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
edited Nov 24 at 19:24
Paolo Leonetti
11.4k21550
asked Nov 24 at 19:13
Israel Meireles Chrisostomo
604413
add a comment |
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
edited Nov 24 at 19:24
Paolo Leonetti
11.4k21550
asked Nov 24 at 19:13
Israel Meireles Chrisostomo
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
add a comment |
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
edited Nov 24 at 19:24
Paolo Leonetti
11.4k21550
asked Nov 24 at 19:13
Israel Meireles Chrisostomo
604413
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
edited Nov 24 at 19:24
Paolo Leonetti
11.4k21550
asked Nov 24 at 19:13
Israel Meireles Chrisostomo
604413
edited Nov 24 at 19:24
Paolo Leonetti
11.4k21550
asked Nov 24 at 19:13
Israel Meireles Chrisostomo
604413
edited Nov 24 at 19:24
Paolo Leonetti
11.4k21550
edited Nov 24 at 19:24
Paolo Leonetti
11.4k21550
edited Nov 24 at 19:24
Paolo Leonetti
11.4k21550
11.4k21550
asked Nov 24 at 19:13
Israel Meireles Chrisostomo
604413
asked Nov 24 at 19:13
Israel Meireles Chrisostomo
604413
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
add a comment |
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
add a comment |
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?
answered Nov 24 at 20:14
Robert Howard
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
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
draft saved
draft discarded
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3011961%2fhow-i-can-solve-the-sistem-of-equation-in-variables-x-y-z%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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?
answered Nov 24 at 20:14
Robert Howard
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
add a comment |
-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?
answered Nov 24 at 20:14
Robert Howard
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
add a comment |
-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?
answered Nov 24 at 20:14
Robert Howard
1,9161822
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?
answered Nov 24 at 20:14
Robert Howard
1,9161822
answered Nov 24 at 20:14
Robert Howard
1,9161822
answered Nov 24 at 20:14
Robert Howard
1,9161822
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
add a comment |
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
add a comment |
draft saved
draft discarded
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
draft saved
draft discarded
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3011961%2fhow-i-can-solve-the-sistem-of-equation-in-variables-x-y-z%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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