Question

The number of real values $\lambda$, such that the system of linear equations
$2 x-3 y+5 z=9 $
$ x+3 y-z=-18 $
$ 3 x-y+\left(\lambda^2-|\lambda|\right) z=16$
has no solution, is :-

• A. 0
• B. 1
• C. 2
• D. 4

Answer 1

Answer: (C)

$ 2 x-3 y+5 z=9 $
$ x+3 y-z=-18$
$ 3 x-y+\left(\lambda^2-|\lambda|\right) z=16$
$ D = \begin{vmatrix} 2 & -3 & 5 \\1 & 3 & -1 \\3 & -1 & \lambda^2-\mid \lambda\end{vmatrix}=0$
$ \Rightarrow 3 \lambda^2-3|\lambda|-11=0$
Clearly one negative and one positive root since $|\lambda|$ is there so negative not possible and two values of $\lambda$ corresponding to positive value
$D _3= \begin{vmatrix} 2 & -3 & 9 \\1 & 3 & -18 \\3 & -1 & 16\end{vmatrix} \neq 0 $ so no solution.