Show that $frac{x}{sqrt{1-a^2}}=frac{y}{sqrt{1-b^2}}=frac{z}{sqrt{1-c^2}}$












1












$begingroup$



Given $3$ equations:





  • $ x-cy-bz=0$


    • $-cx+y-az=0$

    • $-bx-ay+z=0$




Show that $dfrac{x}{sqrt{1-a^2}}=dfrac{y}{sqrt{1-b^2}}=dfrac{z}{sqrt{1-c^2}}$




Now solving the 3 equations I got:



$dfrac{x}{{1-a^2}}=dfrac{y}{{ab+c}}=dfrac{z}{ac+b}$



$dfrac{x}{ac+b}=dfrac{y}{a+bc}=dfrac{z}{1-c^2}$



$dfrac{x}{a+bc}=dfrac{y}{1-b^2}=dfrac{z}{a+bc}$



How to prove the required fact?



Any way to prove it ?






algebra-precalculus number-theory elementary-number-theory ratio







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    You should also not that in the event of linear independence $x,y,z = 0$. Your solution profile does meet that criterion. To repeat my early question, what is known about $a,b,c$??
    $endgroup$
    – DavidG
    Dec 25 '18 at 3:22








  • 1




    $begingroup$
    Here you have a balanced system of linear equations of $x,y,z$. There is a wide spectrum of methods to take your linear system and solve for $x,y.z$. What methods have you been working with at present? Also, there hasn't been any conditions put on $a,b,c$. Here we see that if $a,b,c = pm 1$ that $x,y,z$ do not exist.
    $endgroup$
    – DavidG
    Dec 25 '18 at 3:24










  • $begingroup$
    @DavidG,only thing known is $x,y,zneq pm 1$
    $endgroup$
    – user596656
    Dec 25 '18 at 3:24












  • $begingroup$
    Ok, there are a number of ways you can approach this. Have you learnt about solving Linear Systems using Gaussian Elimination? en.wikipedia.org/wiki/Gaussian_elimination
    $endgroup$
    – DavidG
    Dec 25 '18 at 3:25










  • $begingroup$
    @DavidG,it wont help
    $endgroup$
    – user596656
    Dec 25 '18 at 3:32
















1












$begingroup$



Given $3$ equations:





  • $ x-cy-bz=0$


    • $-cx+y-az=0$

    • $-bx-ay+z=0$




Show that $dfrac{x}{sqrt{1-a^2}}=dfrac{y}{sqrt{1-b^2}}=dfrac{z}{sqrt{1-c^2}}$




Now solving the 3 equations I got:



$dfrac{x}{{1-a^2}}=dfrac{y}{{ab+c}}=dfrac{z}{ac+b}$



$dfrac{x}{ac+b}=dfrac{y}{a+bc}=dfrac{z}{1-c^2}$



$dfrac{x}{a+bc}=dfrac{y}{1-b^2}=dfrac{z}{a+bc}$



How to prove the required fact?



Any way to prove it ?






algebra-precalculus number-theory elementary-number-theory ratio







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    You should also not that in the event of linear independence $x,y,z = 0$. Your solution profile does meet that criterion. To repeat my early question, what is known about $a,b,c$??
    $endgroup$
    – DavidG
    Dec 25 '18 at 3:22








  • 1




    $begingroup$
    Here you have a balanced system of linear equations of $x,y,z$. There is a wide spectrum of methods to take your linear system and solve for $x,y.z$. What methods have you been working with at present? Also, there hasn't been any conditions put on $a,b,c$. Here we see that if $a,b,c = pm 1$ that $x,y,z$ do not exist.
    $endgroup$
    – DavidG
    Dec 25 '18 at 3:24










  • $begingroup$
    @DavidG,only thing known is $x,y,zneq pm 1$
    $endgroup$
    – user596656
    Dec 25 '18 at 3:24












  • $begingroup$
    Ok, there are a number of ways you can approach this. Have you learnt about solving Linear Systems using Gaussian Elimination? en.wikipedia.org/wiki/Gaussian_elimination
    $endgroup$
    – DavidG
    Dec 25 '18 at 3:25










  • $begingroup$
    @DavidG,it wont help
    $endgroup$
    – user596656
    Dec 25 '18 at 3:32














1












1








1


0



$begingroup$



Given $3$ equations:





  • $ x-cy-bz=0$


    • $-cx+y-az=0$

    • $-bx-ay+z=0$




Show that $dfrac{x}{sqrt{1-a^2}}=dfrac{y}{sqrt{1-b^2}}=dfrac{z}{sqrt{1-c^2}}$




Now solving the 3 equations I got:



$dfrac{x}{{1-a^2}}=dfrac{y}{{ab+c}}=dfrac{z}{ac+b}$



$dfrac{x}{ac+b}=dfrac{y}{a+bc}=dfrac{z}{1-c^2}$



$dfrac{x}{a+bc}=dfrac{y}{1-b^2}=dfrac{z}{a+bc}$



How to prove the required fact?



Any way to prove it ?






algebra-precalculus number-theory elementary-number-theory ratio







share|cite|improve this question











$endgroup$





Given $3$ equations:





  • $ x-cy-bz=0$


    • $-cx+y-az=0$

    • $-bx-ay+z=0$




Show that $dfrac{x}{sqrt{1-a^2}}=dfrac{y}{sqrt{1-b^2}}=dfrac{z}{sqrt{1-c^2}}$




Now solving the 3 equations I got:



$dfrac{x}{{1-a^2}}=dfrac{y}{{ab+c}}=dfrac{z}{ac+b}$



$dfrac{x}{ac+b}=dfrac{y}{a+bc}=dfrac{z}{1-c^2}$



$dfrac{x}{a+bc}=dfrac{y}{1-b^2}=dfrac{z}{a+bc}$



How to prove the required fact?



Any way to prove it ?






algebra-precalculus number-theory elementary-number-theory ratio




algebra-precalculus number-theory elementary-number-theory ratio






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 28 '18 at 16:43









Martin Sleziak

44.7k10118272




44.7k10118272










asked Dec 25 '18 at 3:13







user596656















  • 1




    $begingroup$
    You should also not that in the event of linear independence $x,y,z = 0$. Your solution profile does meet that criterion. To repeat my early question, what is known about $a,b,c$??
    $endgroup$
    – DavidG
    Dec 25 '18 at 3:22








  • 1




    $begingroup$
    Here you have a balanced system of linear equations of $x,y,z$. There is a wide spectrum of methods to take your linear system and solve for $x,y.z$. What methods have you been working with at present? Also, there hasn't been any conditions put on $a,b,c$. Here we see that if $a,b,c = pm 1$ that $x,y,z$ do not exist.
    $endgroup$
    – DavidG
    Dec 25 '18 at 3:24










  • $begingroup$
    @DavidG,only thing known is $x,y,zneq pm 1$
    $endgroup$
    – user596656
    Dec 25 '18 at 3:24












  • $begingroup$
    Ok, there are a number of ways you can approach this. Have you learnt about solving Linear Systems using Gaussian Elimination? en.wikipedia.org/wiki/Gaussian_elimination
    $endgroup$
    – DavidG
    Dec 25 '18 at 3:25










  • $begingroup$
    @DavidG,it wont help
    $endgroup$
    – user596656
    Dec 25 '18 at 3:32














  • 1




    $begingroup$
    You should also not that in the event of linear independence $x,y,z = 0$. Your solution profile does meet that criterion. To repeat my early question, what is known about $a,b,c$??
    $endgroup$
    – DavidG
    Dec 25 '18 at 3:22








  • 1




    $begingroup$
    Here you have a balanced system of linear equations of $x,y,z$. There is a wide spectrum of methods to take your linear system and solve for $x,y.z$. What methods have you been working with at present? Also, there hasn't been any conditions put on $a,b,c$. Here we see that if $a,b,c = pm 1$ that $x,y,z$ do not exist.
    $endgroup$
    – DavidG
    Dec 25 '18 at 3:24










  • $begingroup$
    @DavidG,only thing known is $x,y,zneq pm 1$
    $endgroup$
    – user596656
    Dec 25 '18 at 3:24












  • $begingroup$
    Ok, there are a number of ways you can approach this. Have you learnt about solving Linear Systems using Gaussian Elimination? en.wikipedia.org/wiki/Gaussian_elimination
    $endgroup$
    – DavidG
    Dec 25 '18 at 3:25










  • $begingroup$
    @DavidG,it wont help
    $endgroup$
    – user596656
    Dec 25 '18 at 3:32








1




1




$begingroup$
You should also not that in the event of linear independence $x,y,z = 0$. Your solution profile does meet that criterion. To repeat my early question, what is known about $a,b,c$??
$endgroup$
– DavidG
Dec 25 '18 at 3:22






$begingroup$
You should also not that in the event of linear independence $x,y,z = 0$. Your solution profile does meet that criterion. To repeat my early question, what is known about $a,b,c$??
$endgroup$
– DavidG
Dec 25 '18 at 3:22






1




1




$begingroup$
Here you have a balanced system of linear equations of $x,y,z$. There is a wide spectrum of methods to take your linear system and solve for $x,y.z$. What methods have you been working with at present? Also, there hasn't been any conditions put on $a,b,c$. Here we see that if $a,b,c = pm 1$ that $x,y,z$ do not exist.
$endgroup$
– DavidG
Dec 25 '18 at 3:24




$begingroup$
Here you have a balanced system of linear equations of $x,y,z$. There is a wide spectrum of methods to take your linear system and solve for $x,y.z$. What methods have you been working with at present? Also, there hasn't been any conditions put on $a,b,c$. Here we see that if $a,b,c = pm 1$ that $x,y,z$ do not exist.
$endgroup$
– DavidG
Dec 25 '18 at 3:24












$begingroup$
@DavidG,only thing known is $x,y,zneq pm 1$
$endgroup$
– user596656
Dec 25 '18 at 3:24






$begingroup$
@DavidG,only thing known is $x,y,zneq pm 1$
$endgroup$
– user596656
Dec 25 '18 at 3:24














$begingroup$
Ok, there are a number of ways you can approach this. Have you learnt about solving Linear Systems using Gaussian Elimination? en.wikipedia.org/wiki/Gaussian_elimination
$endgroup$
– DavidG
Dec 25 '18 at 3:25




$begingroup$
Ok, there are a number of ways you can approach this. Have you learnt about solving Linear Systems using Gaussian Elimination? en.wikipedia.org/wiki/Gaussian_elimination
$endgroup$
– DavidG
Dec 25 '18 at 3:25












$begingroup$
@DavidG,it wont help
$endgroup$
– user596656
Dec 25 '18 at 3:32




$begingroup$
@DavidG,it wont help
$endgroup$
– user596656
Dec 25 '18 at 3:32










3 Answers
3






active

oldest

votes


















5












$begingroup$

Hint:



Relating to geometry of triangles, looks much like sine rule to me, where $x,y$ and $z$ are sides of triangle and $a, b, c$ are the cosines of the corresponding opposite angles to the respective sides.



But here we restrict that $-1le a, b, cle 1$ because for the denominator to be meaningful in the "real number" sense we need to find the range of $a, b, c$



Edit:



Consider triangle $XYZ$ with sides $x, y, z$ and corresponding opposite angles $X, Y, Z$ . It is quite well known from a little trigonometry of triangles that
$$x=ycos Z +zcos Y$$
$$y=xcos Z +zcos X$$
$$z=ycos X +xcos Y$$



And from sine rule we have $$frac {x}{sin X}=frac {y}{sin Y}=frac {z}{sin Z}$$



Now in your given equations just substitute $a=cos X$,$b=cos Y$,$c=cos Z$



On doing this the given three equations transform to the three equations I have given above. While the equation you need to prove may be simply written as the Sine rule.
I think this clarification might get you visualised what I really mean in my answer.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I am sorry,I am not getting it really
    $endgroup$
    – user596656
    Dec 25 '18 at 3:33










  • $begingroup$
    How can I solve this from above
    $endgroup$
    – user596656
    Dec 25 '18 at 3:34










  • $begingroup$
    Digamma correctly notes that $$sin x =sqrt{1-cos^2(x)}$$
    $endgroup$
    – Rhys Hughes
    Dec 25 '18 at 3:36










  • $begingroup$
    @RhysHughes;if that is the solution,why are we given three equations
    $endgroup$
    – user596656
    Dec 25 '18 at 3:45










  • $begingroup$
    If we assume that they are the sides of the triiangle then there is nothing to prove and no use of the equations
    $endgroup$
    – user596656
    Dec 25 '18 at 3:46



















1












$begingroup$

Hint:



By $(1),(2)$ $$0=x-cy-bz+c(-cx+y-az)=x(1-c^2)-z(b+ca)$$



By $(2),(3)$ $$0=a(-cx+y-az)-bx-ay+z=z(1-a^2)-x(ca+b)$$



$$dfrac{x(1-c^2)}{z(1-a^2)}=dfrac{z(b+ca)}{x(ca+b)}$$



What if $b+cane0$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    HINT:



    Here you have a linear system of 3 equations and 3 unknowns. This is known as a balanced linear system. To verify that your solutions are correct you can either



    (1) Solve the system of equations to derive the values given (Gaussian Elimination, Craymers Rule, LU Decomposition, etc etc (there are many ways)).



    (2) Substitute in your sols of $x,y,z$ into each of your equations to see that they are solutions



    Or



    (3) Give us some background on your mathematical toolkit so we can suggest a method that can be used to solve this system of equations.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I have already told you that I know all the tools that you mentioned but the problem is the required solution is not coming,I dont know what background you are talking about,I have completed my undergraduate degree in Maths
      $endgroup$
      – user596656
      Dec 25 '18 at 3:57










    • $begingroup$
      Using Cramers Rule i got 0 as the solution,Gauss is not giving too,are there any other ways
      $endgroup$
      – user596656
      Dec 25 '18 at 3:58










    • $begingroup$
      If possible do give a way by which you find this solution can be arrived at
      $endgroup$
      – user596656
      Dec 25 '18 at 3:58










    • $begingroup$
      If Gaussian Elimination is not giving you the given values then they are either wrong OR your use of Gaussian Elimination is wrong.
      $endgroup$
      – DavidG
      Dec 25 '18 at 3:59










    • $begingroup$
      Btw, $x,y,z = 0$ is a solution that matches the form given...
      $endgroup$
      – DavidG
      Dec 25 '18 at 4:01











    Your Answer





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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    5












    $begingroup$

    Hint:



    Relating to geometry of triangles, looks much like sine rule to me, where $x,y$ and $z$ are sides of triangle and $a, b, c$ are the cosines of the corresponding opposite angles to the respective sides.



    But here we restrict that $-1le a, b, cle 1$ because for the denominator to be meaningful in the "real number" sense we need to find the range of $a, b, c$



    Edit:



    Consider triangle $XYZ$ with sides $x, y, z$ and corresponding opposite angles $X, Y, Z$ . It is quite well known from a little trigonometry of triangles that
    $$x=ycos Z +zcos Y$$
    $$y=xcos Z +zcos X$$
    $$z=ycos X +xcos Y$$



    And from sine rule we have $$frac {x}{sin X}=frac {y}{sin Y}=frac {z}{sin Z}$$



    Now in your given equations just substitute $a=cos X$,$b=cos Y$,$c=cos Z$



    On doing this the given three equations transform to the three equations I have given above. While the equation you need to prove may be simply written as the Sine rule.
    I think this clarification might get you visualised what I really mean in my answer.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I am sorry,I am not getting it really
      $endgroup$
      – user596656
      Dec 25 '18 at 3:33










    • $begingroup$
      How can I solve this from above
      $endgroup$
      – user596656
      Dec 25 '18 at 3:34










    • $begingroup$
      Digamma correctly notes that $$sin x =sqrt{1-cos^2(x)}$$
      $endgroup$
      – Rhys Hughes
      Dec 25 '18 at 3:36










    • $begingroup$
      @RhysHughes;if that is the solution,why are we given three equations
      $endgroup$
      – user596656
      Dec 25 '18 at 3:45










    • $begingroup$
      If we assume that they are the sides of the triiangle then there is nothing to prove and no use of the equations
      $endgroup$
      – user596656
      Dec 25 '18 at 3:46
















    5












    $begingroup$

    Hint:



    Relating to geometry of triangles, looks much like sine rule to me, where $x,y$ and $z$ are sides of triangle and $a, b, c$ are the cosines of the corresponding opposite angles to the respective sides.



    But here we restrict that $-1le a, b, cle 1$ because for the denominator to be meaningful in the "real number" sense we need to find the range of $a, b, c$



    Edit:



    Consider triangle $XYZ$ with sides $x, y, z$ and corresponding opposite angles $X, Y, Z$ . It is quite well known from a little trigonometry of triangles that
    $$x=ycos Z +zcos Y$$
    $$y=xcos Z +zcos X$$
    $$z=ycos X +xcos Y$$



    And from sine rule we have $$frac {x}{sin X}=frac {y}{sin Y}=frac {z}{sin Z}$$



    Now in your given equations just substitute $a=cos X$,$b=cos Y$,$c=cos Z$



    On doing this the given three equations transform to the three equations I have given above. While the equation you need to prove may be simply written as the Sine rule.
    I think this clarification might get you visualised what I really mean in my answer.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I am sorry,I am not getting it really
      $endgroup$
      – user596656
      Dec 25 '18 at 3:33










    • $begingroup$
      How can I solve this from above
      $endgroup$
      – user596656
      Dec 25 '18 at 3:34










    • $begingroup$
      Digamma correctly notes that $$sin x =sqrt{1-cos^2(x)}$$
      $endgroup$
      – Rhys Hughes
      Dec 25 '18 at 3:36










    • $begingroup$
      @RhysHughes;if that is the solution,why are we given three equations
      $endgroup$
      – user596656
      Dec 25 '18 at 3:45










    • $begingroup$
      If we assume that they are the sides of the triiangle then there is nothing to prove and no use of the equations
      $endgroup$
      – user596656
      Dec 25 '18 at 3:46














    5












    5








    5





    $begingroup$

    Hint:



    Relating to geometry of triangles, looks much like sine rule to me, where $x,y$ and $z$ are sides of triangle and $a, b, c$ are the cosines of the corresponding opposite angles to the respective sides.



    But here we restrict that $-1le a, b, cle 1$ because for the denominator to be meaningful in the "real number" sense we need to find the range of $a, b, c$



    Edit:



    Consider triangle $XYZ$ with sides $x, y, z$ and corresponding opposite angles $X, Y, Z$ . It is quite well known from a little trigonometry of triangles that
    $$x=ycos Z +zcos Y$$
    $$y=xcos Z +zcos X$$
    $$z=ycos X +xcos Y$$



    And from sine rule we have $$frac {x}{sin X}=frac {y}{sin Y}=frac {z}{sin Z}$$



    Now in your given equations just substitute $a=cos X$,$b=cos Y$,$c=cos Z$



    On doing this the given three equations transform to the three equations I have given above. While the equation you need to prove may be simply written as the Sine rule.
    I think this clarification might get you visualised what I really mean in my answer.






    share|cite|improve this answer











    $endgroup$



    Hint:



    Relating to geometry of triangles, looks much like sine rule to me, where $x,y$ and $z$ are sides of triangle and $a, b, c$ are the cosines of the corresponding opposite angles to the respective sides.



    But here we restrict that $-1le a, b, cle 1$ because for the denominator to be meaningful in the "real number" sense we need to find the range of $a, b, c$



    Edit:



    Consider triangle $XYZ$ with sides $x, y, z$ and corresponding opposite angles $X, Y, Z$ . It is quite well known from a little trigonometry of triangles that
    $$x=ycos Z +zcos Y$$
    $$y=xcos Z +zcos X$$
    $$z=ycos X +xcos Y$$



    And from sine rule we have $$frac {x}{sin X}=frac {y}{sin Y}=frac {z}{sin Z}$$



    Now in your given equations just substitute $a=cos X$,$b=cos Y$,$c=cos Z$



    On doing this the given three equations transform to the three equations I have given above. While the equation you need to prove may be simply written as the Sine rule.
    I think this clarification might get you visualised what I really mean in my answer.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 25 '18 at 4:04

























    answered Dec 25 '18 at 3:28









    DarkraiDarkrai

    6,2571441




    6,2571441












    • $begingroup$
      I am sorry,I am not getting it really
      $endgroup$
      – user596656
      Dec 25 '18 at 3:33










    • $begingroup$
      How can I solve this from above
      $endgroup$
      – user596656
      Dec 25 '18 at 3:34










    • $begingroup$
      Digamma correctly notes that $$sin x =sqrt{1-cos^2(x)}$$
      $endgroup$
      – Rhys Hughes
      Dec 25 '18 at 3:36










    • $begingroup$
      @RhysHughes;if that is the solution,why are we given three equations
      $endgroup$
      – user596656
      Dec 25 '18 at 3:45










    • $begingroup$
      If we assume that they are the sides of the triiangle then there is nothing to prove and no use of the equations
      $endgroup$
      – user596656
      Dec 25 '18 at 3:46


















    • $begingroup$
      I am sorry,I am not getting it really
      $endgroup$
      – user596656
      Dec 25 '18 at 3:33










    • $begingroup$
      How can I solve this from above
      $endgroup$
      – user596656
      Dec 25 '18 at 3:34










    • $begingroup$
      Digamma correctly notes that $$sin x =sqrt{1-cos^2(x)}$$
      $endgroup$
      – Rhys Hughes
      Dec 25 '18 at 3:36










    • $begingroup$
      @RhysHughes;if that is the solution,why are we given three equations
      $endgroup$
      – user596656
      Dec 25 '18 at 3:45










    • $begingroup$
      If we assume that they are the sides of the triiangle then there is nothing to prove and no use of the equations
      $endgroup$
      – user596656
      Dec 25 '18 at 3:46
















    $begingroup$
    I am sorry,I am not getting it really
    $endgroup$
    – user596656
    Dec 25 '18 at 3:33




    $begingroup$
    I am sorry,I am not getting it really
    $endgroup$
    – user596656
    Dec 25 '18 at 3:33












    $begingroup$
    How can I solve this from above
    $endgroup$
    – user596656
    Dec 25 '18 at 3:34




    $begingroup$
    How can I solve this from above
    $endgroup$
    – user596656
    Dec 25 '18 at 3:34












    $begingroup$
    Digamma correctly notes that $$sin x =sqrt{1-cos^2(x)}$$
    $endgroup$
    – Rhys Hughes
    Dec 25 '18 at 3:36




    $begingroup$
    Digamma correctly notes that $$sin x =sqrt{1-cos^2(x)}$$
    $endgroup$
    – Rhys Hughes
    Dec 25 '18 at 3:36












    $begingroup$
    @RhysHughes;if that is the solution,why are we given three equations
    $endgroup$
    – user596656
    Dec 25 '18 at 3:45




    $begingroup$
    @RhysHughes;if that is the solution,why are we given three equations
    $endgroup$
    – user596656
    Dec 25 '18 at 3:45












    $begingroup$
    If we assume that they are the sides of the triiangle then there is nothing to prove and no use of the equations
    $endgroup$
    – user596656
    Dec 25 '18 at 3:46




    $begingroup$
    If we assume that they are the sides of the triiangle then there is nothing to prove and no use of the equations
    $endgroup$
    – user596656
    Dec 25 '18 at 3:46











    1












    $begingroup$

    Hint:



    By $(1),(2)$ $$0=x-cy-bz+c(-cx+y-az)=x(1-c^2)-z(b+ca)$$



    By $(2),(3)$ $$0=a(-cx+y-az)-bx-ay+z=z(1-a^2)-x(ca+b)$$



    $$dfrac{x(1-c^2)}{z(1-a^2)}=dfrac{z(b+ca)}{x(ca+b)}$$



    What if $b+cane0$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Hint:



      By $(1),(2)$ $$0=x-cy-bz+c(-cx+y-az)=x(1-c^2)-z(b+ca)$$



      By $(2),(3)$ $$0=a(-cx+y-az)-bx-ay+z=z(1-a^2)-x(ca+b)$$



      $$dfrac{x(1-c^2)}{z(1-a^2)}=dfrac{z(b+ca)}{x(ca+b)}$$



      What if $b+cane0$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Hint:



        By $(1),(2)$ $$0=x-cy-bz+c(-cx+y-az)=x(1-c^2)-z(b+ca)$$



        By $(2),(3)$ $$0=a(-cx+y-az)-bx-ay+z=z(1-a^2)-x(ca+b)$$



        $$dfrac{x(1-c^2)}{z(1-a^2)}=dfrac{z(b+ca)}{x(ca+b)}$$



        What if $b+cane0$






        share|cite|improve this answer









        $endgroup$



        Hint:



        By $(1),(2)$ $$0=x-cy-bz+c(-cx+y-az)=x(1-c^2)-z(b+ca)$$



        By $(2),(3)$ $$0=a(-cx+y-az)-bx-ay+z=z(1-a^2)-x(ca+b)$$



        $$dfrac{x(1-c^2)}{z(1-a^2)}=dfrac{z(b+ca)}{x(ca+b)}$$



        What if $b+cane0$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 25 '18 at 6:41









        lab bhattacharjeelab bhattacharjee

        226k15157275




        226k15157275























            0












            $begingroup$

            HINT:



            Here you have a linear system of 3 equations and 3 unknowns. This is known as a balanced linear system. To verify that your solutions are correct you can either



            (1) Solve the system of equations to derive the values given (Gaussian Elimination, Craymers Rule, LU Decomposition, etc etc (there are many ways)).



            (2) Substitute in your sols of $x,y,z$ into each of your equations to see that they are solutions



            Or



            (3) Give us some background on your mathematical toolkit so we can suggest a method that can be used to solve this system of equations.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I have already told you that I know all the tools that you mentioned but the problem is the required solution is not coming,I dont know what background you are talking about,I have completed my undergraduate degree in Maths
              $endgroup$
              – user596656
              Dec 25 '18 at 3:57










            • $begingroup$
              Using Cramers Rule i got 0 as the solution,Gauss is not giving too,are there any other ways
              $endgroup$
              – user596656
              Dec 25 '18 at 3:58










            • $begingroup$
              If possible do give a way by which you find this solution can be arrived at
              $endgroup$
              – user596656
              Dec 25 '18 at 3:58










            • $begingroup$
              If Gaussian Elimination is not giving you the given values then they are either wrong OR your use of Gaussian Elimination is wrong.
              $endgroup$
              – DavidG
              Dec 25 '18 at 3:59










            • $begingroup$
              Btw, $x,y,z = 0$ is a solution that matches the form given...
              $endgroup$
              – DavidG
              Dec 25 '18 at 4:01
















            0












            $begingroup$

            HINT:



            Here you have a linear system of 3 equations and 3 unknowns. This is known as a balanced linear system. To verify that your solutions are correct you can either



            (1) Solve the system of equations to derive the values given (Gaussian Elimination, Craymers Rule, LU Decomposition, etc etc (there are many ways)).



            (2) Substitute in your sols of $x,y,z$ into each of your equations to see that they are solutions



            Or



            (3) Give us some background on your mathematical toolkit so we can suggest a method that can be used to solve this system of equations.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I have already told you that I know all the tools that you mentioned but the problem is the required solution is not coming,I dont know what background you are talking about,I have completed my undergraduate degree in Maths
              $endgroup$
              – user596656
              Dec 25 '18 at 3:57










            • $begingroup$
              Using Cramers Rule i got 0 as the solution,Gauss is not giving too,are there any other ways
              $endgroup$
              – user596656
              Dec 25 '18 at 3:58










            • $begingroup$
              If possible do give a way by which you find this solution can be arrived at
              $endgroup$
              – user596656
              Dec 25 '18 at 3:58










            • $begingroup$
              If Gaussian Elimination is not giving you the given values then they are either wrong OR your use of Gaussian Elimination is wrong.
              $endgroup$
              – DavidG
              Dec 25 '18 at 3:59










            • $begingroup$
              Btw, $x,y,z = 0$ is a solution that matches the form given...
              $endgroup$
              – DavidG
              Dec 25 '18 at 4:01














            0












            0








            0





            $begingroup$

            HINT:



            Here you have a linear system of 3 equations and 3 unknowns. This is known as a balanced linear system. To verify that your solutions are correct you can either



            (1) Solve the system of equations to derive the values given (Gaussian Elimination, Craymers Rule, LU Decomposition, etc etc (there are many ways)).



            (2) Substitute in your sols of $x,y,z$ into each of your equations to see that they are solutions



            Or



            (3) Give us some background on your mathematical toolkit so we can suggest a method that can be used to solve this system of equations.






            share|cite|improve this answer









            $endgroup$



            HINT:



            Here you have a linear system of 3 equations and 3 unknowns. This is known as a balanced linear system. To verify that your solutions are correct you can either



            (1) Solve the system of equations to derive the values given (Gaussian Elimination, Craymers Rule, LU Decomposition, etc etc (there are many ways)).



            (2) Substitute in your sols of $x,y,z$ into each of your equations to see that they are solutions



            Or



            (3) Give us some background on your mathematical toolkit so we can suggest a method that can be used to solve this system of equations.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 25 '18 at 3:53









            DavidGDavidG

            2,1321724




            2,1321724












            • $begingroup$
              I have already told you that I know all the tools that you mentioned but the problem is the required solution is not coming,I dont know what background you are talking about,I have completed my undergraduate degree in Maths
              $endgroup$
              – user596656
              Dec 25 '18 at 3:57










            • $begingroup$
              Using Cramers Rule i got 0 as the solution,Gauss is not giving too,are there any other ways
              $endgroup$
              – user596656
              Dec 25 '18 at 3:58










            • $begingroup$
              If possible do give a way by which you find this solution can be arrived at
              $endgroup$
              – user596656
              Dec 25 '18 at 3:58










            • $begingroup$
              If Gaussian Elimination is not giving you the given values then they are either wrong OR your use of Gaussian Elimination is wrong.
              $endgroup$
              – DavidG
              Dec 25 '18 at 3:59










            • $begingroup$
              Btw, $x,y,z = 0$ is a solution that matches the form given...
              $endgroup$
              – DavidG
              Dec 25 '18 at 4:01


















            • $begingroup$
              I have already told you that I know all the tools that you mentioned but the problem is the required solution is not coming,I dont know what background you are talking about,I have completed my undergraduate degree in Maths
              $endgroup$
              – user596656
              Dec 25 '18 at 3:57










            • $begingroup$
              Using Cramers Rule i got 0 as the solution,Gauss is not giving too,are there any other ways
              $endgroup$
              – user596656
              Dec 25 '18 at 3:58










            • $begingroup$
              If possible do give a way by which you find this solution can be arrived at
              $endgroup$
              – user596656
              Dec 25 '18 at 3:58










            • $begingroup$
              If Gaussian Elimination is not giving you the given values then they are either wrong OR your use of Gaussian Elimination is wrong.
              $endgroup$
              – DavidG
              Dec 25 '18 at 3:59










            • $begingroup$
              Btw, $x,y,z = 0$ is a solution that matches the form given...
              $endgroup$
              – DavidG
              Dec 25 '18 at 4:01
















            $begingroup$
            I have already told you that I know all the tools that you mentioned but the problem is the required solution is not coming,I dont know what background you are talking about,I have completed my undergraduate degree in Maths
            $endgroup$
            – user596656
            Dec 25 '18 at 3:57




            $begingroup$
            I have already told you that I know all the tools that you mentioned but the problem is the required solution is not coming,I dont know what background you are talking about,I have completed my undergraduate degree in Maths
            $endgroup$
            – user596656
            Dec 25 '18 at 3:57












            $begingroup$
            Using Cramers Rule i got 0 as the solution,Gauss is not giving too,are there any other ways
            $endgroup$
            – user596656
            Dec 25 '18 at 3:58




            $begingroup$
            Using Cramers Rule i got 0 as the solution,Gauss is not giving too,are there any other ways
            $endgroup$
            – user596656
            Dec 25 '18 at 3:58












            $begingroup$
            If possible do give a way by which you find this solution can be arrived at
            $endgroup$
            – user596656
            Dec 25 '18 at 3:58




            $begingroup$
            If possible do give a way by which you find this solution can be arrived at
            $endgroup$
            – user596656
            Dec 25 '18 at 3:58












            $begingroup$
            If Gaussian Elimination is not giving you the given values then they are either wrong OR your use of Gaussian Elimination is wrong.
            $endgroup$
            – DavidG
            Dec 25 '18 at 3:59




            $begingroup$
            If Gaussian Elimination is not giving you the given values then they are either wrong OR your use of Gaussian Elimination is wrong.
            $endgroup$
            – DavidG
            Dec 25 '18 at 3:59












            $begingroup$
            Btw, $x,y,z = 0$ is a solution that matches the form given...
            $endgroup$
            – DavidG
            Dec 25 '18 at 4:01




            $begingroup$
            Btw, $x,y,z = 0$ is a solution that matches the form given...
            $endgroup$
            – DavidG
            Dec 25 '18 at 4:01


















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