### 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 $\sum _{\lambda \in S}(|\lambda|^{2}+|\lambda|)=6$