The number of real values lambda, such that t | vedclass

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Question

$\Delta=\left|\begin{array}{ccc}2 & -3 & 5 \ 1 & 3 & -1 \ 3 & -1 & \lambda^{2}-|\lambda|\end{array}ight|=2\left(3 \lambda^

Vedclass · Accepted Answer

$2$

Vedclass · Answer

$\Delta=\left|\begin{array}{ccc}2 & -3 & 5 \ 1 & 3 & -1 \ 3 & -1 & \lambda^{2}-|\lambda|\end{array}ight|=2\left(3 \lambda^{2}-3|\lambda|-1ight)$

$+3\left(\lambda^{2}-|\lambda|+3ight)$

$+5(-1-9)$

$=9 \lambda^{2}-9|\lambda|-43$

$=9|\lambda|^{2}-9|\lambda|-43$

$\Delta=0$ for 2 values of $|\lambda|$ out of which one is $-ve$ and other is $+ve$

So,$2$ values of $\lambda$ satisfy the system of equations to obtain no solution

The number of real values $\lambda$, such that the system of linear equations $2 x-3 y+5 z=9$ ; $x+3 y-z=-18$ ; $3 x-y+\left(\lambda^{2}-1 \lambda \mid\right) z=16$ has no solution, is :-

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