A) \[5\]
B) \[-2/3\]
C) \[-3\]
D) None of these
Correct Answer: D
Solution :
[d] Let \[\Delta =\left| \begin{matrix} \lambda & 1 & 1 \\ 1 & \lambda & 1 \\ 1 & 1 & \lambda \\ \end{matrix} \right|=\left| \begin{matrix} \lambda +2 & 1 & 1 \\ \lambda +2 & \lambda & 1 \\ \lambda +2 & 1 & \lambda \\ \end{matrix} \right|\] \[[{{C}_{1}}\to {{C}_{1}}+{{C}_{2}}+{{C}_{3}}]\] \[=(\lambda +2)\left| \begin{matrix} 1 & 1 & 1 \\ 1 & \lambda & 1 \\ 1 & 1 & \lambda \\ \end{matrix} \right|=(\lambda +2)\left| \begin{matrix} 1 & 0 & 0 \\ 1 & \lambda -1 & 0 \\ 1 & 0 & \lambda -1 \\ \end{matrix} \right|\] \[=(\lambda +2){{(\lambda -1)}^{2}}\] [using \[{{C}_{2}}\to {{C}_{2}}-{{C}_{1}}\] and \[{{C}_{3}}\to {{C}_{3}}-{{C}_{1}}\]] If \[\Delta =0\], then \[\lambda =-2\] or \[\lambda =1\]. But when \[\lambda =1\], the system of equation becomes \[{{x}_{1}}+{{x}_{2}}+{{x}_{3}}=1\] which has infinite number of solutions. When \[\lambda =-2\], by adding three equations, we obtain \[0=3\] and thus, the system of equations is inconsistent.
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