A) \[k\ne 0\]
B) \[-1<k<1\]
C) \[-2<k<2\]
D) \[k=0\]
Correct Answer: A
Solution :
Since, system \[x+y+z=2\] \[2x+y-z=3\] \[3x+2y+kz=4\] has a unique solution. \[\therefore \] \[D\ne 0\] \[\Rightarrow \] \[\left| \begin{matrix} 1 & 1 & 1 \\ 2 & 1 & -1 \\ 3 & 2 & 5 \\ \end{matrix} \right|\ne 0\] Applying, \[{{R}_{2}}\to {{R}_{2}}-2{{R}_{1}},{{R}_{3}}\to {{R}_{3}}-3{{R}_{1}}\] \[\left| \begin{matrix} 1 & 1 & 1 \\ 0 & -1 & -3 \\ 0 & -1 & k-3 \\ \end{matrix} \right|\ne 0\] \[\Rightarrow \] \[-(k-3)-3\ne 0\] \[\Rightarrow \] \[-k+3-3\ne 0\] \[\Rightarrow \] \[k\ne 0\] Alternate Method The augmented matrix of non-homogeneous system is \[A:B=\left| \begin{matrix} 1 & 1 & 1: & 2 \\ 2 & 1 & -1: & 3 \\ 3 & 2 & k: & 4 \\ \end{matrix} \right|\] Applying\[{{R}_{3}}\to {{R}_{3}}-3{{R}_{1}},{{R}_{2}}\to {{R}_{2}}-2{{R}_{1}}\] \[\left[ \begin{matrix} 1 & 1 & 1 & :2 \\ 0 & -1 & -3 & :-1 \\ 0 & -1 & k-3 & :-2 \\ \end{matrix} \right]\] Applying\[{{R}_{3}}\to {{R}_{3}}-{{R}_{2}}\] \[\left[ \begin{matrix} 1 & 1 & 1 & :2 \\ 0 & -1 & -3 & :-1 \\ 0 & 0 & k & :-1 \\ \end{matrix} \right]\] Here, Rank of \[A:B=3\] and Rank of [A] depends on 'k' If\[k=0,\]then Rank of\[[A]=2\] If\[k\ne 0,\]then Rank of\[[A]=3\] But given the system have unique solution. For this system must be consistent and the condition of consistent is Rank of A:B = Rank of [A] For this,\[(k\ne 0).\] Then system has, unique solution.
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