Question: For the value(s) of k, if any, will the following system have (a) no solution, (b) a unique solution, (c) infinity many solutons: x + y + k z = 1 , x + k y + z = 1 , x k + y + z = − 2.

Question

Question: For the value(s) of   k, if any, will the following system have (a) no solution, (b) a unique solution, (c) infinity many solutons:  x  +  y  +  k  z  =  1  ,    x  +  k  y  +  z  =  1  ,    x  k  +  y  +  z  =  −  2.

lywiau63 · Accepted Answer

Let&#039;s calculate this determinant  det      (                            1                          1                          k                                      1                          k                          1                                      k                          1                          1                      )    =  det      (                            k          +          2                          1                          k                                      k          +          2                          k                          1                                      k          +          2                          1                          1                      )    =      (                            k          +          2                          1                          k                                      0                          k          −          1                          1          −          k                                      0                          0                          1          −          k                      )    =  −  (  k  +  2  )  (  k  −  1      )    2  hence- If   k  ≠  1 and   k  ≠  −  2 there&#039;s exactly one solution.- If   k  =  1 we see easily that the   3 equations aren&#039;t compatible so there is no solution.- If   k  =  −  2 we can see also easily that there&#039;s infinitely many solutions.

Frank Day · Accepted Answer

Just add. This gives   (  k  +  1  )  (  x  +  y  +  z  )  =  0. If   k  =  −  1, then there are an infinite number of solutions.Then if   k  =  0... and so on.