Linear algebra: $Ax=0$ for matrix












0














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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  • Hyperplan equation !
    – Damien
    Dec 2 at 9:42










  • 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.
    – amWhy
    Dec 2 at 19:00
















0














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
























  • Hyperplan equation !
    – Damien
    Dec 2 at 9:42










  • 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.
    – amWhy
    Dec 2 at 19:00














0












0








0







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















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 at 10:17









A.Γ.

21.9k22455




21.9k22455










asked Dec 2 at 9:13









Ella

102




102












  • Hyperplan equation !
    – Damien
    Dec 2 at 9:42










  • 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.
    – amWhy
    Dec 2 at 19:00


















  • Hyperplan equation !
    – Damien
    Dec 2 at 9:42










  • 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.
    – amWhy
    Dec 2 at 19:00
















Hyperplan equation !
– Damien
Dec 2 at 9:42




Hyperplan equation !
– Damien
Dec 2 at 9:42












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.
– amWhy
Dec 2 at 19:00




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.
– amWhy
Dec 2 at 19:00










2 Answers
2






active

oldest

votes


















1














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





















  • 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?
    – Ella
    Dec 2 at 9:46










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










  • 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?
    – Ella
    Dec 2 at 9:59










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










  • Could you help me in finding that?
    – Ella
    Dec 2 at 10:05



















1














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














    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





















    • 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?
      – Ella
      Dec 2 at 9:46










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










    • 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?
      – Ella
      Dec 2 at 9:59










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










    • Could you help me in finding that?
      – Ella
      Dec 2 at 10:05
















    1














    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





















    • 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?
      – Ella
      Dec 2 at 9:46










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










    • 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?
      – Ella
      Dec 2 at 9:59










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










    • Could you help me in finding that?
      – Ella
      Dec 2 at 10:05














    1












    1








    1






    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












    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 at 9:20









    Siong Thye Goh

    99k1464117




    99k1464117












    • 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?
      – Ella
      Dec 2 at 9:46










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










    • 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?
      – Ella
      Dec 2 at 9:59










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










    • Could you help me in finding that?
      – Ella
      Dec 2 at 10:05


















    • 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?
      – Ella
      Dec 2 at 9:46










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










    • 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?
      – Ella
      Dec 2 at 9:59










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










    • Could you help me in finding that?
      – Ella
      Dec 2 at 10:05
















    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?
    – Ella
    Dec 2 at 9:46




    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?
    – Ella
    Dec 2 at 9:46












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




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












    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?
    – Ella
    Dec 2 at 9:59




    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?
    – Ella
    Dec 2 at 9:59












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




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












    Could you help me in finding that?
    – Ella
    Dec 2 at 10:05




    Could you help me in finding that?
    – Ella
    Dec 2 at 10:05











    1














    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


























      1














      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
























        1












        1








        1






        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












        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 at 9:20









        José Carlos Santos

        149k22119221




        149k22119221






























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