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  • Let S be the set of all  for which the system of the linear equation. <img alt="\\2x-y+2z=2\\ x-2y+\lambda z=4\\

Let S be the set of all \lambda\; \epsilon \;R for which the system of the linear equation.
\\2x-y+2z=2\\ x-2y+\lambda z=4\\ x+\lambda y+z=4 has no solution. Then the set S is
 
Option: 1 contains more than two elements.
 
Option: 2 is an empty set.  
Option: 3 is a singleton.
 
Option: 4 contains exactly two elements.

Answers (1)

best_answer

\begin{aligned} &2 x-y+2 z=2\\ &x-2 y+\lambda z=-4\\ &x+\lambda y+z=4\\ &\text { For no solution: }\\ &\mathrm{D}=\left|\begin{array}{ccc} 2 & -1 & 2 \\ 1 & -2 & \lambda \\ 1 & \lambda & 1 \end{array}\right|=0 \end{aligned}

\begin{array}{l} \Rightarrow 2\left(-2-\lambda^{2}\right)+1(1-\lambda)+2(\lambda+2)=0 \\ \Rightarrow-2 \lambda^{2}+\lambda+1=0 \\ =\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) \end{array}

Which is not equal to zeros for

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

Posted by

himanshu.meshram

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