneous linear system of equations has non-trivial solution

∴ ∣ ∣ 21 k 3 k - 12 55 - 14 ∣ ∣ = 0

\Rightarrow 2 (- 14 k + 60) - 3 (- 14 - 5 k) + 5 (- 12 - k^{2}) = 0

\Rightarrow 5 k^{2} + 13 k - 102 = 0 \Rightarrow (5 k - 17) (k + 6) = 0

\Rightarrow k = - 6, \frac{17}{5}

Q. If the system of equations $2 x + 3 y + 5 = 0$ , $x + k y + 5 = 0$ , $k x - 12 y - 14 = 0$ has non-trivial solution, then the value of $k$ is

$- 2, \frac{12}{5}$

$- 1, \frac{1}{5}$

$- 6, \frac{17}{5}$

$- 6, \frac{12}{5}$

The homogeneous linear system of equations has non-trivial solution
$∴ ∣ ∣ 21 k 3 k - 12 55 - 14 ∣ ∣ = 0$
$\Rightarrow 2 (- 14 k + 60) - 3 (- 14 - 5 k) + 5 (- 12 - k^{2}) = 0$
$\Rightarrow 5 k^{2} + 13 k - 102 = 0 \Rightarrow (5 k - 17) (k + 6) = 0$
$\Rightarrow k = - 6, \frac{17}{5}$

Q. If the system of equations 2x+3y+5=0, x+ky+5=0, kx−12y−14=0 has non-trivial solution, then the value of k is

−2,512​

−1,51​

−6,517​

−6,512​