Question

If the system of equations, $x-ky-z=0$, $kx-y-z=0$, $x+y+z=0$, has a non zero solution, then the possible values of $k$ are

• A. $-1,2$
• B. $1,2$
• C. $0,1$
• D. $-1,1$

Answer 1

The correct option is D $-1,1$

$x-ky-z=0$ (1)
$kx-y-z=0$ (2)
$x+y+z=0$ (3)
Add (2) and (3) equation,
$kx+x=0$
Thus, $k=-1$
Subtract (1) and (2) equation,
$x-kx-ky+y=0$
$(1-k)(x+y)=0$
$k=1$
Thus, possible values of $k=-1,1$