S denote the set of all real values of lambda | vedclass

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Question

$\left|\begin{array}{lll}\lambda & 1 & 1 \ 1 & \lambda & 1 \ 1 & 1 & \lambda\end{array}ight|=0$

$(\lambda+2)\left|\begi

Vedclass · Accepted Answer

$6$

Vedclass · Answer

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

$S$ 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

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