Question

If the system of equations
$kx+y+2z=1$
$3x-y-2z=2$
$-2x-2y-4z=3$
has intinitely many solutions, then $k$ is equal to

Answer 1

$D=0$
$\Rightarrow \left|\begin{array}{ccc}k& 1& 2\\ 3& -1& -2\\ -2& -2& -4\end{array}\right|$
$\Rightarrow k(4-4)-1(-12-4)+2(-6-2)$
$\Rightarrow 16-16=0$
Also. $D_1=D_2=D_3=0$
$\Rightarrow D_2=\left|\begin{array}{ccc}k& 1& 2\\ 3& 2& -2\\ -2& 3& -4\end{array}\right|$
$\Rightarrow k(-8+6)-1(-12-4)+2(9+4)=0$
$\Rightarrow -2k+16+26=0$
$\Rightarrow 2k=42$
$\Rightarrow k=21$