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

The Correct Option is A

$2x-3y+5z=9$
$x+3y-z=-18$
$3x-y+\left({\lambda }^2-|\lambda |\right)z=16$
$D=\left|\begin{array}{ccc}2& -3& 5\\ 1& 3& -1\\ 3& -1& {\lambda }^2-\mid \lambda \end{array}\right|=0$
$\Rightarrow 3{\lambda }^2-3|\lambda |-11=0$
Clearly one negative and one positive root since $|\lambda |$ is there so negative not possible and two values of $\lambda$ corresponding to positive value
$D_3=\left|\begin{array}{ccc}2& -3& 9\\ 1& 3& -18\\ 3& -1& 16\end{array}\right|\ne 0$ so no solution.