Question

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 $\displaystyle \sum_{\lambda \in S}\left(|\lambda|^2+|\lambda|\right)$ is equal to

• A. 12
• B. 4
• C. 2
• D. 6

Answer 1

Answer: (D)

$\begin{vmatrix}\lambda & 1 & 1 \\ 1 & \lambda & 1 \\ 1 & 1 & \lambda\end{vmatrix}=0$
$(\lambda+2)\begin{vmatrix}1 & 1 & 1 \\ 1 & \lambda & 1 \\ 1 & 1 & \lambda\end{vmatrix}=0$
$ (\lambda+2)\left[1\left(\lambda^2-1\right)-1(\lambda-1)+(1-\lambda)\right]=0 $
$(\lambda+2)\left[\left(\lambda^2-2 \lambda+1\right)=0\right.$
$ (\lambda+2)(\lambda-1)^2=0 \Rightarrow \lambda=-2, \lambda=1$
at $\lambda=1$ system has infinite solution, for inconsistent $\lambda=-2$
so $\sum\left(|-2|^2+|-2|\right)=6$