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

Question

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

Infinity Learn · Accepted Answer

The augmented matrix of the system of equations is $$k1111k1k11kk2R2$$→$$kR2$$−$$R1$$,$$R3$$→$$kR3$$−$$R1$$                                                                               ≈$$k1110k2$$−$$1k$$−$$1k2$$−$$11k$$−$$1k2$$−$$1k3$$−$$1$$                                                                              ≈k−$$12k1110k+11k+101k+1k2+k+1R3$$→$$k+1R3$$−$$R2$$                                                                              ≈k−$$12k1110k+11k+100k2+2kkk2+2k+1$$                                                                               ≈kk−$$12k1110k+11k+100k+2k2+2k+1Third$$ equation of the system of equations is $$k+2z=k+12$$   $$z=k+12k+2If$$ $$k+2=0$$, system has no solution,  it means for k $$=$$ $$-2$$ , system has no solution.

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

Detailed Solution

Get Expert Academic Guidance – Connect with a Counselor Today!