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The number of real values of $\lambda$ , such that the system of linear equations

$\mathrm{2 x-3 y+5 z=9} \\$

$\mathrm{x+3 y-z=-18} \\$

$\mathrm{3 x-y+\left(\lambda^{2}-|\lambda|\right) z=16 }$

has no solutions, is
Option: 1
$0$

Option: 2
$1$

Option: 3
$2$

Option: 4
$4$

Answers (1)

best_answer

For no solution

$\mathrm{\Delta=0} \\$

$\mathrm{\left|\begin{array}{ccc} 2 & -3 & 5 \\ 1 & 3 & -1 \\ 3 & -1 & \lambda^{2}-1 \lambda \mid \end{array}\right|=0 }$

$\mathrm{\Rightarrow 2\left(3 \lambda^{2}-3|\lambda|-1\right)+3\left(\lambda^{2}-|\lambda|+3\right)+5(-1-9)=0} \\$

$\mathrm{\Rightarrow 9 \lambda^{2}-9|\lambda|-43=0} \\$

$\mathrm{ \text { As } f(0)<0}$

$\Rightarrow \left |\lambda \right|$ has a positive ad a negative root,which gives z real values of $\lambda$ .

Hence correct option is 3

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