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Question

The system of equations

kx+y+z=1x+ky+z=kx+y+kz=k2
have no solution,if k equals ?

A
0
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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 D -2
Δ=∣ ∣k111k111k∣ ∣
Δ=(k1)2(k+2)
Taking Δ=0
(k1)2(k+2)=0
k=1,k=2
At these values of k, system can have either no solution or infinitely many solution.
For k=1, equations takes the form x+y+z=1
Hence, infinitely many solution for k=1.
For k=-2,
D1=∣ ∣111221412∣ ∣
D1=90
So, D=0, at least one D10
Hence, system has no solution for k=-2

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