JEE MAIN - Mathematics (2023 - 1st February Morning Shift - No. 5)
Let $$S$$ denote the set of all real values of $$\lambda$$ such that the system of equations
$$\lambda x+y+z=1$$
$$x+\lambda y+z=1$$
$$x+y+\lambda z=1$$
is inconsistent, then $$\sum_\limits{\lambda \in S}\left(|\lambda|^{2}+|\lambda|\right)$$ is equal to
12
2
4
6
Explanation
$\left|\begin{array}{lll}\lambda & 1 & 1 \\ 1 & \lambda & 1 \\ 1 & 1 & \lambda\end{array}\right|=0$
$$ \begin{aligned} & \lambda\left(\lambda^{2}-1\right)-1(\lambda-1)+1(1-\lambda)=0 \\\\ & \Rightarrow \lambda^{3}-\lambda-\lambda+1+1-\lambda=0 \\\\ & \Rightarrow \lambda^{3}-3 \lambda+2=0 \\\\ & \Rightarrow (\lambda-1)\left(\lambda^{2}+\lambda-2\right)=0 \end{aligned} $$
$\Rightarrow$ $\lambda=1,-2$
For $\lambda=1 \Rightarrow \infty$ solution
$\lambda=-2 \Rightarrow$ no solution
$\sum\limits_{\lambda \in S}|\lambda|^{2}+|\lambda|=6$
$$ \begin{aligned} & \lambda\left(\lambda^{2}-1\right)-1(\lambda-1)+1(1-\lambda)=0 \\\\ & \Rightarrow \lambda^{3}-\lambda-\lambda+1+1-\lambda=0 \\\\ & \Rightarrow \lambda^{3}-3 \lambda+2=0 \\\\ & \Rightarrow (\lambda-1)\left(\lambda^{2}+\lambda-2\right)=0 \end{aligned} $$
$\Rightarrow$ $\lambda=1,-2$
For $\lambda=1 \Rightarrow \infty$ solution
$\lambda=-2 \Rightarrow$ no solution
$\sum\limits_{\lambda \in S}|\lambda|^{2}+|\lambda|=6$
Comments (0)
