Mixing
How could this this equation system be transformed to a X-Y-Z linear equation system?
$begingroup$
I'm trying to solve this problem, which basically states this:
Based on the system of equations of the variables $x, y$ and $z$:
$$begin{cases}
x y - 2sqrt{y} + 3 y z = 8 \
2 x y - 3sqrt{y} + 2 y z = 7 \
- x y + sqrt{y} + 2 y z = 4 \
end{cases}$$please tell which of the next alternatives are correct:
- The system is possible and determined
- The range of the extended matrix of the system is 2.
- On the transposed matrix of the coefficients, associated with the system: $a_{12} = -3$
- The coefficient matrix associated to the system is non invertible.
So, what I did was construct the extended matrix from the equation system given in the problem statement, replacing the variables as this:
- $a = xy$
- $b = sqrt{y}$
- $c = yz$
The matrix constructed is this:
begin{bmatrix}
1 & -2 & 3 & 8 \
2 & -3 & 2 & 7 \
-1 & 1 & 2 & 4
end{bmatrix}
Solving by gaussian method, the matrix is reduced to:
begin{bmatrix}
1 & -2 & 3 & 8 \
0 & 1 & -4 & -9 \
0 & 0 & 1 & 3
end{bmatrix}
So finally:
- $a = 5$
- $b = 3$
- $c = 3$
The values of $x, y$ and $z$ could be calculated using the previous equations, giving this values as results:
- $x = 5/9$
- $y = 9$
- $z = 1/3$
This is the far as I can go on solving this problem, my question basically is, how to transform this to a linear equation system of x, y and z, because I do have the values of $x, y$ and $z$ but the statements that I have to indicate if they are correct or not, are based on that matrix that I don't know how to build.
I was thinking that maybe this problem doesn't look for me to build such a matrix, instead, given that there exist a solution for the system proposed initially, I should deduct the veracity of the statements:
- First statement is correct as each variable has just one unique value.
- Second statement is incorrect because for each variable having one single value, the range of the matrix should be 3.
Second and third statement I have no idea how to answer to.
Please give me a hand on this.
Thank you a lot.
linear-algebra systems-of-equations
asked Jan 13 at 0:49
Carlos Córdova S.Carlos Córdova S.
134
$endgroup$
add a comment |
$begingroup$
I'm trying to solve this problem, which basically states this:
Based on the system of equations of the variables $x, y$ and $z$:
$$begin{cases}
x y - 2sqrt{y} + 3 y z = 8 \
2 x y - 3sqrt{y} + 2 y z = 7 \
- x y + sqrt{y} + 2 y z = 4 \
end{cases}$$please tell which of the next alternatives are correct:
- The system is possible and determined
- The range of the extended matrix of the system is 2.
- On the transposed matrix of the coefficients, associated with the system: $a_{12} = -3$
- The coefficient matrix associated to the system is non invertible.
So, what I did was construct the extended matrix from the equation system given in the problem statement, replacing the variables as this:
- $a = xy$
- $b = sqrt{y}$
- $c = yz$
The matrix constructed is this:
begin{bmatrix}
1 & -2 & 3 & 8 \
2 & -3 & 2 & 7 \
-1 & 1 & 2 & 4
end{bmatrix}
Solving by gaussian method, the matrix is reduced to:
begin{bmatrix}
1 & -2 & 3 & 8 \
0 & 1 & -4 & -9 \
0 & 0 & 1 & 3
end{bmatrix}
So finally:
- $a = 5$
- $b = 3$
- $c = 3$
The values of $x, y$ and $z$ could be calculated using the previous equations, giving this values as results:
- $x = 5/9$
- $y = 9$
- $z = 1/3$
This is the far as I can go on solving this problem, my question basically is, how to transform this to a linear equation system of x, y and z, because I do have the values of $x, y$ and $z$ but the statements that I have to indicate if they are correct or not, are based on that matrix that I don't know how to build.
I was thinking that maybe this problem doesn't look for me to build such a matrix, instead, given that there exist a solution for the system proposed initially, I should deduct the veracity of the statements:
- First statement is correct as each variable has just one unique value.
- Second statement is incorrect because for each variable having one single value, the range of the matrix should be 3.
Second and third statement I have no idea how to answer to.
Please give me a hand on this.
Thank you a lot.
linear-algebra systems-of-equations
asked Jan 13 at 0:49
Carlos Córdova S.Carlos Córdova S.
134
$endgroup$
$begingroup$
It is important to include the entire question as text, instead of as images. I added the necessary MathJax for you.
$endgroup$
– Nominal Animal
Jan 13 at 0:56
$begingroup$
Thank you for including it!
$endgroup$
– Carlos Córdova S.
Jan 13 at 0:58
1
$begingroup$
I suspect that the matrix-related items refer to the coefficients of the linear "$a$-$b$-$c$" system you created by assigning $a=xy$, $b=sqrt{y}$, $c=yz$.
$endgroup$
– Blue
Jan 13 at 1:13
$begingroup$
It is not, the problem said "Based on the equation system of the variables "x-y-z"
$endgroup$
– Carlos Córdova S.
Jan 13 at 1:15
3
$begingroup$
@CarlosCórdovaS.: Well, the $x$-$y$-$z$ system is hopelessly non-linear, so there's no appropriate interpretation of "matrix of coefficients" for it. (The work you did in assigning $a$, $b$, $c$ seems exactly the correct approach.)
$endgroup$
– Blue
Jan 13 at 1:20
add a comment |
$begingroup$
I'm trying to solve this problem, which basically states this:
Based on the system of equations of the variables $x, y$ and $z$:
$$begin{cases}
x y - 2sqrt{y} + 3 y z = 8 \
2 x y - 3sqrt{y} + 2 y z = 7 \
- x y + sqrt{y} + 2 y z = 4 \
end{cases}$$please tell which of the next alternatives are correct:
- The system is possible and determined
- The range of the extended matrix of the system is 2.
- On the transposed matrix of the coefficients, associated with the system: $a_{12} = -3$
- The coefficient matrix associated to the system is non invertible.
So, what I did was construct the extended matrix from the equation system given in the problem statement, replacing the variables as this:
- $a = xy$
- $b = sqrt{y}$
- $c = yz$
The matrix constructed is this:
begin{bmatrix}
1 & -2 & 3 & 8 \
2 & -3 & 2 & 7 \
-1 & 1 & 2 & 4
end{bmatrix}
Solving by gaussian method, the matrix is reduced to:
begin{bmatrix}
1 & -2 & 3 & 8 \
0 & 1 & -4 & -9 \
0 & 0 & 1 & 3
end{bmatrix}
So finally:
- $a = 5$
- $b = 3$
- $c = 3$
The values of $x, y$ and $z$ could be calculated using the previous equations, giving this values as results:
- $x = 5/9$
- $y = 9$
- $z = 1/3$
This is the far as I can go on solving this problem, my question basically is, how to transform this to a linear equation system of x, y and z, because I do have the values of $x, y$ and $z$ but the statements that I have to indicate if they are correct or not, are based on that matrix that I don't know how to build.
I was thinking that maybe this problem doesn't look for me to build such a matrix, instead, given that there exist a solution for the system proposed initially, I should deduct the veracity of the statements:
- First statement is correct as each variable has just one unique value.
- Second statement is incorrect because for each variable having one single value, the range of the matrix should be 3.
Second and third statement I have no idea how to answer to.
Please give me a hand on this.
Thank you a lot.
linear-algebra systems-of-equations
asked Jan 13 at 0:49
Carlos Córdova S.Carlos Córdova S.
134
$endgroup$
I'm trying to solve this problem, which basically states this:
Based on the system of equations of the variables $x, y$ and $z$:
$$begin{cases}
x y - 2sqrt{y} + 3 y z = 8 \
2 x y - 3sqrt{y} + 2 y z = 7 \
- x y + sqrt{y} + 2 y z = 4 \
end{cases}$$please tell which of the next alternatives are correct:
- The system is possible and determined
- The range of the extended matrix of the system is 2.
- On the transposed matrix of the coefficients, associated with the system: $a_{12} = -3$
- The coefficient matrix associated to the system is non invertible.
So, what I did was construct the extended matrix from the equation system given in the problem statement, replacing the variables as this:
- $a = xy$
- $b = sqrt{y}$
- $c = yz$
The matrix constructed is this:
begin{bmatrix}
1 & -2 & 3 & 8 \
2 & -3 & 2 & 7 \
-1 & 1 & 2 & 4
end{bmatrix}
Solving by gaussian method, the matrix is reduced to:
begin{bmatrix}
1 & -2 & 3 & 8 \
0 & 1 & -4 & -9 \
0 & 0 & 1 & 3
end{bmatrix}
So finally:
- $a = 5$
- $b = 3$
- $c = 3$
The values of $x, y$ and $z$ could be calculated using the previous equations, giving this values as results:
- $x = 5/9$
- $y = 9$
- $z = 1/3$
This is the far as I can go on solving this problem, my question basically is, how to transform this to a linear equation system of x, y and z, because I do have the values of $x, y$ and $z$ but the statements that I have to indicate if they are correct or not, are based on that matrix that I don't know how to build.
I was thinking that maybe this problem doesn't look for me to build such a matrix, instead, given that there exist a solution for the system proposed initially, I should deduct the veracity of the statements:
- First statement is correct as each variable has just one unique value.
- Second statement is incorrect because for each variable having one single value, the range of the matrix should be 3.
Second and third statement I have no idea how to answer to.
Please give me a hand on this.
Thank you a lot.
linear-algebra systems-of-equations
linear-algebra systems-of-equations
asked Jan 13 at 0:49
Carlos Córdova S.Carlos Córdova S.
134
asked Jan 13 at 0:49
Carlos Córdova S.Carlos Córdova S.
134
edited Jan 13 at 1:15
Carlos Córdova S.
asked Jan 13 at 0:49
Carlos Córdova S.Carlos Córdova S.
134
asked Jan 13 at 0:49
Carlos Córdova S.Carlos Córdova S.
134
asked Jan 13 at 0:49
Carlos Córdova S.Carlos Córdova S.
134
134
$begingroup$
It is important to include the entire question as text, instead of as images. I added the necessary MathJax for you.
$endgroup$
– Nominal Animal
Jan 13 at 0:56
$begingroup$
Thank you for including it!
$endgroup$
– Carlos Córdova S.
Jan 13 at 0:58
1
$begingroup$
I suspect that the matrix-related items refer to the coefficients of the linear "$a$-$b$-$c$" system you created by assigning $a=xy$, $b=sqrt{y}$, $c=yz$.
$endgroup$
– Blue
Jan 13 at 1:13
$begingroup$
It is not, the problem said "Based on the equation system of the variables "x-y-z"
$endgroup$
– Carlos Córdova S.
Jan 13 at 1:15
3
$begingroup$
@CarlosCórdovaS.: Well, the $x$-$y$-$z$ system is hopelessly non-linear, so there's no appropriate interpretation of "matrix of coefficients" for it. (The work you did in assigning $a$, $b$, $c$ seems exactly the correct approach.)
$endgroup$
– Blue
Jan 13 at 1:20
add a comment |
$begingroup$
It is important to include the entire question as text, instead of as images. I added the necessary MathJax for you.
$endgroup$
– Nominal Animal
Jan 13 at 0:56
$begingroup$
Thank you for including it!
$endgroup$
– Carlos Córdova S.
Jan 13 at 0:58
1
$begingroup$
I suspect that the matrix-related items refer to the coefficients of the linear "$a$-$b$-$c$" system you created by assigning $a=xy$, $b=sqrt{y}$, $c=yz$.
$endgroup$
– Blue
Jan 13 at 1:13
$begingroup$
It is not, the problem said "Based on the equation system of the variables "x-y-z"
$endgroup$
– Carlos Córdova S.
Jan 13 at 1:15
3
$begingroup$
@CarlosCórdovaS.: Well, the $x$-$y$-$z$ system is hopelessly non-linear, so there's no appropriate interpretation of "matrix of coefficients" for it. (The work you did in assigning $a$, $b$, $c$ seems exactly the correct approach.)
$endgroup$
– Blue
Jan 13 at 1:20
$begingroup$
It is important to include the entire question as text, instead of as images. I added the necessary MathJax for you.
$endgroup$
– Nominal Animal
Jan 13 at 0:56
$begingroup$
It is important to include the entire question as text, instead of as images. I added the necessary MathJax for you.
$endgroup$
– Nominal Animal
Jan 13 at 0:56
$begingroup$
Thank you for including it!
$endgroup$
– Carlos Córdova S.
Jan 13 at 0:58
$begingroup$
Thank you for including it!
$endgroup$
– Carlos Córdova S.
Jan 13 at 0:58
1
1
$begingroup$
I suspect that the matrix-related items refer to the coefficients of the linear "$a$-$b$-$c$" system you created by assigning $a=xy$, $b=sqrt{y}$, $c=yz$.
$endgroup$
– Blue
Jan 13 at 1:13
$begingroup$
I suspect that the matrix-related items refer to the coefficients of the linear "$a$-$b$-$c$" system you created by assigning $a=xy$, $b=sqrt{y}$, $c=yz$.
$endgroup$
– Blue
Jan 13 at 1:13
$begingroup$
It is not, the problem said "Based on the equation system of the variables "x-y-z"
$endgroup$
– Carlos Córdova S.
Jan 13 at 1:15
$begingroup$
It is not, the problem said "Based on the equation system of the variables "x-y-z"
$endgroup$
– Carlos Córdova S.
Jan 13 at 1:15
3
3
$begingroup$
@CarlosCórdovaS.: Well, the $x$-$y$-$z$ system is hopelessly non-linear, so there's no appropriate interpretation of "matrix of coefficients" for it. (The work you did in assigning $a$, $b$, $c$ seems exactly the correct approach.)
$endgroup$
– Blue
Jan 13 at 1:20
$begingroup$
@CarlosCórdovaS.: Well, the $x$-$y$-$z$ system is hopelessly non-linear, so there's no appropriate interpretation of "matrix of coefficients" for it. (The work you did in assigning $a$, $b$, $c$ seems exactly the correct approach.)
$endgroup$
– Blue
Jan 13 at 1:20
add a comment |
0
active
oldest
votes
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
});
}
});
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%2f3071580%2fhow-could-this-this-equation-system-be-transformed-to-a-x-y-z-linear-equation-sy%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
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%2f3071580%2fhow-could-this-this-equation-system-be-transformed-to-a-x-y-z-linear-equation-sy%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
$begingroup$
It is important to include the entire question as text, instead of as images. I added the necessary MathJax for you.
$endgroup$
– Nominal Animal
Jan 13 at 0:56
$begingroup$
Thank you for including it!
$endgroup$
– Carlos Córdova S.
Jan 13 at 0:58
1
$begingroup$
I suspect that the matrix-related items refer to the coefficients of the linear "$a$-$b$-$c$" system you created by assigning $a=xy$, $b=sqrt{y}$, $c=yz$.
$endgroup$
– Blue
Jan 13 at 1:13
$begingroup$
It is not, the problem said "Based on the equation system of the variables "x-y-z"
$endgroup$
– Carlos Córdova S.
Jan 13 at 1:15
3
$begingroup$
@CarlosCórdovaS.: Well, the $x$-$y$-$z$ system is hopelessly non-linear, so there's no appropriate interpretation of "matrix of coefficients" for it. (The work you did in assigning $a$, $b$, $c$ seems exactly the correct approach.)
$endgroup$
– Blue
Jan 13 at 1:20