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Question

Mathematics Question on Determinants

Let SS denote the set of all real values of λ\lambda such that the system of equations
λx+y+z=1\lambda x+y+z=1
x+λy+z=1x+\lambda y+z=1
x+y+λz=1x+y+\lambda z=1
is inconsistent, then λS(λ2+λ)\displaystyle \sum_{\lambda \in S}\left(|\lambda|^2+|\lambda|\right) is equal to

A

12

B

4

C

2

D

6

Answer

6

Explanation

Solution

The correct answer is (D) : 6
λ11 1λ1 11λ=0\begin{vmatrix} λ&1&1\\\ 1&λ&1\\\ 1&1&λ\end{vmatrix}=0
(λ+2)λ11 1λ1 11λ=0(λ+2)\begin{vmatrix} λ&1&1\\\ 1&λ&1\\\ 1&1&λ\end{vmatrix}=0
(λ+2)[1(λ21)1(λ1)+(1λ)]=0(λ+2)[1(λ^2−1)−1(λ−1)+(1−λ)]=0
(λ+2)[(λ22λ+1)=0(λ+2)[(λ^2−2λ+1)=0
(λ+2)(λ1)2=0λ=2,λ=1(λ+2)(λ−1)^2=0 ⇒λ=−2,λ=1
at λ=1λ=1 system has infinite solution, for inconsistent λ=2λ=-2
so (22+2)=6∑(|−2|^2+|−2|)=6