Question

The value of $\lambda $ for which the system of equations $2x-y-z=12$ , $x-2y+z=-4$ , $x+y+\lambda x=4$ has no solution is
(A) $3$
(B) $-3$
(C) $2$
(D) $-2$

Answer 1

Hint: In this question we have been asked to find the value of $\lambda $ for which the given system of equations $2x-y-z=12$ , $x-2y+z=-4$ , $x+y+\lambda x=4$ has no solution. From the basic concepts we will use the concept of solving system of equations which says that there are 3 possible solutions having an unique solution given by $x=\dfrac{{{\Delta }_{x}}}{\Delta },y=\dfrac{{{\Delta }_{y}}}{\Delta },z=\dfrac{{{\Delta }_{z}}}{\Delta }$, having infinite dependent solutions and no solution when $\Delta =0$ .

Complete step-by-step solution:
Now considering from the questions we have been asked to find the value of $\lambda $ for which the given system of equations $2x-y-z=12$ , $x-2y+z=-4$ , $x+y+\lambda x=4$ has no solution.
From the basic concepts we will use the concept of solving system of equations which says that there are 3 possible solutions having an unique solution given by $x=\dfrac{{{\Delta }_{x}}}{\Delta } , y=\dfrac{{{\Delta }_{y}}}{\Delta } , z=\dfrac{{{\Delta }_{z}}}{\Delta }$, having infinite dependent solutions and no solution when $\Delta =0$ .
We can say that $\Delta =\left| \begin{matrix}
   2 & -1 & -1 \\
   1 & -2 & 1 \\
   \lambda +1 & 1 & 0 \\
\end{matrix} \right|$ , ${{\Delta }_{x}}=\left| \begin{matrix}
   12 & -1 & -1 \\
   -4 & -2 & 1 \\
   4 & 1 & 0 \\
\end{matrix} \right|$ , ${{\Delta }_{y}}=\left| \begin{matrix}
   2 & 12 & -1 \\
   1 & -4 & 1 \\
   \lambda +1 & 4 & 0 \\
\end{matrix} \right|$ and ${{\Delta }_{z}}=\left| \begin{matrix}
   2 & -1 & 12 \\
   1 & -2 & -4 \\
   \lambda +1 & 1 & 4 \\
\end{matrix} \right|$
As it is given that the system of equations has no solutions we can say that $\Delta =0$ .
By using this we can say that
$\begin{align}
  & \Delta =\left| \begin{matrix}
   2 & -1 & -1 \\
   1 & -2 & 1 \\
   \lambda +1 & 1 & 0 \\
\end{matrix} \right|\Rightarrow 2\left| \begin{matrix}
   -2 & 1 \\
   1 & 0 \\
\end{matrix} \right|-\left( -1 \right)\left| \begin{matrix}
   1 & 1 \\
   \lambda +1 & 0 \\
\end{matrix} \right|+\left( -1 \right)\left| \begin{matrix}
   1 & -2 \\
   \lambda +1 & 1 \\
\end{matrix} \right| \\
 & \Rightarrow 2\left( 0-1 \right)+\left( 0-\lambda -1 \right)-\left( 1-\left( -2 \right)\left( \lambda +1 \right) \right)=0 \\
 & \Rightarrow -2-\lambda -1-3-2\lambda =0 \\
 & \Rightarrow -6-3\lambda =0 \\
 & \Rightarrow 3\lambda =-6 \\
 & \Rightarrow \lambda =-2 \\
\end{align}$
Hence we can conclude that the value of $\lambda $ is given as $-2$ for which the given system of equations $2x-y-z=12$ , $x-2y+z=-4$ , $x+y+\lambda x=4$ has no solution.
Hence we will mark the option “D” as correct.

Note: While answering questions of this type we should be sure with the concepts that we are going to apply and calculations that we are going to perform in between the steps. If someone has made a calculation mistake unintentionally then they will end up having a wrong conclusion.