A) \[k=0\]
B) \[-1<k<1\]
C) \[-2<k<2\]
D) \[k\ne 0\]
Correct Answer: D
Solution :
[d] The given system of equations is \[x+y+z=2\] ... (i) \[2x+y-z=3\] ... (ii) and \[3x+2y+kz=4\] ... (iii) This system has a unique solution if \[\left| \begin{matrix} 1 & 1 & 1 \\ 2 & 1 & -1 \\ 3 & 2 & k \\ \end{matrix} \right|\ne 0\] Applying \[{{C}_{2}}\to {{C}_{2}}-{{C}_{1}}\] and \[{{C}_{3}}\to {{C}_{3}}-{{C}_{1}}\] We get \[\left| \begin{matrix} 1 & 0 & 0 \\ 3 & -1 & -3 \\ 3 & -1 & k-3 \\ \end{matrix} \right|\ne 0\] \[\Rightarrow \,\,\,-1(k-3)-3\ne 0\] or \[-k+3-3\ne 0\Rightarrow k\ne 0\]
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