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Question

If the system of equations xαyαz=0,βxy+βz=0,γx+γyz=0, where α,β,γ1 have only non-trivial solutions, 11+α+11+β+11+γ=

A
2
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B
1
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C
1
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D
2
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Solution

The correct option is C 2
We have,
xα(y+z)=0(1+α)xα(x+y+z)=0 ...(1)
yβ(x+z)=0(1+β)yβ(x+y+z)=0 ...(2)
zγ(x+y)=0(1+γ)zγ(x+y+z)=0 ...(3)
Since the given system of equation have only non-trivial solutions, (x+y+z)0 [ if x+y+z=0, then in view of (1),(2) and (3), we have x=0,y=0,z=0, a trivial solution]
α1+α=xx+y+z,β1+β=yx+y+z
and, γ1+γ=zx+y+z
On adding, we get α1+α+β1+β+γ1+γ=1
(1α1+α)+(1β1+β)+(1γ1+γ)=31
11+α+11+β+11+γ=2.

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