If the system of equations
$k x+y+2 z=1$
$3 x-y-2 z=2$
$-2 x-2 y-4 z=3$
has infinitely many solutions, then $\mathrm{k}$ is equal to
$D=0$
$\Rightarrow\left|\begin{array}{ccc}\mathrm{k} & 1 & 2 \\ 3 & -1 & -2 \\ -2 & -2 & -4\end{array}\right|=0$
$\Rightarrow \mathrm{k}(4-4)-1(-12-4)+2(-6-2)$
$\Rightarrow 16-16=0$
Also. $\quad \mathrm{D}_{1}=\mathrm{D}_{2}=\mathrm{D}_{3}=0$
$\Rightarrow \mathrm{D}_{2}=\left|\begin{array}{ccc}\mathrm{k} & 1 & 2 \\ 3 & 2 & -2 \\ -2 & 3 & -4\end{array}\right|=0$
$\Rightarrow \mathrm{k}(-8+6)-1(-12-4)+2(9+4)=0$
$\Rightarrow-2 \mathrm{k}+16+26=0$
$\Rightarrow 2 \mathrm{k}=42$
$\Rightarrow \mathrm{k}=21$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.