Determinants Question 7
Question 7 - 01 February - Shift 1
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 $\sum _{\lambda \in S}(|\lambda|^{2}+|\lambda|)$ is equal to
(1) 2
(2) 12
(3) 4
(4) 6
Show Answer
Answer: (4)
Solution:
Formula: System of equations with 3 variables, consistency of solutions: inconsistent system
$ \begin{vmatrix} \lambda & 1 & 1 \\ 1 & \lambda & 1 \\ 1 & 1 & \lambda\end{vmatrix} =0$
$(\lambda+2) \begin{vmatrix} 1 & 1 & 1 \\ 1 & \lambda & 1 \\ 1 & 1 & \lambda\end{vmatrix} =0$
$(\lambda+2)[1(\lambda^{2}-1)-1(\lambda-1)+(1-\lambda)]=0$
$(\lambda+2)[(\lambda^{2}-2 \lambda+1)=0.$
$(\lambda+2)(\lambda-1)^{2}=0 \Rightarrow \lambda=-2, \lambda=1$
at $\lambda=1$ system has infinite solution, for inconsistent $\lambda=-2$
So $\sum _{\lambda \in S}(|\lambda|^{2}+|\lambda|)=6$
