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

• A. 12
• B. 4
• C. 2
• D. 6

Answer 1

Answer: (D)

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