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NDA 2017
Exam
NDA
Year
2017
Subject
Mathematics
Chapter
Determinants

NDA 2017 · Mathematics · Determinants

The system of equations k x+y+z=1, x+k y+z=k and x+y+k z=k^2 has no...

The system of equations kx+y+z=1,x+ky+z=kk x+y+z=1, x+k y+z=k and x+y+kz=k2x+y+k z=k^2 has no solution if kk equals

Options

  1. A00
  2. B11
  3. C−1-1
  4. D−2-2

Correct answer

D. −2-2

Step-by-step solution

kx+y+z=1x+ky+z=kx+y+kz=k2\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|k111k111k\begin{array}{lll}\mathrm{k} & 1 & 1 \\ 1 & \mathrm{k} & 1 \\ 1 & 1 & \mathrm{k}\end{array}\right|=0. ⇒k(k2−1)−1(k−1)+(1−k)=0⇒k(k+1)(k−1)−1(k−1)−(k−1)=0⇒(k−1)[k(k+1)−1−1]=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}⇒(k−1)(k2+k−2)=0&⇒k=1 or  −2\begin{aligned}&\Rightarrow(\mathrm{k}-1)\left(\mathrm{k}^2+\mathrm{k}-2\right)=0 \& \Rightarrow \mathrm{k}=1 \text{\text{ or } }-2 \end{aligned}ForFor\mathrm{k}=1,allequationsaresameandhaveinfinitesolution.So,for, all equations are same and have infinite solution. So, for\mathrm{k}=-2$, equations have no solution.
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