A) Infinite solution
B) No solution
C) Unique solution
D) None of these
Correct Answer: C
Solution :
We have, \[{{x}_{1}}+2{{x}_{2}}+3{{x}_{3}}=c\] \[2a{{x}_{1}}+3{{x}_{2}}+{{x}_{3}}=c\] \[3b{{x}_{1}}+{{x}_{2}}+2{{x}_{3}}=c\] Let\[a=b=c=1\]. Then \[D=\left| \,\begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 3 & 1 & 2 \\ \end{matrix}\, \right|\] = \[1\,(5)-2\,(1)+3\,(-7)=-18\ne 0\] \[{{D}_{x}}=\left| \,\begin{matrix} 1 & 2 & 3 \\ 1 & 3 & 1 \\ 1 & 1 & 2 \\ \end{matrix}\, \right|=-3\] Similarly\[{{D}_{y}}={{D}_{z}}=-3\]. Now,\[\Delta =(2+i)\,\left| \,\begin{matrix} 1 & 1 & i \\ 1 & 1+2i & 1+i \\ 1 & 2 & 1-i \\ \end{matrix}\, \right|\,\] \[y=z=\frac{1}{6}\] Hence\[D\ne 0\], \[x=y=z\], i.e., unique solution.
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