Let λ be a real number for which system of linear equations x + y + z = 6 ; 4x + λ y -

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3 and 4 .Determinants and Matrices

hard

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

A

${\lambda ^2} - \lambda - 6\, = 0$

B

${\lambda ^2} - 3\lambda - 4 = 0$

C

${\lambda ^2} + 3\lambda - 4 = 0$

D

${\lambda ^2} + \lambda - 6 = 0$

(JEE MAIN-2019)

Solution

$D=0$

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

Standard 12

Mathematics

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(JEE MAIN-2017)