A) - 1
B) 0
C) 1
D) None of these
Correct Answer: A
Solution :
The system will have a non-zero solution, if \[\Delta \equiv \left| \,\begin{matrix} {{a}^{3}} & {{(a+1)}^{3}} & {{(a+2)}^{3}} \\ a & a+1 & a+2 \\ 1 & 1 & 1 \\ \end{matrix}\, \right|=0\] \[\Rightarrow \left| \,\begin{matrix} {{a}^{3}} & 3{{a}^{2}}+3a+1 & 3{{(a+1)}^{2}}+3(a+1)+1 \\ {{a}^{2}} & 1 & 1 \\ 1 & 0 & 0 \\ \end{matrix}\, \right|=0\]by \[\begin{align} & {{C}_{2}}\to {{C}_{2}}-{{C}_{1}} \\ & {{C}_{3}}\to {{C}_{3}}-{{C}_{2}} \\ \end{align}\] Þ \[3{{a}^{2}}+3a+1-\{3{{(a+1)}^{2}}+3(a+1)+1\}\](expanding along \[{{R}_{3}}\]) Þ \[-6(a+1)=0\Rightarrow a=-1\].
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