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Question

The system of equations kx+y+z=1,x+ky+z=k and x+y+zk=k2 has no solution if k is equal to

A
2
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B
1
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C
1
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D
0
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Solution

The correct option is A 2
Given system of equations
kx+y+z=1x+ky+z=kx+y+zk=k2
Now,
D=∣ ∣k111k111k∣ ∣=0k(k21)(k1)+(1k)=0(k1)(k2+k11)=0(k1)(k2+k2)(k1)(k1)(k+2)=0k=1,k=2
For k=1 equations becomes identical, so k=2 for no solution.

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