A) \[\lambda \ne -2\]
B) \[\lambda \ne 2\]
C) \[\lambda \ne 3\]
D) \[\lambda \ne -3\]
Correct Answer: A
Solution :
The given matrix \[A=\left[ \begin{matrix} 2 & \lambda & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \\ \end{matrix} \right]\] is nonsingular, if \[|A|\,\ne 0\] \[|A|\,\,=\left| \,\begin{matrix} 2 & \lambda & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \\ \end{matrix}\, \right|\,\]=\[\left| \,\begin{matrix} 1 & \lambda +3 & 0 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \\ \end{matrix}\, \right|\,\],\[[{{R}_{1}}\to {{R}_{2}}+{{R}_{1}}]\] = \[\left| \,\begin{matrix} 1 & \lambda +3 & 0 \\ 0 & 1 & 1 \\ 0 & -\lambda -5 & -3 \\ \end{matrix}\, \right|\] \[\left[ \begin{matrix} {{R}_{2}}\to {{R}_{2}}+{{R}_{3}} \\ {{R}_{3}}\to {{R}_{3}}-{{R}_{1}} \\ \end{matrix} \right]\] = \[1\,(-3+\lambda +5)\ne 0\] \[\Rightarrow \lambda +2\ne 0\]\[\Rightarrow \lambda \,\,\ne -2.\]
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