How i can solve the sistem of equation in variables x,y,z












0














How i can solve the sistem of equation in variables x,y,z:



$xy'=x'y$



$yz'=y'z$



$zx'=z'x$



where x',y' and z' are constants.






linear-algebra







share|cite|improve this question




















  • 4




    It might be better to use different letters for the constants.
    – Anurag A
    Nov 24 at 19:17












  • The set of solutions is a subspace of $mathbf{R}^3$, which depends on the values of the constants $x^prime, y^prime, z^prime$. What is exatly the question? "Find explicitly, as a function of these constants, the subspace?"
    – Paolo Leonetti
    Nov 24 at 19:30












  • I forget about this.solve in R.
    – Israel Meireles Chrisostomo
    Nov 24 at 19:42
















0














How i can solve the sistem of equation in variables x,y,z:



$xy'=x'y$



$yz'=y'z$



$zx'=z'x$



where x',y' and z' are constants.






linear-algebra







share|cite|improve this question




















  • 4




    It might be better to use different letters for the constants.
    – Anurag A
    Nov 24 at 19:17












  • The set of solutions is a subspace of $mathbf{R}^3$, which depends on the values of the constants $x^prime, y^prime, z^prime$. What is exatly the question? "Find explicitly, as a function of these constants, the subspace?"
    – Paolo Leonetti
    Nov 24 at 19:30












  • I forget about this.solve in R.
    – Israel Meireles Chrisostomo
    Nov 24 at 19:42














0












0








0







How i can solve the sistem of equation in variables x,y,z:



$xy'=x'y$



$yz'=y'z$



$zx'=z'x$



where x',y' and z' are constants.






linear-algebra







share|cite|improve this question















How i can solve the sistem of equation in variables x,y,z:



$xy'=x'y$



$yz'=y'z$



$zx'=z'x$



where x',y' and z' are constants.






linear-algebra




linear-algebra






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 24 at 19:24









Paolo Leonetti

11.4k21550




11.4k21550










asked Nov 24 at 19:13









Israel Meireles Chrisostomo

604413




604413








  • 4




    It might be better to use different letters for the constants.
    – Anurag A
    Nov 24 at 19:17












  • The set of solutions is a subspace of $mathbf{R}^3$, which depends on the values of the constants $x^prime, y^prime, z^prime$. What is exatly the question? "Find explicitly, as a function of these constants, the subspace?"
    – Paolo Leonetti
    Nov 24 at 19:30












  • I forget about this.solve in R.
    – Israel Meireles Chrisostomo
    Nov 24 at 19:42














  • 4




    It might be better to use different letters for the constants.
    – Anurag A
    Nov 24 at 19:17












  • The set of solutions is a subspace of $mathbf{R}^3$, which depends on the values of the constants $x^prime, y^prime, z^prime$. What is exatly the question? "Find explicitly, as a function of these constants, the subspace?"
    – Paolo Leonetti
    Nov 24 at 19:30












  • I forget about this.solve in R.
    – Israel Meireles Chrisostomo
    Nov 24 at 19:42








4




4




It might be better to use different letters for the constants.
– Anurag A
Nov 24 at 19:17






It might be better to use different letters for the constants.
– Anurag A
Nov 24 at 19:17














The set of solutions is a subspace of $mathbf{R}^3$, which depends on the values of the constants $x^prime, y^prime, z^prime$. What is exatly the question? "Find explicitly, as a function of these constants, the subspace?"
– Paolo Leonetti
Nov 24 at 19:30






The set of solutions is a subspace of $mathbf{R}^3$, which depends on the values of the constants $x^prime, y^prime, z^prime$. What is exatly the question? "Find explicitly, as a function of these constants, the subspace?"
– Paolo Leonetti
Nov 24 at 19:30














I forget about this.solve in R.
– Israel Meireles Chrisostomo
Nov 24 at 19:42




I forget about this.solve in R.
– Israel Meireles Chrisostomo
Nov 24 at 19:42










1 Answer
1






active

oldest

votes


















-1














Let $x'=a$, $y'=b$, and $z'=c$, just for simplicity. Then the system becomes



$$begin{align}
bx-ayphantom{.+cz}&=0 \
cy-bz&=0 \
-cxphantom{.+by}+az&=0
end{align}$$



We can represent this system with the augmented matrix



$$left[
begin{array}{ccc|c}
b&-a&0&0 \
0&c&-b&0 \
-c&0&a&0
end{array}
right]$$



Can you take it from here?






share|cite|improve this answer





















  • can you desenvolved this, please!!!
    – Israel Meireles Chrisostomo
    Nov 24 at 21:37










  • How familiar are you with augmented matrices? Also, could you please add some information to your question about where this problem came from, as well as any thoughts you might have had about how to start it?
    – Robert Howard
    Nov 24 at 21:51










  • I don't know how to solve this for scheduling.It's impossible, because zero column
    – Israel Meireles Chrisostomo
    Nov 24 at 22:35












  • It's not impossible. There's a trivial solution, where $x=y=z=0$, but there are also infinitely many other solutions. You can obtain the relations $y=dfrac{bx}{a}$ and $z=dfrac{cx}{a}$, so that any value of $x$ you choose will correspond to a unique solution.
    – Robert Howard
    Nov 25 at 2:46











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









-1














Let $x'=a$, $y'=b$, and $z'=c$, just for simplicity. Then the system becomes



$$begin{align}
bx-ayphantom{.+cz}&=0 \
cy-bz&=0 \
-cxphantom{.+by}+az&=0
end{align}$$



We can represent this system with the augmented matrix



$$left[
begin{array}{ccc|c}
b&-a&0&0 \
0&c&-b&0 \
-c&0&a&0
end{array}
right]$$



Can you take it from here?






share|cite|improve this answer





















  • can you desenvolved this, please!!!
    – Israel Meireles Chrisostomo
    Nov 24 at 21:37










  • How familiar are you with augmented matrices? Also, could you please add some information to your question about where this problem came from, as well as any thoughts you might have had about how to start it?
    – Robert Howard
    Nov 24 at 21:51










  • I don't know how to solve this for scheduling.It's impossible, because zero column
    – Israel Meireles Chrisostomo
    Nov 24 at 22:35












  • It's not impossible. There's a trivial solution, where $x=y=z=0$, but there are also infinitely many other solutions. You can obtain the relations $y=dfrac{bx}{a}$ and $z=dfrac{cx}{a}$, so that any value of $x$ you choose will correspond to a unique solution.
    – Robert Howard
    Nov 25 at 2:46
















-1














Let $x'=a$, $y'=b$, and $z'=c$, just for simplicity. Then the system becomes



$$begin{align}
bx-ayphantom{.+cz}&=0 \
cy-bz&=0 \
-cxphantom{.+by}+az&=0
end{align}$$



We can represent this system with the augmented matrix



$$left[
begin{array}{ccc|c}
b&-a&0&0 \
0&c&-b&0 \
-c&0&a&0
end{array}
right]$$



Can you take it from here?






share|cite|improve this answer





















  • can you desenvolved this, please!!!
    – Israel Meireles Chrisostomo
    Nov 24 at 21:37










  • How familiar are you with augmented matrices? Also, could you please add some information to your question about where this problem came from, as well as any thoughts you might have had about how to start it?
    – Robert Howard
    Nov 24 at 21:51










  • I don't know how to solve this for scheduling.It's impossible, because zero column
    – Israel Meireles Chrisostomo
    Nov 24 at 22:35












  • It's not impossible. There's a trivial solution, where $x=y=z=0$, but there are also infinitely many other solutions. You can obtain the relations $y=dfrac{bx}{a}$ and $z=dfrac{cx}{a}$, so that any value of $x$ you choose will correspond to a unique solution.
    – Robert Howard
    Nov 25 at 2:46














-1












-1








-1






Let $x'=a$, $y'=b$, and $z'=c$, just for simplicity. Then the system becomes



$$begin{align}
bx-ayphantom{.+cz}&=0 \
cy-bz&=0 \
-cxphantom{.+by}+az&=0
end{align}$$



We can represent this system with the augmented matrix



$$left[
begin{array}{ccc|c}
b&-a&0&0 \
0&c&-b&0 \
-c&0&a&0
end{array}
right]$$



Can you take it from here?






share|cite|improve this answer












Let $x'=a$, $y'=b$, and $z'=c$, just for simplicity. Then the system becomes



$$begin{align}
bx-ayphantom{.+cz}&=0 \
cy-bz&=0 \
-cxphantom{.+by}+az&=0
end{align}$$



We can represent this system with the augmented matrix



$$left[
begin{array}{ccc|c}
b&-a&0&0 \
0&c&-b&0 \
-c&0&a&0
end{array}
right]$$



Can you take it from here?







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 24 at 20:14









Robert Howard

1,9161822




1,9161822












  • can you desenvolved this, please!!!
    – Israel Meireles Chrisostomo
    Nov 24 at 21:37










  • How familiar are you with augmented matrices? Also, could you please add some information to your question about where this problem came from, as well as any thoughts you might have had about how to start it?
    – Robert Howard
    Nov 24 at 21:51










  • I don't know how to solve this for scheduling.It's impossible, because zero column
    – Israel Meireles Chrisostomo
    Nov 24 at 22:35












  • It's not impossible. There's a trivial solution, where $x=y=z=0$, but there are also infinitely many other solutions. You can obtain the relations $y=dfrac{bx}{a}$ and $z=dfrac{cx}{a}$, so that any value of $x$ you choose will correspond to a unique solution.
    – Robert Howard
    Nov 25 at 2:46


















  • can you desenvolved this, please!!!
    – Israel Meireles Chrisostomo
    Nov 24 at 21:37










  • How familiar are you with augmented matrices? Also, could you please add some information to your question about where this problem came from, as well as any thoughts you might have had about how to start it?
    – Robert Howard
    Nov 24 at 21:51










  • I don't know how to solve this for scheduling.It's impossible, because zero column
    – Israel Meireles Chrisostomo
    Nov 24 at 22:35












  • It's not impossible. There's a trivial solution, where $x=y=z=0$, but there are also infinitely many other solutions. You can obtain the relations $y=dfrac{bx}{a}$ and $z=dfrac{cx}{a}$, so that any value of $x$ you choose will correspond to a unique solution.
    – Robert Howard
    Nov 25 at 2:46
















can you desenvolved this, please!!!
– Israel Meireles Chrisostomo
Nov 24 at 21:37




can you desenvolved this, please!!!
– Israel Meireles Chrisostomo
Nov 24 at 21:37












How familiar are you with augmented matrices? Also, could you please add some information to your question about where this problem came from, as well as any thoughts you might have had about how to start it?
– Robert Howard
Nov 24 at 21:51




How familiar are you with augmented matrices? Also, could you please add some information to your question about where this problem came from, as well as any thoughts you might have had about how to start it?
– Robert Howard
Nov 24 at 21:51












I don't know how to solve this for scheduling.It's impossible, because zero column
– Israel Meireles Chrisostomo
Nov 24 at 22:35






I don't know how to solve this for scheduling.It's impossible, because zero column
– Israel Meireles Chrisostomo
Nov 24 at 22:35














It's not impossible. There's a trivial solution, where $x=y=z=0$, but there are also infinitely many other solutions. You can obtain the relations $y=dfrac{bx}{a}$ and $z=dfrac{cx}{a}$, so that any value of $x$ you choose will correspond to a unique solution.
– Robert Howard
Nov 25 at 2:46




It's not impossible. There's a trivial solution, where $x=y=z=0$, but there are also infinitely many other solutions. You can obtain the relations $y=dfrac{bx}{a}$ and $z=dfrac{cx}{a}$, so that any value of $x$ you choose will correspond to a unique solution.
– Robert Howard
Nov 25 at 2:46


















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