MCQ
Let $\lambda $ be a real number for which the system of linear equations $x + y + z = 6$
 ; $4x + \lambda y - \lambda z = \lambda - 2$ ; $3x + 2y -4z = -5$ Has indefinitely many solutions. Then $\lambda $ is a root of the quadratic equation
  • ${\lambda ^2} - \lambda  - 6\, = 0$
  • B
    ${\lambda ^2} - 3\lambda  - 4 = 0$
  • C
    ${\lambda ^2} + 3\lambda  - 4 = 0$
  • D
    ${\lambda ^2} + \lambda  - 6 = 0$

Answer

Correct option: A.
${\lambda ^2} - \lambda  - 6\, = 0$
a
$D=0$

$\left| {\begin{array}{*{20}{c}}
1&1&1\\
4&\lambda &\lambda \\
3&2&{ - 4}
\end{array}} \right| = 0 \Rightarrow \lambda  = 3$

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