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. $-2, \frac{12}{5}$
• B. $-1, \frac{1}{5}$
• C. $-6, \frac{17}{5}$
• D. $-6, \frac{12}{5}$

Answer 1

Answer: (C)

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

Answer 2

The homogeneous linear system of equations has non-trivial solution $\therefore\quad\left|\begin{matrix}2&3&5\\\ 1&k&5\\\ k&-12&-14\end{matrix}\right|=0$ $\Rightarrow\quad2\left(-14k+60\right)-3\left(-14-5k\right)+5\left(-12-k^{2}\right)=0$ $\Rightarrow\quad5k^{2}+13k-102=0\,\Rightarrow\quad\left(5k-17\right)\left(k+6\right)=0$ $\Rightarrow\quad k=-6, \frac{17}{5}$