| \[x+y+z=6\] |
| \[4x+\lambda y\lambda z=\lambda 2\] |
| \[3x+2y4z=5\] |
| has infinitely many solutions. Then\[\lambda \]is a root of the quadratic equation. |
A) \[{{\lambda }^{2}}-3\lambda -4=0\]
B) \[{{\lambda }^{2}}-\lambda -6=0\]
C) \[{{\lambda }^{2}}+3\lambda -4=0\]
D) \[{{\lambda }^{2}}+\lambda -6=0\]
Correct Answer: B
Solution :
D = 0 \[\left| \begin{matrix} 1 & 1 & 1 \\ 4 & \lambda & -\lambda \\ 3 & 2 & -4 \\ \end{matrix} \right|=0\Rightarrow \lambda =3\]
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