NDA2017MathematicsDeterminantsActual
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
- A0
- B1
- C-1
- 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.