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Question

The system of equation αx+y+z=α1,x+αy+z=α1,x+y+αz=α1 has no solution if α is

A
either 2 or 1
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B
2
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C
1
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D
2
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Solution

The correct option is B 2
Here, D=∣ ∣α111α111α∣ ∣
D=(α1)2(α+2)
Taking D=0
(α1)2(α+2)=0
α=1,α=2
At these values of α, system can have either no solution or infinitely many solution.
For α=1, equations takes the form x+y+z=0
Hence, infinitely many solution for α=1.
For α=2,
D1=∣ ∣311321312∣ ∣
D1=270
So, D=0, at least one D10
Hence, system has no solution for α=2

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