Q177 / 270 · NDA 2017 (Phase 2) · Mathematics

The system of equations $k x+y+z=1, x+k y+z=k$ and $x+y+k z=k^2$ has no solution if $k$ equals

  • A
    $0$
  • B
    $1$
  • C
    $-1$
  • Correct answer
    D
    $-2$

Solution

$\begin{aligned} & k x+y+z=1 \\ & x+k y+z=k \\ & x+y+k z=k^2\end{aligned}$ These equations will have no solution of $\left|\begin{array}{lll}\mathrm{k} & 1 & 1 \\ 1 & \mathrm{k} & 1 \\ 1 & 1 & \mathrm{k}\end{array}\right|=0$. $\begin{aligned} & \Rightarrow \mathrm{k}\left(\mathrm{k}^2-1\right)-1(\mathrm{k}-1)+(1-\mathrm{k})=0 \\ & \Rightarrow \mathrm{k}(\mathrm{k}+1)(\mathrm{k}-1)-1(\mathrm{k}-1)-(\mathrm{k}-1)=0 \\ & \Rightarrow(\mathrm{k}-1)[\mathrm{k}(\mathrm{k}+1)-1-1]=0\end{aligned}$ $\begin{aligned}&\Rightarrow(\mathrm{k}-1)\left(\mathrm{k}^2+\mathrm{k}-2\right)=0 \\& \Rightarrow \mathrm{k}=1 \text { or }-2 \end{aligned}$ For $\mathrm{k}=1$, all equations are same and have infinite solution. So, for $\mathrm{k}=-2$, equations have no solution.

Hence, the correct answer is option: 'D'.

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