Solve the following system of equations using Gaussian Elimination Method












0












$begingroup$


Solve the following system of equations using Gaussian Elimination Method
$$x+2y+3z=2$$
$$x+y-z=1$$
$$2x+3y+2z=3$$.



My Attempt :
$$x+2y+3z=2………(1)$$
$$x+y-z=1…………(2)$$
$$2x+3y+2z=3………(3)$$
Subtracting equation $(1)$ from equation $(2)$, we have
$$y+4z=1………(4)$$
Multiplying equation $(1)$ by $2$ and then Subtracting from equation $(3)$, we have
$$y+4z=1………(5)$$
Subtracting equation $(4)$ from equation $(5)$, we have
$$0=0$$.



How do I proceed now?






systems-of-equations gaussian-elimination







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$endgroup$








  • 1




    $begingroup$
    You just throw out the third equation, and keep the 2 first ones. Either you conclude that there is an infinity of solutions constituting a one dimensional subspace $E$ or your instructor wants you to give details about a basis of $E$...
    $endgroup$
    – Jean Marie
    Dec 9 '18 at 18:04
















0












$begingroup$


Solve the following system of equations using Gaussian Elimination Method
$$x+2y+3z=2$$
$$x+y-z=1$$
$$2x+3y+2z=3$$.



My Attempt :
$$x+2y+3z=2………(1)$$
$$x+y-z=1…………(2)$$
$$2x+3y+2z=3………(3)$$
Subtracting equation $(1)$ from equation $(2)$, we have
$$y+4z=1………(4)$$
Multiplying equation $(1)$ by $2$ and then Subtracting from equation $(3)$, we have
$$y+4z=1………(5)$$
Subtracting equation $(4)$ from equation $(5)$, we have
$$0=0$$.



How do I proceed now?






systems-of-equations gaussian-elimination







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    You just throw out the third equation, and keep the 2 first ones. Either you conclude that there is an infinity of solutions constituting a one dimensional subspace $E$ or your instructor wants you to give details about a basis of $E$...
    $endgroup$
    – Jean Marie
    Dec 9 '18 at 18:04














0












0








0





$begingroup$


Solve the following system of equations using Gaussian Elimination Method
$$x+2y+3z=2$$
$$x+y-z=1$$
$$2x+3y+2z=3$$.



My Attempt :
$$x+2y+3z=2………(1)$$
$$x+y-z=1…………(2)$$
$$2x+3y+2z=3………(3)$$
Subtracting equation $(1)$ from equation $(2)$, we have
$$y+4z=1………(4)$$
Multiplying equation $(1)$ by $2$ and then Subtracting from equation $(3)$, we have
$$y+4z=1………(5)$$
Subtracting equation $(4)$ from equation $(5)$, we have
$$0=0$$.



How do I proceed now?






systems-of-equations gaussian-elimination







share|cite|improve this question









$endgroup$




Solve the following system of equations using Gaussian Elimination Method
$$x+2y+3z=2$$
$$x+y-z=1$$
$$2x+3y+2z=3$$.



My Attempt :
$$x+2y+3z=2………(1)$$
$$x+y-z=1…………(2)$$
$$2x+3y+2z=3………(3)$$
Subtracting equation $(1)$ from equation $(2)$, we have
$$y+4z=1………(4)$$
Multiplying equation $(1)$ by $2$ and then Subtracting from equation $(3)$, we have
$$y+4z=1………(5)$$
Subtracting equation $(4)$ from equation $(5)$, we have
$$0=0$$.



How do I proceed now?






systems-of-equations gaussian-elimination




systems-of-equations gaussian-elimination






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




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asked Dec 9 '18 at 17:58









blue_eyed_...blue_eyed_...

3,25921646




3,25921646








  • 1




    $begingroup$
    You just throw out the third equation, and keep the 2 first ones. Either you conclude that there is an infinity of solutions constituting a one dimensional subspace $E$ or your instructor wants you to give details about a basis of $E$...
    $endgroup$
    – Jean Marie
    Dec 9 '18 at 18:04














  • 1




    $begingroup$
    You just throw out the third equation, and keep the 2 first ones. Either you conclude that there is an infinity of solutions constituting a one dimensional subspace $E$ or your instructor wants you to give details about a basis of $E$...
    $endgroup$
    – Jean Marie
    Dec 9 '18 at 18:04








1




1




$begingroup$
You just throw out the third equation, and keep the 2 first ones. Either you conclude that there is an infinity of solutions constituting a one dimensional subspace $E$ or your instructor wants you to give details about a basis of $E$...
$endgroup$
– Jean Marie
Dec 9 '18 at 18:04




$begingroup$
You just throw out the third equation, and keep the 2 first ones. Either you conclude that there is an infinity of solutions constituting a one dimensional subspace $E$ or your instructor wants you to give details about a basis of $E$...
$endgroup$
– Jean Marie
Dec 9 '18 at 18:04










2 Answers
2






active

oldest

votes


















2












$begingroup$

Since adding first two you are getting last one the system has infinite solutions.



From $ y=1-4z$ you get, by puting it in (2) $x =5z$. So your system has a solution $$(x,y,z) = (5t,1-4t,t)$$
where $t$ is an arbitray real number.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    You will get the system:
    $$x+2y+3z=2$$
    $$-y-4z=-1$$
    Substituting $$z=t$$ you will get $$y=1-4t$$ and $$x=2-2(1-4t)-3t$$
    The system has infinity many solutions.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Is this the process of Gaussian-elimination?
      $endgroup$
      – blue_eyed_...
      Dec 9 '18 at 18:07










    • $begingroup$
      Yes this is the so-called Gaussian- elimination.
      $endgroup$
      – Dr. Sonnhard Graubner
      Dec 9 '18 at 18:15











    Your Answer





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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    Since adding first two you are getting last one the system has infinite solutions.



    From $ y=1-4z$ you get, by puting it in (2) $x =5z$. So your system has a solution $$(x,y,z) = (5t,1-4t,t)$$
    where $t$ is an arbitray real number.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Since adding first two you are getting last one the system has infinite solutions.



      From $ y=1-4z$ you get, by puting it in (2) $x =5z$. So your system has a solution $$(x,y,z) = (5t,1-4t,t)$$
      where $t$ is an arbitray real number.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Since adding first two you are getting last one the system has infinite solutions.



        From $ y=1-4z$ you get, by puting it in (2) $x =5z$. So your system has a solution $$(x,y,z) = (5t,1-4t,t)$$
        where $t$ is an arbitray real number.






        share|cite|improve this answer









        $endgroup$



        Since adding first two you are getting last one the system has infinite solutions.



        From $ y=1-4z$ you get, by puting it in (2) $x =5z$. So your system has a solution $$(x,y,z) = (5t,1-4t,t)$$
        where $t$ is an arbitray real number.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 9 '18 at 18:05









        greedoidgreedoid

        39.2k114797




        39.2k114797























            1












            $begingroup$

            You will get the system:
            $$x+2y+3z=2$$
            $$-y-4z=-1$$
            Substituting $$z=t$$ you will get $$y=1-4t$$ and $$x=2-2(1-4t)-3t$$
            The system has infinity many solutions.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Is this the process of Gaussian-elimination?
              $endgroup$
              – blue_eyed_...
              Dec 9 '18 at 18:07










            • $begingroup$
              Yes this is the so-called Gaussian- elimination.
              $endgroup$
              – Dr. Sonnhard Graubner
              Dec 9 '18 at 18:15
















            1












            $begingroup$

            You will get the system:
            $$x+2y+3z=2$$
            $$-y-4z=-1$$
            Substituting $$z=t$$ you will get $$y=1-4t$$ and $$x=2-2(1-4t)-3t$$
            The system has infinity many solutions.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Is this the process of Gaussian-elimination?
              $endgroup$
              – blue_eyed_...
              Dec 9 '18 at 18:07










            • $begingroup$
              Yes this is the so-called Gaussian- elimination.
              $endgroup$
              – Dr. Sonnhard Graubner
              Dec 9 '18 at 18:15














            1












            1








            1





            $begingroup$

            You will get the system:
            $$x+2y+3z=2$$
            $$-y-4z=-1$$
            Substituting $$z=t$$ you will get $$y=1-4t$$ and $$x=2-2(1-4t)-3t$$
            The system has infinity many solutions.






            share|cite|improve this answer









            $endgroup$



            You will get the system:
            $$x+2y+3z=2$$
            $$-y-4z=-1$$
            Substituting $$z=t$$ you will get $$y=1-4t$$ and $$x=2-2(1-4t)-3t$$
            The system has infinity many solutions.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 9 '18 at 18:06









            Dr. Sonnhard GraubnerDr. Sonnhard Graubner

            74.2k42865




            74.2k42865












            • $begingroup$
              Is this the process of Gaussian-elimination?
              $endgroup$
              – blue_eyed_...
              Dec 9 '18 at 18:07










            • $begingroup$
              Yes this is the so-called Gaussian- elimination.
              $endgroup$
              – Dr. Sonnhard Graubner
              Dec 9 '18 at 18:15


















            • $begingroup$
              Is this the process of Gaussian-elimination?
              $endgroup$
              – blue_eyed_...
              Dec 9 '18 at 18:07










            • $begingroup$
              Yes this is the so-called Gaussian- elimination.
              $endgroup$
              – Dr. Sonnhard Graubner
              Dec 9 '18 at 18:15
















            $begingroup$
            Is this the process of Gaussian-elimination?
            $endgroup$
            – blue_eyed_...
            Dec 9 '18 at 18:07




            $begingroup$
            Is this the process of Gaussian-elimination?
            $endgroup$
            – blue_eyed_...
            Dec 9 '18 at 18:07












            $begingroup$
            Yes this is the so-called Gaussian- elimination.
            $endgroup$
            – Dr. Sonnhard Graubner
            Dec 9 '18 at 18:15




            $begingroup$
            Yes this is the so-called Gaussian- elimination.
            $endgroup$
            – Dr. Sonnhard Graubner
            Dec 9 '18 at 18:15


















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