Question

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 infinitely many solutions. Then $\lambda$ is a root of the quadratic equation :

• A. ${\lambda }^2+3\lambda -4=0$
• B. ${\lambda }^2-3\lambda -4=0$
• C. ${\lambda }^2+\lambda -6=0$
• D. ${\lambda }^2-\lambda -6=0$

Answer 1

The correct option is D ${\lambda }^2-\lambda -6=0$

Given,
$x+y+z=6...\left(1\right)$
$4x+\lambda y-\lambda z=\lambda -2...\left(2\right)$
$3x+2y-4z=-5...\left(3\right)$

has infinitely many solutions, then
$\Delta =0$
$\left|\begin{array}{ccc}1& 1& 1\\ 4& \lambda & -\lambda \\ 3& 2& -4\end{array}\right|=0$
$\Rightarrow -8\lambda +24=0\Rightarrow \lambda =3$