Linear algebra: $Ax=0$ for matrix












0












$begingroup$


I have a matrix which after doing row reduced echelon form is as follows:$$A=begin{bmatrix}1&1&-1&-1\0&0&0&0\0&0&0&0\0&0&0&0end{bmatrix}.$$
When I try to solve $Ax = 0$, I am confused. $x_2,x_3,x_4$ are free variables because they have no pivot and only $x_1$ is pivoted variable. Following is the equation I get:
$$x_1+x_2-x_3-x_4 = 0$$
What is the best way to find all the solutions to $Ax=0$?






linear-algebra matrices







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  • $begingroup$
    Hyperplan equation !
    $endgroup$
    – Damien
    Dec 2 '18 at 9:42










  • $begingroup$
    In case the real issue is the X-Y Problem, please add the original matrix and your steps in row reduction, so we can verify your matrix, or correct it so you can move on.
    $endgroup$
    – amWhy
    Dec 2 '18 at 19:00
















0












$begingroup$


I have a matrix which after doing row reduced echelon form is as follows:$$A=begin{bmatrix}1&1&-1&-1\0&0&0&0\0&0&0&0\0&0&0&0end{bmatrix}.$$
When I try to solve $Ax = 0$, I am confused. $x_2,x_3,x_4$ are free variables because they have no pivot and only $x_1$ is pivoted variable. Following is the equation I get:
$$x_1+x_2-x_3-x_4 = 0$$
What is the best way to find all the solutions to $Ax=0$?






linear-algebra matrices







share|cite|improve this question











$endgroup$












  • $begingroup$
    Hyperplan equation !
    $endgroup$
    – Damien
    Dec 2 '18 at 9:42










  • $begingroup$
    In case the real issue is the X-Y Problem, please add the original matrix and your steps in row reduction, so we can verify your matrix, or correct it so you can move on.
    $endgroup$
    – amWhy
    Dec 2 '18 at 19:00














0












0








0





$begingroup$


I have a matrix which after doing row reduced echelon form is as follows:$$A=begin{bmatrix}1&1&-1&-1\0&0&0&0\0&0&0&0\0&0&0&0end{bmatrix}.$$
When I try to solve $Ax = 0$, I am confused. $x_2,x_3,x_4$ are free variables because they have no pivot and only $x_1$ is pivoted variable. Following is the equation I get:
$$x_1+x_2-x_3-x_4 = 0$$
What is the best way to find all the solutions to $Ax=0$?






linear-algebra matrices







share|cite|improve this question











$endgroup$




I have a matrix which after doing row reduced echelon form is as follows:$$A=begin{bmatrix}1&1&-1&-1\0&0&0&0\0&0&0&0\0&0&0&0end{bmatrix}.$$
When I try to solve $Ax = 0$, I am confused. $x_2,x_3,x_4$ are free variables because they have no pivot and only $x_1$ is pivoted variable. Following is the equation I get:
$$x_1+x_2-x_3-x_4 = 0$$
What is the best way to find all the solutions to $Ax=0$?






linear-algebra matrices




linear-algebra matrices






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 2 '18 at 10:17









A.Γ.

22.7k32656




22.7k32656










asked Dec 2 '18 at 9:13









EllaElla

102




102












  • $begingroup$
    Hyperplan equation !
    $endgroup$
    – Damien
    Dec 2 '18 at 9:42










  • $begingroup$
    In case the real issue is the X-Y Problem, please add the original matrix and your steps in row reduction, so we can verify your matrix, or correct it so you can move on.
    $endgroup$
    – amWhy
    Dec 2 '18 at 19:00


















  • $begingroup$
    Hyperplan equation !
    $endgroup$
    – Damien
    Dec 2 '18 at 9:42










  • $begingroup$
    In case the real issue is the X-Y Problem, please add the original matrix and your steps in row reduction, so we can verify your matrix, or correct it so you can move on.
    $endgroup$
    – amWhy
    Dec 2 '18 at 19:00
















$begingroup$
Hyperplan equation !
$endgroup$
– Damien
Dec 2 '18 at 9:42




$begingroup$
Hyperplan equation !
$endgroup$
– Damien
Dec 2 '18 at 9:42












$begingroup$
In case the real issue is the X-Y Problem, please add the original matrix and your steps in row reduction, so we can verify your matrix, or correct it so you can move on.
$endgroup$
– amWhy
Dec 2 '18 at 19:00




$begingroup$
In case the real issue is the X-Y Problem, please add the original matrix and your steps in row reduction, so we can verify your matrix, or correct it so you can move on.
$endgroup$
– amWhy
Dec 2 '18 at 19:00










2 Answers
2






active

oldest

votes


















1












$begingroup$

We have $x_1= -x_2+x_3+x_4$.



Let $x_2=s$, $x_3=u$, $x_4=v$ (the free variables), then we have



$$x_1=-s+u+v.$$



$$begin{bmatrix} x_1 \ x_2 \ x_3 \ x_4 end{bmatrix}=sbegin{bmatrix} -1 \ 1 \ 0 \ 0 end{bmatrix} + ubegin{bmatrix} 1 \ 0 \ 1 \ 0 end{bmatrix} + vbegin{bmatrix} 1 \ 0 \ 0 \ 1 end{bmatrix}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you for the answer. Also, for Ax=b, if I do row reduced echolen form for "b" size too, I get "2".. so same way we will do it for Ax=b? I am wondering what will be the particular solution and special solution?
    $endgroup$
    – Ella
    Dec 2 '18 at 9:46










  • $begingroup$
    I am guessing you mean $x_1=2-x_2+x_3+x_4$. then $(2,0,0,0)$ is a particular solution.
    $endgroup$
    – Siong Thye Goh
    Dec 2 '18 at 9:52










  • $begingroup$
    No, for instance after row reduced echelon from I have "b" as 2 0 0 0 Then as x2,x3,x4 are free variables so we can set any value for them. This way we will have so many solutions? How to find all the solutions for Ax=b then?
    $endgroup$
    – Ella
    Dec 2 '18 at 9:59










  • $begingroup$
    We find a particualr solution plus the general homogeneous solution part.
    $endgroup$
    – Siong Thye Goh
    Dec 2 '18 at 10:00










  • $begingroup$
    Could you help me in finding that?
    $endgroup$
    – Ella
    Dec 2 '18 at 10:05



















1












$begingroup$

You have already found all solutions: it's the set of all $(x_1,x_2,x_3,x_4)$ such that $x_1+x_2-x_3-x_4=0$.






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









    1












    $begingroup$

    We have $x_1= -x_2+x_3+x_4$.



    Let $x_2=s$, $x_3=u$, $x_4=v$ (the free variables), then we have



    $$x_1=-s+u+v.$$



    $$begin{bmatrix} x_1 \ x_2 \ x_3 \ x_4 end{bmatrix}=sbegin{bmatrix} -1 \ 1 \ 0 \ 0 end{bmatrix} + ubegin{bmatrix} 1 \ 0 \ 1 \ 0 end{bmatrix} + vbegin{bmatrix} 1 \ 0 \ 0 \ 1 end{bmatrix}$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you for the answer. Also, for Ax=b, if I do row reduced echolen form for "b" size too, I get "2".. so same way we will do it for Ax=b? I am wondering what will be the particular solution and special solution?
      $endgroup$
      – Ella
      Dec 2 '18 at 9:46










    • $begingroup$
      I am guessing you mean $x_1=2-x_2+x_3+x_4$. then $(2,0,0,0)$ is a particular solution.
      $endgroup$
      – Siong Thye Goh
      Dec 2 '18 at 9:52










    • $begingroup$
      No, for instance after row reduced echelon from I have "b" as 2 0 0 0 Then as x2,x3,x4 are free variables so we can set any value for them. This way we will have so many solutions? How to find all the solutions for Ax=b then?
      $endgroup$
      – Ella
      Dec 2 '18 at 9:59










    • $begingroup$
      We find a particualr solution plus the general homogeneous solution part.
      $endgroup$
      – Siong Thye Goh
      Dec 2 '18 at 10:00










    • $begingroup$
      Could you help me in finding that?
      $endgroup$
      – Ella
      Dec 2 '18 at 10:05
















    1












    $begingroup$

    We have $x_1= -x_2+x_3+x_4$.



    Let $x_2=s$, $x_3=u$, $x_4=v$ (the free variables), then we have



    $$x_1=-s+u+v.$$



    $$begin{bmatrix} x_1 \ x_2 \ x_3 \ x_4 end{bmatrix}=sbegin{bmatrix} -1 \ 1 \ 0 \ 0 end{bmatrix} + ubegin{bmatrix} 1 \ 0 \ 1 \ 0 end{bmatrix} + vbegin{bmatrix} 1 \ 0 \ 0 \ 1 end{bmatrix}$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you for the answer. Also, for Ax=b, if I do row reduced echolen form for "b" size too, I get "2".. so same way we will do it for Ax=b? I am wondering what will be the particular solution and special solution?
      $endgroup$
      – Ella
      Dec 2 '18 at 9:46










    • $begingroup$
      I am guessing you mean $x_1=2-x_2+x_3+x_4$. then $(2,0,0,0)$ is a particular solution.
      $endgroup$
      – Siong Thye Goh
      Dec 2 '18 at 9:52










    • $begingroup$
      No, for instance after row reduced echelon from I have "b" as 2 0 0 0 Then as x2,x3,x4 are free variables so we can set any value for them. This way we will have so many solutions? How to find all the solutions for Ax=b then?
      $endgroup$
      – Ella
      Dec 2 '18 at 9:59










    • $begingroup$
      We find a particualr solution plus the general homogeneous solution part.
      $endgroup$
      – Siong Thye Goh
      Dec 2 '18 at 10:00










    • $begingroup$
      Could you help me in finding that?
      $endgroup$
      – Ella
      Dec 2 '18 at 10:05














    1












    1








    1





    $begingroup$

    We have $x_1= -x_2+x_3+x_4$.



    Let $x_2=s$, $x_3=u$, $x_4=v$ (the free variables), then we have



    $$x_1=-s+u+v.$$



    $$begin{bmatrix} x_1 \ x_2 \ x_3 \ x_4 end{bmatrix}=sbegin{bmatrix} -1 \ 1 \ 0 \ 0 end{bmatrix} + ubegin{bmatrix} 1 \ 0 \ 1 \ 0 end{bmatrix} + vbegin{bmatrix} 1 \ 0 \ 0 \ 1 end{bmatrix}$$






    share|cite|improve this answer









    $endgroup$



    We have $x_1= -x_2+x_3+x_4$.



    Let $x_2=s$, $x_3=u$, $x_4=v$ (the free variables), then we have



    $$x_1=-s+u+v.$$



    $$begin{bmatrix} x_1 \ x_2 \ x_3 \ x_4 end{bmatrix}=sbegin{bmatrix} -1 \ 1 \ 0 \ 0 end{bmatrix} + ubegin{bmatrix} 1 \ 0 \ 1 \ 0 end{bmatrix} + vbegin{bmatrix} 1 \ 0 \ 0 \ 1 end{bmatrix}$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 2 '18 at 9:20









    Siong Thye GohSiong Thye Goh

    100k1466117




    100k1466117












    • $begingroup$
      Thank you for the answer. Also, for Ax=b, if I do row reduced echolen form for "b" size too, I get "2".. so same way we will do it for Ax=b? I am wondering what will be the particular solution and special solution?
      $endgroup$
      – Ella
      Dec 2 '18 at 9:46










    • $begingroup$
      I am guessing you mean $x_1=2-x_2+x_3+x_4$. then $(2,0,0,0)$ is a particular solution.
      $endgroup$
      – Siong Thye Goh
      Dec 2 '18 at 9:52










    • $begingroup$
      No, for instance after row reduced echelon from I have "b" as 2 0 0 0 Then as x2,x3,x4 are free variables so we can set any value for them. This way we will have so many solutions? How to find all the solutions for Ax=b then?
      $endgroup$
      – Ella
      Dec 2 '18 at 9:59










    • $begingroup$
      We find a particualr solution plus the general homogeneous solution part.
      $endgroup$
      – Siong Thye Goh
      Dec 2 '18 at 10:00










    • $begingroup$
      Could you help me in finding that?
      $endgroup$
      – Ella
      Dec 2 '18 at 10:05


















    • $begingroup$
      Thank you for the answer. Also, for Ax=b, if I do row reduced echolen form for "b" size too, I get "2".. so same way we will do it for Ax=b? I am wondering what will be the particular solution and special solution?
      $endgroup$
      – Ella
      Dec 2 '18 at 9:46










    • $begingroup$
      I am guessing you mean $x_1=2-x_2+x_3+x_4$. then $(2,0,0,0)$ is a particular solution.
      $endgroup$
      – Siong Thye Goh
      Dec 2 '18 at 9:52










    • $begingroup$
      No, for instance after row reduced echelon from I have "b" as 2 0 0 0 Then as x2,x3,x4 are free variables so we can set any value for them. This way we will have so many solutions? How to find all the solutions for Ax=b then?
      $endgroup$
      – Ella
      Dec 2 '18 at 9:59










    • $begingroup$
      We find a particualr solution plus the general homogeneous solution part.
      $endgroup$
      – Siong Thye Goh
      Dec 2 '18 at 10:00










    • $begingroup$
      Could you help me in finding that?
      $endgroup$
      – Ella
      Dec 2 '18 at 10:05
















    $begingroup$
    Thank you for the answer. Also, for Ax=b, if I do row reduced echolen form for "b" size too, I get "2".. so same way we will do it for Ax=b? I am wondering what will be the particular solution and special solution?
    $endgroup$
    – Ella
    Dec 2 '18 at 9:46




    $begingroup$
    Thank you for the answer. Also, for Ax=b, if I do row reduced echolen form for "b" size too, I get "2".. so same way we will do it for Ax=b? I am wondering what will be the particular solution and special solution?
    $endgroup$
    – Ella
    Dec 2 '18 at 9:46












    $begingroup$
    I am guessing you mean $x_1=2-x_2+x_3+x_4$. then $(2,0,0,0)$ is a particular solution.
    $endgroup$
    – Siong Thye Goh
    Dec 2 '18 at 9:52




    $begingroup$
    I am guessing you mean $x_1=2-x_2+x_3+x_4$. then $(2,0,0,0)$ is a particular solution.
    $endgroup$
    – Siong Thye Goh
    Dec 2 '18 at 9:52












    $begingroup$
    No, for instance after row reduced echelon from I have "b" as 2 0 0 0 Then as x2,x3,x4 are free variables so we can set any value for them. This way we will have so many solutions? How to find all the solutions for Ax=b then?
    $endgroup$
    – Ella
    Dec 2 '18 at 9:59




    $begingroup$
    No, for instance after row reduced echelon from I have "b" as 2 0 0 0 Then as x2,x3,x4 are free variables so we can set any value for them. This way we will have so many solutions? How to find all the solutions for Ax=b then?
    $endgroup$
    – Ella
    Dec 2 '18 at 9:59












    $begingroup$
    We find a particualr solution plus the general homogeneous solution part.
    $endgroup$
    – Siong Thye Goh
    Dec 2 '18 at 10:00




    $begingroup$
    We find a particualr solution plus the general homogeneous solution part.
    $endgroup$
    – Siong Thye Goh
    Dec 2 '18 at 10:00












    $begingroup$
    Could you help me in finding that?
    $endgroup$
    – Ella
    Dec 2 '18 at 10:05




    $begingroup$
    Could you help me in finding that?
    $endgroup$
    – Ella
    Dec 2 '18 at 10:05











    1












    $begingroup$

    You have already found all solutions: it's the set of all $(x_1,x_2,x_3,x_4)$ such that $x_1+x_2-x_3-x_4=0$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      You have already found all solutions: it's the set of all $(x_1,x_2,x_3,x_4)$ such that $x_1+x_2-x_3-x_4=0$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        You have already found all solutions: it's the set of all $(x_1,x_2,x_3,x_4)$ such that $x_1+x_2-x_3-x_4=0$.






        share|cite|improve this answer









        $endgroup$



        You have already found all solutions: it's the set of all $(x_1,x_2,x_3,x_4)$ such that $x_1+x_2-x_3-x_4=0$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 2 '18 at 9:20









        José Carlos SantosJosé Carlos Santos

        156k22126227




        156k22126227






























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