Gauss Jordan elimination problem. Is this right?

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so I immediately converted the equations to an augmented matrix and I want to know if I did this correctly:



beginmatrix
2 & 2 & 1 & 0
\2 & -3 & -4 & 0
\4 & -1 & -3 & 0
endmatrix



I added some multiple of the first row to the second and third:



beginmatrix
2 & 2 & 1 & 0
\0 & -5 & -5 & 0
\0 & -5 & -5 & 0
endmatrix



are these reductions legit? Can I divide by -5 and can I just eliminate a row if it's identical?



beginmatrix
2 & 2 & 1 & 0
\0 & 1 & 1 & 0
endmatrix



divide row 1 by 2



beginmatrix
1 & 1 & 0.5 & 0
\0 & 1 & 1 & 0
endmatrix



finally:



beginmatrix
1 & 0 & -0.5 & 0
\0 & 1 & 1 & 0
endmatrix



Is that okay?




linear-algebra




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  • 1




    The equation in the row you eliminated became 0=0, telling you that the system is underdetermined.
    – ncmathsadist
    Aug 1 at 15:00











  • So there's no solution?
    – Jwan622
    Aug 1 at 15:18














up vote
0
down vote

favorite












enter image description here



so I immediately converted the equations to an augmented matrix and I want to know if I did this correctly:



beginmatrix
2 & 2 & 1 & 0
\2 & -3 & -4 & 0
\4 & -1 & -3 & 0
endmatrix



I added some multiple of the first row to the second and third:



beginmatrix
2 & 2 & 1 & 0
\0 & -5 & -5 & 0
\0 & -5 & -5 & 0
endmatrix



are these reductions legit? Can I divide by -5 and can I just eliminate a row if it's identical?



beginmatrix
2 & 2 & 1 & 0
\0 & 1 & 1 & 0
endmatrix



divide row 1 by 2



beginmatrix
1 & 1 & 0.5 & 0
\0 & 1 & 1 & 0
endmatrix



finally:



beginmatrix
1 & 0 & -0.5 & 0
\0 & 1 & 1 & 0
endmatrix



Is that okay?




linear-algebra




share|cite|improve this question















  • 1




    The equation in the row you eliminated became 0=0, telling you that the system is underdetermined.
    – ncmathsadist
    Aug 1 at 15:00











  • So there's no solution?
    – Jwan622
    Aug 1 at 15:18












up vote
0
down vote

favorite









up vote
0
down vote

favorite











enter image description here



so I immediately converted the equations to an augmented matrix and I want to know if I did this correctly:



beginmatrix
2 & 2 & 1 & 0
\2 & -3 & -4 & 0
\4 & -1 & -3 & 0
endmatrix



I added some multiple of the first row to the second and third:



beginmatrix
2 & 2 & 1 & 0
\0 & -5 & -5 & 0
\0 & -5 & -5 & 0
endmatrix



are these reductions legit? Can I divide by -5 and can I just eliminate a row if it's identical?



beginmatrix
2 & 2 & 1 & 0
\0 & 1 & 1 & 0
endmatrix



divide row 1 by 2



beginmatrix
1 & 1 & 0.5 & 0
\0 & 1 & 1 & 0
endmatrix



finally:



beginmatrix
1 & 0 & -0.5 & 0
\0 & 1 & 1 & 0
endmatrix



Is that okay?




linear-algebra




share|cite|improve this question











enter image description here



so I immediately converted the equations to an augmented matrix and I want to know if I did this correctly:



beginmatrix
2 & 2 & 1 & 0
\2 & -3 & -4 & 0
\4 & -1 & -3 & 0
endmatrix



I added some multiple of the first row to the second and third:



beginmatrix
2 & 2 & 1 & 0
\0 & -5 & -5 & 0
\0 & -5 & -5 & 0
endmatrix



are these reductions legit? Can I divide by -5 and can I just eliminate a row if it's identical?



beginmatrix
2 & 2 & 1 & 0
\0 & 1 & 1 & 0
endmatrix



divide row 1 by 2



beginmatrix
1 & 1 & 0.5 & 0
\0 & 1 & 1 & 0
endmatrix



finally:



beginmatrix
1 & 0 & -0.5 & 0
\0 & 1 & 1 & 0
endmatrix



Is that okay?





linear-algebra





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Aug 1 at 14:54









Jwan622

1,60111224




1,60111224







  • 1




    The equation in the row you eliminated became 0=0, telling you that the system is underdetermined.
    – ncmathsadist
    Aug 1 at 15:00











  • So there's no solution?
    – Jwan622
    Aug 1 at 15:18












  • 1




    The equation in the row you eliminated became 0=0, telling you that the system is underdetermined.
    – ncmathsadist
    Aug 1 at 15:00











  • So there's no solution?
    – Jwan622
    Aug 1 at 15:18







1




1




The equation in the row you eliminated became 0=0, telling you that the system is underdetermined.
– ncmathsadist
Aug 1 at 15:00





The equation in the row you eliminated became 0=0, telling you that the system is underdetermined.
– ncmathsadist
Aug 1 at 15:00













So there's no solution?
– Jwan622
Aug 1 at 15:18




So there's no solution?
– Jwan622
Aug 1 at 15:18










2 Answers
2






active

oldest

votes

















up vote
2
down vote



accepted










So far so good. Of ourse, you can delete one of two identical rows. The simple explanation is that it makes no difference if a condition (in this case, an equation) is written down once or twice. But you can also argue using Gaussian elimination: subtract one from the other, and you can definitely delete the all zero row that occurs.



But you haven't finished the proof. The question was to provide the answer parametrically.
The way you carried out the Gaussian elimination process, the variable $z$ is a free variable.
So $zin mathbbR$ is a parameter (it can be an arbitrary number).
Now, you can easily express the rest of the variables using $z$.
From the first equation, you obtain $x= 0.5z$, and from the second, you obtain $y=-z$.



So the set of all solutions of the system is the set of triples $(0.5z, -z,z)$, where $zin mathbbR$ is arbitrary.






share|cite|improve this answer




























    up vote
    0
    down vote













    What does this $$ beginmatrix
    1 & 0 & -0.5 & 0
    \0 & 1 & 1 & 0
    endmatrix$$



    tell you about the solutions of your system?



    The first one says $$x- 0.5z=0$$ and the second one says $$y+z=0$$



    What you need to do is finding parametric solution for your system.



    For example if you let $$z=t$$ then solve for $x$ and $y$ in terms of $t$ you will be done.






    share|cite|improve this answer





















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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      2
      down vote



      accepted










      So far so good. Of ourse, you can delete one of two identical rows. The simple explanation is that it makes no difference if a condition (in this case, an equation) is written down once or twice. But you can also argue using Gaussian elimination: subtract one from the other, and you can definitely delete the all zero row that occurs.



      But you haven't finished the proof. The question was to provide the answer parametrically.
      The way you carried out the Gaussian elimination process, the variable $z$ is a free variable.
      So $zin mathbbR$ is a parameter (it can be an arbitrary number).
      Now, you can easily express the rest of the variables using $z$.
      From the first equation, you obtain $x= 0.5z$, and from the second, you obtain $y=-z$.



      So the set of all solutions of the system is the set of triples $(0.5z, -z,z)$, where $zin mathbbR$ is arbitrary.






      share|cite|improve this answer

























        up vote
        2
        down vote



        accepted










        So far so good. Of ourse, you can delete one of two identical rows. The simple explanation is that it makes no difference if a condition (in this case, an equation) is written down once or twice. But you can also argue using Gaussian elimination: subtract one from the other, and you can definitely delete the all zero row that occurs.



        But you haven't finished the proof. The question was to provide the answer parametrically.
        The way you carried out the Gaussian elimination process, the variable $z$ is a free variable.
        So $zin mathbbR$ is a parameter (it can be an arbitrary number).
        Now, you can easily express the rest of the variables using $z$.
        From the first equation, you obtain $x= 0.5z$, and from the second, you obtain $y=-z$.



        So the set of all solutions of the system is the set of triples $(0.5z, -z,z)$, where $zin mathbbR$ is arbitrary.






        share|cite|improve this answer























          up vote
          2
          down vote



          accepted







          up vote
          2
          down vote



          accepted






          So far so good. Of ourse, you can delete one of two identical rows. The simple explanation is that it makes no difference if a condition (in this case, an equation) is written down once or twice. But you can also argue using Gaussian elimination: subtract one from the other, and you can definitely delete the all zero row that occurs.



          But you haven't finished the proof. The question was to provide the answer parametrically.
          The way you carried out the Gaussian elimination process, the variable $z$ is a free variable.
          So $zin mathbbR$ is a parameter (it can be an arbitrary number).
          Now, you can easily express the rest of the variables using $z$.
          From the first equation, you obtain $x= 0.5z$, and from the second, you obtain $y=-z$.



          So the set of all solutions of the system is the set of triples $(0.5z, -z,z)$, where $zin mathbbR$ is arbitrary.






          share|cite|improve this answer













          So far so good. Of ourse, you can delete one of two identical rows. The simple explanation is that it makes no difference if a condition (in this case, an equation) is written down once or twice. But you can also argue using Gaussian elimination: subtract one from the other, and you can definitely delete the all zero row that occurs.



          But you haven't finished the proof. The question was to provide the answer parametrically.
          The way you carried out the Gaussian elimination process, the variable $z$ is a free variable.
          So $zin mathbbR$ is a parameter (it can be an arbitrary number).
          Now, you can easily express the rest of the variables using $z$.
          From the first equation, you obtain $x= 0.5z$, and from the second, you obtain $y=-z$.



          So the set of all solutions of the system is the set of triples $(0.5z, -z,z)$, where $zin mathbbR$ is arbitrary.







          share|cite|improve this answer













          share|cite|improve this answer



          share|cite|improve this answer











          answered Aug 1 at 15:00









          A. Pongrácz

          1,294115




          1,294115




















              up vote
              0
              down vote













              What does this $$ beginmatrix
              1 & 0 & -0.5 & 0
              \0 & 1 & 1 & 0
              endmatrix$$



              tell you about the solutions of your system?



              The first one says $$x- 0.5z=0$$ and the second one says $$y+z=0$$



              What you need to do is finding parametric solution for your system.



              For example if you let $$z=t$$ then solve for $x$ and $y$ in terms of $t$ you will be done.






              share|cite|improve this answer

























                up vote
                0
                down vote













                What does this $$ beginmatrix
                1 & 0 & -0.5 & 0
                \0 & 1 & 1 & 0
                endmatrix$$



                tell you about the solutions of your system?



                The first one says $$x- 0.5z=0$$ and the second one says $$y+z=0$$



                What you need to do is finding parametric solution for your system.



                For example if you let $$z=t$$ then solve for $x$ and $y$ in terms of $t$ you will be done.






                share|cite|improve this answer























                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  What does this $$ beginmatrix
                  1 & 0 & -0.5 & 0
                  \0 & 1 & 1 & 0
                  endmatrix$$



                  tell you about the solutions of your system?



                  The first one says $$x- 0.5z=0$$ and the second one says $$y+z=0$$



                  What you need to do is finding parametric solution for your system.



                  For example if you let $$z=t$$ then solve for $x$ and $y$ in terms of $t$ you will be done.






                  share|cite|improve this answer













                  What does this $$ beginmatrix
                  1 & 0 & -0.5 & 0
                  \0 & 1 & 1 & 0
                  endmatrix$$



                  tell you about the solutions of your system?



                  The first one says $$x- 0.5z=0$$ and the second one says $$y+z=0$$



                  What you need to do is finding parametric solution for your system.



                  For example if you let $$z=t$$ then solve for $x$ and $y$ in terms of $t$ you will be done.







                  share|cite|improve this answer













                  share|cite|improve this answer



                  share|cite|improve this answer











                  answered Aug 1 at 15:22









                  Mohammad Riazi-Kermani

                  27.3k41851




                  27.3k41851






















                       

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