Solve the following system of linear equations for any values of real parameter a { x + y + 2 z = 1 2 x + a y − z = 4 3 x + y + 3 z = 1 Calculating the value of determinant I found out, after I equalled it with zero, that a has the value − 4 3 . So I thought that if it's required to solve this equation for any a, then first of all I had to suppose that D = 0, this means that a = − 4 3 . For this value I found out that there is an infinite number of solutions for this system. But now my question is: what do I have to do with the case when D ≠ 0?

Question

Solve the following system of linear equations for any values of real parameter   a      {                            x          +          y          +          2          z          =          1                                      2          x          +          a          y          −          z          =          4                                      3          x          +          y          +          3          z          =          1                        Calculating the value of determinant I found out, after I equalled it with zero, that   a has the value             −      4        3  . So I thought that if it&#039;s required to solve this equation for any   a, then first of all I had to suppose that   D  =  0, this means that   a  =            −      4        3  . For this value I found out that there is an infinite number of solutions for this system. But now my question is: what do I have to do with the case when   D  ≠  0?

trantegisis · Accepted Answer

What you have is a linear system   D  x  =  b, thus for   a  ≠  −      4    3  , the solution is given by:  x  =      D          −      1        b, and the solution will be in terms of   a.

Lucia Grimes · Accepted Answer

Why not simply using Gaussian elimination for the remaining case? You should notice that some steps are only valid if   3  a  +  4  ≠  0.Try to avoid dividing by expressions containing parameters (or any other things including parameters) as long as possible. That&#039;s why I have started by working with the first and the third row, since these two rows do not contain the parameter   a.      (                            1                          1                          2                          1                                      2                          a                          −          1                          4                                      3                          1                          3                          1                      )    ∼      (                            1                          1                          2                          1                                      2                          a                          −          1                          4                                      0                          −          2                          −          3                          −          2                      )    ∼      (                            1                          1                          2                          1                                      2                          a                          −          1                          4                                      0                          1                                      3            2                                    1                      )    ∼      (                            1                          0                                      1            2                                    0                                      2                          a                          −          1                          4                                      0                          1                                      3            2                                    1                      )    ∼      (                            1                          0                                      1            2                                    0                                      0                          a                          −          2                          4                                      0                          1                                      3            2                                    1                      )    ∼      (                            1                          0                                      1            2                                    0                                      0                          a          +                      4            3                                    0                                      16            3                                                0                          1                                      3            2                                    1                      )    ∼      (                            1                          0                                      1            2                                    0                                      0                          1                          0                                      16                          3              a              +              4                                                            0                          1                                      3            2                                    1                      )    ∼      (                            1                          0                                      1            2                                    0                                      0                          1                          0                                      16                          3              a              +              4                                                            0                          0                                      3            2                                                              3              a              −              12                                      3              a              +              4                                            )    ∼      (                            1                          0                                      1            2                                    0                                      0                          1                          0                                      16                          3              a              +              4                                                            0                          0                          1                                                    2              a              −              8                                      3              a              +              4                                            )    ∼      (                            1                          0                          0                                                    4              −              a                                      3              a              +              4                                                            0                          1                          0                                      16                          3              a              +              4                                                            0                          0                          1                                                    2              a              −              8                                      3              a              +              4                                            )

Solve the following system of linear equations for any values of real parameter a {

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