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

Question

Accepted Answer

"$\left|\begin{array}{lll}\lambda & 1 & 1 \\ 1 & \lambda & 1 \\ 1 & 1 & \lambda\end{array}\right|=0$

$$ \begin{aligned} & \lambda\left(\lambda^{2}-1\right)-1(\lambda-1)+1(1-\lambda)=0 \\\\ & \Rightarrow \lambda^{3}-\lambda-\lambda+1+1-\lambda=0 \\\\ & \Rightarrow \lambda^{3}-3 \lambda+2=0 \\\\ & \Rightarrow (\lambda-1)\left(\lambda^{2}+\lambda-2\right)=0 \end{aligned} $$

$\Rightarrow$ $\lambda=1,-2$

For $\lambda=1 \Rightarrow \infty$ solution

$\lambda=-2 \Rightarrow$ no solution

$\sum\limits_{\lambda \in S}|\lambda|^{2}+|\lambda|=6$"

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

Solution

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

Solution

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More Matrices and Determinants questions

Drill 25 more like these. Every day. Free.

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