Let S be the set of all

Question:

Let $S$ be the set of all $\lambda \in R$ for which the system of linear equations

$2 x-y+2 z=2$

$x-2 y+\lambda z=-4$

$x+\lambda y+z=4$

has no solution. Then the set $\mathrm{S}$

  1. contains more than two elements.

  2. is a singleton.

  3. contains exactly two elements.

  4. is an empty set.


Correct Option: , 3

Solution:

$2 x-y+2 z=2$

$x-2 y+\lambda z=-4$

$x+\lambda y+z=4$

For no solution :

$D=\left|\begin{array}{ccc}2 & -1 & 2 \\ 1 & -2 & \lambda \\ 1 & \lambda & 1\end{array}\right|=0$

$\Rightarrow 2\left(-2-\lambda^{2}\right)+1(1-\lambda)+2(\lambda+2)=0$

$\Rightarrow-2 \lambda^{2}+\lambda+1=0$

$\Rightarrow \lambda=1,-\frac{1}{2}$

$\mathrm{D}_{\mathrm{x}}=\left|\begin{array}{ccc}2 & -1 & 2 \\ -4 & 2 & \lambda \\ 4 & \lambda & 1\end{array}\right|=2\left|\begin{array}{ccc}1 & -1 & 2 \\ -2 & -2 & \lambda \\ \lambda & \lambda & 1\end{array}\right|$

$=2(1+\lambda)$

whichis not equal to zero for

$\lambda=1,-\frac{1}{2}$

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