For what values of k does this system of equations have a unique solution? { y + 2 k z = 0 x + 2 y + 6 z = 2 k x + 2 z = 1 I have [ 1 2 6 2 0 1 2 k 0 k 0 2 1 ] When I row reduce, I get: [ 1 0 6 − 4 k 2 0 1 2 k 0 0 0 2 − 6 k − 4 k 2 1 − 2 k ]

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For what values of k does this system of equations have a unique solution?      {                            y          +          2          k          z          =          0                                      x          +          2          y          +          6          z          =          2                                      k          x          +          2          z          =          1                        I have       [                            1                          2                          6                          2                                      0                          1                          2          k                          0                                      k                          0                          2                          1                      ]  When I row reduce, I get:      [                            1                          0                          6          −          4          k                          2                                      0                          1                          2          k                          0                                      0                          0                          2          −          6          k          −          4                      k            2                                    1          −          2          k                      ]

Lamont Adkins · Accepted Answer

In that case, recall that your system will be inconsistent if, after row reduction, you have a row of the form   (  0     0     0  ∣  1  ) since this row would correspond to the equation   0  x  +  0  y  +  0  z  =  1 which clearly has no solutions.On the other hand, you also don&#039;t want a row of of the form   (  0     0     0  ∣  0  ), which would give you a free variable and hence infinitely many solutions.Thus, you should find the values of   k for which   2  −  6  k  −  4      k    2    =  0. By our discussion, we can see that as long as you avoid those values of   k, your system will have a solution, and this solution will be unique.You can find these &quot;bad&quot; values of   k by methods from high school algebra, e.g. the quadratic formula.

Cory Patrick · Accepted Answer

Now the last row could be rewritten as      (                            1                          0                          6          −          4          k                          2                                      0                          1                          2          k                          0                                      0                          0                          2          (          2          k          −          1          )          (          k          −          1          )                          −          (          2          k          −          1          )                      )  Now we have the following cases1.   k  =  1: the system has no solution.      (                            1                          0                          2                          2                                      0                          1                          2                          0                                      0                          0                          0                          −          1                      )  2.   k  =  1  /  2: the system has infinitely many solution.      (                            1                          0                          4                          2                                      0                          1                          1                          0                                      0                          0                          0                          0                      )  3.   k  ∈  R  ,  k  ≠  1  ,  1  /  2: the system has unique solution.

For what values of k does this system of equations have a unique solution? { <mtable co

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