# If the system of equations

Question:

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

Solution:

$D=0$

$\Rightarrow\left|\begin{array}{ccc}\mathrm{k} & 1 & 2 \\ 3 & -1 & -2 \\ -2 & -2 & -4\end{array}\right|=0$

$\Rightarrow \mathrm{k}(4-4)-1(-12-4)+2(-6-2)$

$\Rightarrow 16-16=0$

Also. $\quad \mathrm{D}_{1}=\mathrm{D}_{2}=\mathrm{D}_{3}=0$

$\Rightarrow \mathrm{D}_{2}=\left|\begin{array}{ccc}\mathrm{k} & 1 & 2 \\ 3 & 2 & -2 \\ -2 & 3 & -4\end{array}\right|=0$

$\Rightarrow \mathrm{k}(-8+6)-1(-12-4)+2(9+4)=0$

$\Rightarrow-2 \mathrm{k}+16+26=0$

$\Rightarrow 2 \mathrm{k}=42$

$\Rightarrow \mathrm{k}=21$