A) \[K=-\,2~\]
B) \[K=-1\]
C) \[K=0\]
D) \[~K=1\]
Correct Answer: B
Solution :
The system of given equations are \[Kx+2y-z=1\] ?(i) \[(K-1)y-2z=2\] ?(ii) and \[(K+2)z=3\] ?(iii) This system of equations has a unique solution, it \[\left| \begin{matrix} K & 2 & -1 \\ 0 & K-1 & -2 \\ 0 & 0 & K+2 \\ \end{matrix} \right|\ne 0\] \[\Rightarrow \] \[(K+2)\left| \begin{matrix} K & 2 \\ 0 & K-1 \\ \end{matrix} \right|\ne 0\] \[\Rightarrow \] \[(K+2)(K)(K-1)\ne 0\] \[\Rightarrow \] \[K\ne -2,\,\,0,\,\,1\] i.e., \[K=-1,\]is a required answer.
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