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The set of values of $k$ for which the system of simultaneous equations $\mathrm{x}+\mathrm{y}+\mathrm{kz}=1,2 \mathrm{x}+2 \mathrm{y}=3$ and $\mathrm{x}+2 \mathrm{y}+2 \mathrm{kz}=$ $\mathrm{k}$ has no real solution is
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The correct answer is:
$\{0\}$
Given system of equation
$\begin{aligned} & x+y+k z=1...(i) \\ & 2 x+2 y=3 ...(ii)\\ & x+2 y+2 k z=k...(iii)\end{aligned}$
has no real solution
$\begin{aligned} & \therefore\left|\begin{array}{lll}1 & 1 & \mathrm{~K} \\ 2 & 2 & 0 \\ 1 & 2 & 2 \mathrm{~K}\end{array}\right|=0 \\ & \Rightarrow 1(4 \mathrm{~K}-0)-1(4 \mathrm{~K}-0)+\mathrm{K}(4-2)=0 \\ & \Rightarrow 4 \mathrm{~K}-4 \mathrm{~K}+2 \mathrm{~K}=0 \Rightarrow \mathrm{K}=0\end{aligned}$
$\begin{aligned} & x+y+k z=1...(i) \\ & 2 x+2 y=3 ...(ii)\\ & x+2 y+2 k z=k...(iii)\end{aligned}$
has no real solution
$\begin{aligned} & \therefore\left|\begin{array}{lll}1 & 1 & \mathrm{~K} \\ 2 & 2 & 0 \\ 1 & 2 & 2 \mathrm{~K}\end{array}\right|=0 \\ & \Rightarrow 1(4 \mathrm{~K}-0)-1(4 \mathrm{~K}-0)+\mathrm{K}(4-2)=0 \\ & \Rightarrow 4 \mathrm{~K}-4 \mathrm{~K}+2 \mathrm{~K}=0 \Rightarrow \mathrm{K}=0\end{aligned}$
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