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The system of equations k x+y+z=1, x+k y+z=k and x+y+k z=k^2 has no solution if k equals

Options

  1. A0
  2. B1
  3. C-1
  4. D-2

Correct answer

D. -2

Step-by-step solution

aligned & k x+y+z=1 & x+k y+z=k & x+y+k z=k^2 aligned These equations will have no solution of | array lll k & 1 & 1 1 & k & 1 1 & 1 & k array |=0 . aligned & k ( k ^2-1 )-1( k -1)+(1- k )=0 & k ( k +1)( k -1)-1( k -1)-( k -1)=0 & ( k -1)[ k ( k +1)-1-1]=0 aligned aligned & ( k -1) ( k ^2+ k -2 )=0 & k =1 or -2 aligned For k =1 , all equations are same and have infinite solution. So, for k =-2 , equations have no solution.

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