The system of equations Kx+y+z=1, x+Ky+z=K and x+y+Kz=K^{2} have no solution if K is equal to?

  • Option 1)

    0

  • Option 2)

    1

  • Option 3)

    -1

  • Option 4)

    -2

 

Answers (1)

\begin{vmatrix} k&1&1\\1&k&1\\1&1&k \end{vmatrix}=0       for no solution 

 

K(K2-1)-1(K-1)+1(1-K)=0

K(K-1)(K+1)-2(K-1)=0

(K-1)(K2+K-2)=0

(K-1) (K+2) (K-1) =0

(K-1)^{2}\neq 0\: So\:K\neq 1\\ \therefore K=-2 


Option 1)

0

This solution is incorrect 

Option 2)

1

This solution is incorrect 

Option 3)

-1

This solution is incorrect 

Option 4)

-2

This solution is correct 

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