Find the values of k for which the simultaneous equations do not have a unique solution for x , y and z.Also show that when k = − 2 the equations are inconsistent k x + 2 y + z = 0 3 x + 0 y − 2 z = 4 3 x − 6 k y − 4 z = 14Using a determinant and setting to zero, then solving the quadratic gives | k 2 1 3 0 − 2 3 − 6 k − 4 | = 0 k = − 2 , k = 1 2 So far so good, but when subbing − 2 for k − 2 x + 2 y + z = 0 . . . e q n 1 3 x + 0 y − 2 z = 4 . . . e q n 2 3 x + 12 y − 4 z = 14 . . . e q n 3subbing eqn 2 from eqn 3 gives 12 y − 2 z = 10 ⇒ y = 5 + z 6 Subbing for y into eqn 1 − 2 x + 5 + z 3 − z = 0 ⇒ x = 5 − 2 z 6 Subbing for y and x into eqn 3 5 − 2 z 2 + 10 − 2 z = 14 ⇒ z = − 1 2 , x = 1 a n d y = 3 4 These values for x , y and z seem to prove unique solutions for these equations, yet from the determinant and also the question in the text book should they not be inconsistent?

Question

Find the values of   k for which the simultaneous equations do not have a unique solution for   x  ,  y and   z.Also show that when   k  =  −  2 the equations are inconsistent  k  x  +  2  y  +  z  =  0  3  x  +  0  y  −  2  z  =  4  3  x  −  6  k  y  −  4  z  =  14Using a determinant and setting to zero, then solving the quadratic gives      |                            k                          2                          1                                      3                          0                          −          2                                      3                          −          6          k                          −          4                      |    =  0  k  =  −  2  ,  k  =      1    2  So far so good, but when subbing   −  2 for   k  −  2  x  +  2  y  +  z  =  0    .  .  .    e  q  n    1  3  x  +  0  y  −  2  z  =  4    .  .  .    e  q  n    2  3  x  +  12  y  −  4  z  =  14    .  .  .    e  q  n    3subbing eqn   2 from eqn   3 gives  12  y  −  2  z  =  10  ⇒  y  =            5      +      z        6  Subbing for   y into eqn   1  −  2  x  +            5      +      z        3    −  z  =  0  ⇒  x  =            5      −      2      z        6  Subbing for   y and   x into eqn   3            5      −      2      z        2    +  10  −  2  z  =  14  ⇒  z  =            −      1        2    ,    x  =  1    a  n  d    y  =      3    4  These values for   x  ,  y and   z seem to prove unique solutions for these equations, yet from the determinant and also the question in the text book should they not be inconsistent?

karburitc · Accepted Answer

Firstly, a non-zero determinant doesn&#039;t imply there are no solutions. The system of equations      [                            1                          1                          1                          1                                      2                          2                          2                          2                                      3                          3                          3                          3                      ]  has an infinite number of solutions, for example, and the relevant determinant will be zero. So, we can&#039;t be sure the system will be inconsistent from a zero determinant alone.The matrix      [                            −          2                          2                          1                          0                                      3                          0                          −          2                          4                                      3                          12                          −          4                          14                      ]  has row echelon form                                            ∼                                    R              3                        ←                          R              3                        −                          R              2                                                [                                                    −                2                                            2                                            1                                            0                                                                    3                                            0                                            −                2                                            4                                                                    0                                            12                                            −                2                                            10                                              ]                                                            ∼                                    R              2                        ←                          R              2                        +                          3              2                                      R              1                                                [                                                    −                2                                            2                                            1                                            0                                                                    0                                            3                                            −                0.5                                            4                                                                    0                                            12                                            −                2                                            10                                              ]                                                            ∼                                    R              3                        ←                          R              3                        −            4                          R              2                                                [                                                    −                2                                            2                                            1                                            0                                                                    0                                            3                                            −                0.5                                            4                                                                    0                                            0                                            0                                            −                6                                              ]                    which is inconsistent. So, no solutions exist.

Find the values of k for which the simultaneous equations do not have a unique solution for x

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