I have the following linear system: x + y + 2 z + 2 = 0 3 x − y + 14 z − 6 = 0 x + 2 y + 5 = 0 I immediately noticed that there was no z term in the last equation and thus determined that I will end with 0 z = some number number and therefore, came to the conclusion that there is no solution to the linear system. To see if I was right, I checked with echelon form and that also suggested that there was no solution. L 1 : x + y + 2 z + 2 = 0 L 2 : 3 x − y + 14 z − 6 = 0 L 3 : x + 2 y + 5 = 0 Then − 3 L 1 + L 2 → L 2 x + y + 2 z + 2 = 0 0 x + 3 y − 42 z − 18 = 0 x + 2 y + 0 z + 5 = 0 Then − L 1 + L 3 → L 3 x + y + 2 z + 2 = 0 0 x + 3 y − 42 z − 18 = 0 0 x + 1 y + 0 z + 3 = 0 Then − L 2 / 3 + L 3 → L 3 x + y + 2 z + 2 = 0 0 x + 3 y − 42 z − 18 = 0 0 x + 0 y + 0 z + 3 = 0 Firstly, is my answer correct?

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I have the following linear system:                          x        +        y        +        2        z        +        2        =        0                                  3        x        −        y        +        14        z        −        6        =        0                                  x        +        2        y        +        5        =        0            I immediately noticed that there was no   z term in the last equation and thus determined that I will end with   0  z  =  some number number and therefore, came to the conclusion that there is no solution to the linear system. To see if I was right, I checked with echelon form and that also suggested that there was no solution.                                    L          1                :        x        +        y        +        2        z        +        2        =        0                                            L          2                :        3        x        −        y        +        14        z        −        6        =        0                                            L          3                :        x        +        2        y        +        5        =        0            Then   −  3      L    1    +      L    2    →      L    2                            x        +        y        +        2        z        +        2        =        0                                  0        x        +        3        y        −        42        z        −        18        =        0                                  x        +        2        y        +        0        z        +        5        =        0            Then   −      L    1    +      L    3    →      L    3                            x        +        y        +        2        z        +        2        =        0                                  0        x        +        3        y        −        42        z        −        18        =        0                                  0        x        +        1        y        +        0        z        +        3        =        0            Then   −      L    2    /  3  +      L    3    →      L    3                            x        +        y        +        2        z        +        2        =        0                                  0        x        +        3        y        −        42        z        −        18        =        0                                  0        x        +        0        y        +        0        z        +        3        =        0            Firstly, is my answer correct?

Ismael Blackwell · Accepted Answer

First of all, you &quot;jumped to&quot; an erroneous conclusion, based on inspection. In a system of equations, one or more variables may fail to be present in one or more equations.A most extreme example would be the three equations in three unknowns:                          x        −        1        =        0                                  y        −        1        =        0                                  z        −        1        =        0            From which we can &quot;read off&quot; the unique solution:      (                            x                                      y                                      z                      )    =      (                            1                                      1                                      1                      )  There would be a problem (and no solution would exist) if you had the following (say, reduced) linear system of equations:                          x        +        y        +        2        z        +        2        =        0                                  3        x        −        y        +        14        z        −        6        =        0                                  0        x        +        0        y        +        0        z        +        5        =        0            Note that in the above system, we have the absurd equation   5  =  0: such a system is called inconsistent, and clearly, no solution exits.Finally, if you end up with an equation of all &quot;zeros&quot;:   0  x  +  0  y  +  0  z  =  0, then infinitely many solutions exist, and depending on how many such equations exist in your system, you might have a system with one parameter, or two, which then, while an infinite number of solutions exist, there would constraints which limit exactly which solutions are valid; i.e., the parameter(s) would define a &quot;family&quot; of infinitely many solutions.In your last elementary row operation, note that you didn&#039;t operate on the   −  42  z  −  18 of       L    2  :We go from:                          x        +        y        +        2        z        +        2        =        0                                  0        x        +        3        y        −        42        z        −        18        =        0                                  0        x        +        1        y        +        0        z        +        3        =        0            Then applying, correctly,   −      L    2    /  3  +      L    3    →      L    3                            x        +        y        +        2        z        +        2        =        0                                  0        x        +        3        y        −        42        z        −        18        =        0                                  0        x        +        0        y        +        14        z        +        9        =        0            Now, you&#039;ll see that a unique solution exists:   14  z  =  −  9    ⟹  To solve the system at this point, you might want to write your equations as follows:                          x        +        y        +        2        z        =        −        2                                  0        x        +        3        y        −        42        z        =        18                                  0        x        +        0        y        +        14        z        =        −        9

aniawnua · Accepted Answer

Your first idea is off, because, for example the system  x  +  y  +  z  =  1  ,  x  +  2  y  =  1  ,  x  +  y  +  2  z  =  1still has a solution even though   z is absent from the second equation.On the other hand, your manipulation of equations (you call it echelon form) is one of possible correct methods of finding solutions/proving their absence.

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