Solving the homogeneous system associated with a matrix
So i have this matrix
$$
begin{pmatrix}
0 & 0 & 0 \
3 & 3 & -6 \
0 & 0 & 0 \
end{pmatrix}
$$
and I want to solve the homogeneous system associated with it
I re-write the matrix as
$$
begin{pmatrix}
0 & 0 & 0 \
3 & 3 & -6 \
0 & 0 & 0 \
end{pmatrix} · (x,y,z)= $$
begin{pmatrix}
0 & 0 & 0 \
0 & 0 & 0 \
0 & 0 & 0 \
end{pmatrix}
$$
$$
However, I do not know how to proceed from here as I've started studying matrices as a solo student from a couple of days.
The solutions are the linear combinations of (1,1,1) and (-1,1,0).
I am confused on how to get to these solutions which are the linear combinations of (1,1,1) and (-1,1,0) and if anybody could show me the steps the simplest way possible it would be great ! Thanks!
linear-algebra matrices matrix-equations
asked Dec 1 '18 at 14:08
BM97
758
add a comment |
So i have this matrix
$$
begin{pmatrix}
0 & 0 & 0 \
3 & 3 & -6 \
0 & 0 & 0 \
end{pmatrix}
$$
and I want to solve the homogeneous system associated with it
I re-write the matrix as
$$
begin{pmatrix}
0 & 0 & 0 \
3 & 3 & -6 \
0 & 0 & 0 \
end{pmatrix} · (x,y,z)= $$
begin{pmatrix}
0 & 0 & 0 \
0 & 0 & 0 \
0 & 0 & 0 \
end{pmatrix}
$$
$$
However, I do not know how to proceed from here as I've started studying matrices as a solo student from a couple of days.
The solutions are the linear combinations of (1,1,1) and (-1,1,0).
I am confused on how to get to these solutions which are the linear combinations of (1,1,1) and (-1,1,0) and if anybody could show me the steps the simplest way possible it would be great ! Thanks!
linear-algebra matrices matrix-equations
asked Dec 1 '18 at 14:08
BM97
758
Matrices have no "solutions": they are just matrices, so what do you really mean?
– DonAntonio
Dec 1 '18 at 14:14
I'm sorry. I want to solving the homogeneous system associated with the matrix which gives me the linear combinations of (1,0,0) and (-1,1,0)
– BM97
Dec 1 '18 at 14:21
The r.h.s. of your system should be a column vector.
– Bernard
Dec 1 '18 at 14:36
Yes, they are, i just couldn't write them in the right for
– BM97
Dec 1 '18 at 14:39
add a comment |
So i have this matrix
$$
begin{pmatrix}
0 & 0 & 0 \
3 & 3 & -6 \
0 & 0 & 0 \
end{pmatrix}
$$
and I want to solve the homogeneous system associated with it
I re-write the matrix as
$$
begin{pmatrix}
0 & 0 & 0 \
3 & 3 & -6 \
0 & 0 & 0 \
end{pmatrix} · (x,y,z)= $$
begin{pmatrix}
0 & 0 & 0 \
0 & 0 & 0 \
0 & 0 & 0 \
end{pmatrix}
$$
$$
However, I do not know how to proceed from here as I've started studying matrices as a solo student from a couple of days.
The solutions are the linear combinations of (1,1,1) and (-1,1,0).
I am confused on how to get to these solutions which are the linear combinations of (1,1,1) and (-1,1,0) and if anybody could show me the steps the simplest way possible it would be great ! Thanks!
linear-algebra matrices matrix-equations
asked Dec 1 '18 at 14:08
BM97
758
So i have this matrix
$$
begin{pmatrix}
0 & 0 & 0 \
3 & 3 & -6 \
0 & 0 & 0 \
end{pmatrix}
$$
and I want to solve the homogeneous system associated with it
I re-write the matrix as
$$
begin{pmatrix}
0 & 0 & 0 \
3 & 3 & -6 \
0 & 0 & 0 \
end{pmatrix} · (x,y,z)= $$
begin{pmatrix}
0 & 0 & 0 \
0 & 0 & 0 \
0 & 0 & 0 \
end{pmatrix}
$$
$$
However, I do not know how to proceed from here as I've started studying matrices as a solo student from a couple of days.
The solutions are the linear combinations of (1,1,1) and (-1,1,0).
I am confused on how to get to these solutions which are the linear combinations of (1,1,1) and (-1,1,0) and if anybody could show me the steps the simplest way possible it would be great ! Thanks!
linear-algebra matrices matrix-equations
linear-algebra matrices matrix-equations
asked Dec 1 '18 at 14:08
BM97
758
asked Dec 1 '18 at 14:08
BM97
758
edited Dec 1 '18 at 14:20
asked Dec 1 '18 at 14:08
BM97
758
asked Dec 1 '18 at 14:08
BM97
758
asked Dec 1 '18 at 14:08
BM97
758
758
Matrices have no "solutions": they are just matrices, so what do you really mean?
– DonAntonio
Dec 1 '18 at 14:14
I'm sorry. I want to solving the homogeneous system associated with the matrix which gives me the linear combinations of (1,0,0) and (-1,1,0)
– BM97
Dec 1 '18 at 14:21
The r.h.s. of your system should be a column vector.
– Bernard
Dec 1 '18 at 14:36
Yes, they are, i just couldn't write them in the right for
– BM97
Dec 1 '18 at 14:39
add a comment |
Matrices have no "solutions": they are just matrices, so what do you really mean?
– DonAntonio
Dec 1 '18 at 14:14
I'm sorry. I want to solving the homogeneous system associated with the matrix which gives me the linear combinations of (1,0,0) and (-1,1,0)
– BM97
Dec 1 '18 at 14:21
The r.h.s. of your system should be a column vector.
– Bernard
Dec 1 '18 at 14:36
Yes, they are, i just couldn't write them in the right for
– BM97
Dec 1 '18 at 14:39
Matrices have no "solutions": they are just matrices, so what do you really mean?
– DonAntonio
Dec 1 '18 at 14:14
Matrices have no "solutions": they are just matrices, so what do you really mean?
– DonAntonio
Dec 1 '18 at 14:14
I'm sorry. I want to solving the homogeneous system associated with the matrix which gives me the linear combinations of (1,0,0) and (-1,1,0)
– BM97
Dec 1 '18 at 14:21
I'm sorry. I want to solving the homogeneous system associated with the matrix which gives me the linear combinations of (1,0,0) and (-1,1,0)
– BM97
Dec 1 '18 at 14:21
The r.h.s. of your system should be a column vector.
– Bernard
Dec 1 '18 at 14:36
The r.h.s. of your system should be a column vector.
– Bernard
Dec 1 '18 at 14:36
Yes, they are, i just couldn't write them in the right for
– BM97
Dec 1 '18 at 14:39
Yes, they are, i just couldn't write them in the right for
– BM97
Dec 1 '18 at 14:39
add a comment |
2 Answers
2
active
oldest
votes
The homogeneous system of linear equations whose coefficients matrix is your matrix is equivalent with the homogeneous system whose coefficients matrix is
$$begin{pmatrix}
0&0&0\0&0&0\
1&1&-2end{pmatrix}implies x+y-2z=0implies text{Span},left{;begin{pmatrix}1\-1\0end{pmatrix};,;;begin{pmatrix}2\0\1end{pmatrix};right}$$
Observe that both vectors you mention at the end of your question belong to the solution space (the span) above, and they both are linearly independents, so you can choose those two as basis or else you can take the basis I wrote above.
answered Dec 1 '18 at 14:37
DonAntonio
177k1491225
I am sorry but I didn't understand the last Part. I understood where you get x+y-2z=0 and that is really great. However, i am a little confused from here. With this equation x+y−2z=0, how do i proceed to get the exact results (1,1,1) and (-1,0,0)?
– BM97
Dec 1 '18 at 14:48
My confusion arises since I've been studying vectors only from a couple of days and i don't really grasp what you wrote above, i'm sorry
– BM97
Dec 1 '18 at 14:50
@BM97 I really don't know then how to explain this to you. Perhaps someone else...This kind of exercises, as far as I know, is given when one has already studied the basics of linear (or vector) spaces, linear combinations, dependency and etc., and certainly not after just two days...
– DonAntonio
Dec 1 '18 at 15:04
Okay , i got that the span is equal to (1,-1,0), (2,0,1).... would that be okay as an answer?? Its as fair as the vectors (1,1,1) and (-1,1,0) correct?
– BM97
Dec 1 '18 at 16:06
Meaning that if i write (1,1,1)(-1,1,0) or (1,-1,0)(2,0,1) its the same thing correct?
– BM97
Dec 1 '18 at 16:08
|
show 1 more comment
It's quite simple: your system is equivalent to the single equation
$$x+y-2z=0iff z=tfrac12(x+y)$$
so the subspace $S$ of solutions is isomorphic to $K^2$ ($K$ is the base field) via the isomorphism:
begin{align}
K^2&longrightarrow Ssubset K^3\
(x,y)&longmapsto (x,y,tfrac12(x+y))
end{align}
Now an isomorphism maps a basis onto a basis, so determine a basis of $K^2$ that maps onto the indicated basis.
answered Dec 1 '18 at 14:44
Bernard
118k639112
I am sorry but I didn't understand the last Part. I understood where you get x+y-2z=0 and that is really great. However, i am a little confused from here. With this equation x+y−2z=0, how do i proceed to get the exact results (1,1,1) and (-1,0,0)?
– BM97
Dec 1 '18 at 14:51
Which values of $(x,y)$ map to $(1,1,1)$ and $(-1,10)$? Is the set of these values a basis of $K^2$?
– Bernard
Dec 1 '18 at 14:55
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
});
}
});
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%2f3021382%2fsolving-the-homogeneous-system-associated-with-a-matrix%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
The homogeneous system of linear equations whose coefficients matrix is your matrix is equivalent with the homogeneous system whose coefficients matrix is
$$begin{pmatrix}
0&0&0\0&0&0\
1&1&-2end{pmatrix}implies x+y-2z=0implies text{Span},left{;begin{pmatrix}1\-1\0end{pmatrix};,;;begin{pmatrix}2\0\1end{pmatrix};right}$$
Observe that both vectors you mention at the end of your question belong to the solution space (the span) above, and they both are linearly independents, so you can choose those two as basis or else you can take the basis I wrote above.
answered Dec 1 '18 at 14:37
DonAntonio
177k1491225
I am sorry but I didn't understand the last Part. I understood where you get x+y-2z=0 and that is really great. However, i am a little confused from here. With this equation x+y−2z=0, how do i proceed to get the exact results (1,1,1) and (-1,0,0)?
– BM97
Dec 1 '18 at 14:48
My confusion arises since I've been studying vectors only from a couple of days and i don't really grasp what you wrote above, i'm sorry
– BM97
Dec 1 '18 at 14:50
@BM97 I really don't know then how to explain this to you. Perhaps someone else...This kind of exercises, as far as I know, is given when one has already studied the basics of linear (or vector) spaces, linear combinations, dependency and etc., and certainly not after just two days...
– DonAntonio
Dec 1 '18 at 15:04
Okay , i got that the span is equal to (1,-1,0), (2,0,1).... would that be okay as an answer?? Its as fair as the vectors (1,1,1) and (-1,1,0) correct?
– BM97
Dec 1 '18 at 16:06
Meaning that if i write (1,1,1)(-1,1,0) or (1,-1,0)(2,0,1) its the same thing correct?
– BM97
Dec 1 '18 at 16:08
|
show 1 more comment
The homogeneous system of linear equations whose coefficients matrix is your matrix is equivalent with the homogeneous system whose coefficients matrix is
$$begin{pmatrix}
0&0&0\0&0&0\
1&1&-2end{pmatrix}implies x+y-2z=0implies text{Span},left{;begin{pmatrix}1\-1\0end{pmatrix};,;;begin{pmatrix}2\0\1end{pmatrix};right}$$
Observe that both vectors you mention at the end of your question belong to the solution space (the span) above, and they both are linearly independents, so you can choose those two as basis or else you can take the basis I wrote above.
answered Dec 1 '18 at 14:37
DonAntonio
177k1491225
I am sorry but I didn't understand the last Part. I understood where you get x+y-2z=0 and that is really great. However, i am a little confused from here. With this equation x+y−2z=0, how do i proceed to get the exact results (1,1,1) and (-1,0,0)?
– BM97
Dec 1 '18 at 14:48
My confusion arises since I've been studying vectors only from a couple of days and i don't really grasp what you wrote above, i'm sorry
– BM97
Dec 1 '18 at 14:50
@BM97 I really don't know then how to explain this to you. Perhaps someone else...This kind of exercises, as far as I know, is given when one has already studied the basics of linear (or vector) spaces, linear combinations, dependency and etc., and certainly not after just two days...
– DonAntonio
Dec 1 '18 at 15:04
Okay , i got that the span is equal to (1,-1,0), (2,0,1).... would that be okay as an answer?? Its as fair as the vectors (1,1,1) and (-1,1,0) correct?
– BM97
Dec 1 '18 at 16:06
Meaning that if i write (1,1,1)(-1,1,0) or (1,-1,0)(2,0,1) its the same thing correct?
– BM97
Dec 1 '18 at 16:08
|
show 1 more comment
The homogeneous system of linear equations whose coefficients matrix is your matrix is equivalent with the homogeneous system whose coefficients matrix is
$$begin{pmatrix}
0&0&0\0&0&0\
1&1&-2end{pmatrix}implies x+y-2z=0implies text{Span},left{;begin{pmatrix}1\-1\0end{pmatrix};,;;begin{pmatrix}2\0\1end{pmatrix};right}$$
Observe that both vectors you mention at the end of your question belong to the solution space (the span) above, and they both are linearly independents, so you can choose those two as basis or else you can take the basis I wrote above.
answered Dec 1 '18 at 14:37
DonAntonio
177k1491225
The homogeneous system of linear equations whose coefficients matrix is your matrix is equivalent with the homogeneous system whose coefficients matrix is
$$begin{pmatrix}
0&0&0\0&0&0\
1&1&-2end{pmatrix}implies x+y-2z=0implies text{Span},left{;begin{pmatrix}1\-1\0end{pmatrix};,;;begin{pmatrix}2\0\1end{pmatrix};right}$$
Observe that both vectors you mention at the end of your question belong to the solution space (the span) above, and they both are linearly independents, so you can choose those two as basis or else you can take the basis I wrote above.
answered Dec 1 '18 at 14:37
DonAntonio
177k1491225
answered Dec 1 '18 at 14:37
DonAntonio
177k1491225
answered Dec 1 '18 at 14:37
DonAntonio
177k1491225
answered Dec 1 '18 at 14:37
DonAntonio
177k1491225
177k1491225
I am sorry but I didn't understand the last Part. I understood where you get x+y-2z=0 and that is really great. However, i am a little confused from here. With this equation x+y−2z=0, how do i proceed to get the exact results (1,1,1) and (-1,0,0)?
– BM97
Dec 1 '18 at 14:48
My confusion arises since I've been studying vectors only from a couple of days and i don't really grasp what you wrote above, i'm sorry
– BM97
Dec 1 '18 at 14:50
@BM97 I really don't know then how to explain this to you. Perhaps someone else...This kind of exercises, as far as I know, is given when one has already studied the basics of linear (or vector) spaces, linear combinations, dependency and etc., and certainly not after just two days...
– DonAntonio
Dec 1 '18 at 15:04
Okay , i got that the span is equal to (1,-1,0), (2,0,1).... would that be okay as an answer?? Its as fair as the vectors (1,1,1) and (-1,1,0) correct?
– BM97
Dec 1 '18 at 16:06
Meaning that if i write (1,1,1)(-1,1,0) or (1,-1,0)(2,0,1) its the same thing correct?
– BM97
Dec 1 '18 at 16:08
|
show 1 more comment
I am sorry but I didn't understand the last Part. I understood where you get x+y-2z=0 and that is really great. However, i am a little confused from here. With this equation x+y−2z=0, how do i proceed to get the exact results (1,1,1) and (-1,0,0)?
– BM97
Dec 1 '18 at 14:48
My confusion arises since I've been studying vectors only from a couple of days and i don't really grasp what you wrote above, i'm sorry
– BM97
Dec 1 '18 at 14:50
@BM97 I really don't know then how to explain this to you. Perhaps someone else...This kind of exercises, as far as I know, is given when one has already studied the basics of linear (or vector) spaces, linear combinations, dependency and etc., and certainly not after just two days...
– DonAntonio
Dec 1 '18 at 15:04
Okay , i got that the span is equal to (1,-1,0), (2,0,1).... would that be okay as an answer?? Its as fair as the vectors (1,1,1) and (-1,1,0) correct?
– BM97
Dec 1 '18 at 16:06
Meaning that if i write (1,1,1)(-1,1,0) or (1,-1,0)(2,0,1) its the same thing correct?
– BM97
Dec 1 '18 at 16:08
I am sorry but I didn't understand the last Part. I understood where you get x+y-2z=0 and that is really great. However, i am a little confused from here. With this equation x+y−2z=0, how do i proceed to get the exact results (1,1,1) and (-1,0,0)?
– BM97
Dec 1 '18 at 14:48
I am sorry but I didn't understand the last Part. I understood where you get x+y-2z=0 and that is really great. However, i am a little confused from here. With this equation x+y−2z=0, how do i proceed to get the exact results (1,1,1) and (-1,0,0)?
– BM97
Dec 1 '18 at 14:48
My confusion arises since I've been studying vectors only from a couple of days and i don't really grasp what you wrote above, i'm sorry
– BM97
Dec 1 '18 at 14:50
My confusion arises since I've been studying vectors only from a couple of days and i don't really grasp what you wrote above, i'm sorry
– BM97
Dec 1 '18 at 14:50
@BM97 I really don't know then how to explain this to you. Perhaps someone else...This kind of exercises, as far as I know, is given when one has already studied the basics of linear (or vector) spaces, linear combinations, dependency and etc., and certainly not after just two days...
– DonAntonio
Dec 1 '18 at 15:04
@BM97 I really don't know then how to explain this to you. Perhaps someone else...This kind of exercises, as far as I know, is given when one has already studied the basics of linear (or vector) spaces, linear combinations, dependency and etc., and certainly not after just two days...
– DonAntonio
Dec 1 '18 at 15:04
Okay , i got that the span is equal to (1,-1,0), (2,0,1).... would that be okay as an answer?? Its as fair as the vectors (1,1,1) and (-1,1,0) correct?
– BM97
Dec 1 '18 at 16:06
Okay , i got that the span is equal to (1,-1,0), (2,0,1).... would that be okay as an answer?? Its as fair as the vectors (1,1,1) and (-1,1,0) correct?
– BM97
Dec 1 '18 at 16:06
Meaning that if i write (1,1,1)(-1,1,0) or (1,-1,0)(2,0,1) its the same thing correct?
– BM97
Dec 1 '18 at 16:08
Meaning that if i write (1,1,1)(-1,1,0) or (1,-1,0)(2,0,1) its the same thing correct?
– BM97
Dec 1 '18 at 16:08
|
show 1 more comment
It's quite simple: your system is equivalent to the single equation
$$x+y-2z=0iff z=tfrac12(x+y)$$
so the subspace $S$ of solutions is isomorphic to $K^2$ ($K$ is the base field) via the isomorphism:
begin{align}
K^2&longrightarrow Ssubset K^3\
(x,y)&longmapsto (x,y,tfrac12(x+y))
end{align}
Now an isomorphism maps a basis onto a basis, so determine a basis of $K^2$ that maps onto the indicated basis.
answered Dec 1 '18 at 14:44
Bernard
118k639112
I am sorry but I didn't understand the last Part. I understood where you get x+y-2z=0 and that is really great. However, i am a little confused from here. With this equation x+y−2z=0, how do i proceed to get the exact results (1,1,1) and (-1,0,0)?
– BM97
Dec 1 '18 at 14:51
Which values of $(x,y)$ map to $(1,1,1)$ and $(-1,10)$? Is the set of these values a basis of $K^2$?
– Bernard
Dec 1 '18 at 14:55
add a comment |
It's quite simple: your system is equivalent to the single equation
$$x+y-2z=0iff z=tfrac12(x+y)$$
so the subspace $S$ of solutions is isomorphic to $K^2$ ($K$ is the base field) via the isomorphism:
begin{align}
K^2&longrightarrow Ssubset K^3\
(x,y)&longmapsto (x,y,tfrac12(x+y))
end{align}
Now an isomorphism maps a basis onto a basis, so determine a basis of $K^2$ that maps onto the indicated basis.
answered Dec 1 '18 at 14:44
Bernard
118k639112
I am sorry but I didn't understand the last Part. I understood where you get x+y-2z=0 and that is really great. However, i am a little confused from here. With this equation x+y−2z=0, how do i proceed to get the exact results (1,1,1) and (-1,0,0)?
– BM97
Dec 1 '18 at 14:51
Which values of $(x,y)$ map to $(1,1,1)$ and $(-1,10)$? Is the set of these values a basis of $K^2$?
– Bernard
Dec 1 '18 at 14:55
add a comment |
It's quite simple: your system is equivalent to the single equation
$$x+y-2z=0iff z=tfrac12(x+y)$$
so the subspace $S$ of solutions is isomorphic to $K^2$ ($K$ is the base field) via the isomorphism:
begin{align}
K^2&longrightarrow Ssubset K^3\
(x,y)&longmapsto (x,y,tfrac12(x+y))
end{align}
Now an isomorphism maps a basis onto a basis, so determine a basis of $K^2$ that maps onto the indicated basis.
answered Dec 1 '18 at 14:44
Bernard
118k639112
It's quite simple: your system is equivalent to the single equation
$$x+y-2z=0iff z=tfrac12(x+y)$$
so the subspace $S$ of solutions is isomorphic to $K^2$ ($K$ is the base field) via the isomorphism:
begin{align}
K^2&longrightarrow Ssubset K^3\
(x,y)&longmapsto (x,y,tfrac12(x+y))
end{align}
Now an isomorphism maps a basis onto a basis, so determine a basis of $K^2$ that maps onto the indicated basis.
answered Dec 1 '18 at 14:44
Bernard
118k639112
answered Dec 1 '18 at 14:44
Bernard
118k639112
answered Dec 1 '18 at 14:44
Bernard
118k639112
answered Dec 1 '18 at 14:44
Bernard
118k639112
118k639112
I am sorry but I didn't understand the last Part. I understood where you get x+y-2z=0 and that is really great. However, i am a little confused from here. With this equation x+y−2z=0, how do i proceed to get the exact results (1,1,1) and (-1,0,0)?
– BM97
Dec 1 '18 at 14:51
Which values of $(x,y)$ map to $(1,1,1)$ and $(-1,10)$? Is the set of these values a basis of $K^2$?
– Bernard
Dec 1 '18 at 14:55
add a comment |
I am sorry but I didn't understand the last Part. I understood where you get x+y-2z=0 and that is really great. However, i am a little confused from here. With this equation x+y−2z=0, how do i proceed to get the exact results (1,1,1) and (-1,0,0)?
– BM97
Dec 1 '18 at 14:51
Which values of $(x,y)$ map to $(1,1,1)$ and $(-1,10)$? Is the set of these values a basis of $K^2$?
– Bernard
Dec 1 '18 at 14:55
I am sorry but I didn't understand the last Part. I understood where you get x+y-2z=0 and that is really great. However, i am a little confused from here. With this equation x+y−2z=0, how do i proceed to get the exact results (1,1,1) and (-1,0,0)?
– BM97
Dec 1 '18 at 14:51
I am sorry but I didn't understand the last Part. I understood where you get x+y-2z=0 and that is really great. However, i am a little confused from here. With this equation x+y−2z=0, how do i proceed to get the exact results (1,1,1) and (-1,0,0)?
– BM97
Dec 1 '18 at 14:51
Which values of $(x,y)$ map to $(1,1,1)$ and $(-1,10)$? Is the set of these values a basis of $K^2$?
– Bernard
Dec 1 '18 at 14:55
Which values of $(x,y)$ map to $(1,1,1)$ and $(-1,10)$? Is the set of these values a basis of $K^2$?
– Bernard
Dec 1 '18 at 14:55
add a comment |
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.
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%2f3021382%2fsolving-the-homogeneous-system-associated-with-a-matrix%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
Matrices have no "solutions": they are just matrices, so what do you really mean?
– DonAntonio
Dec 1 '18 at 14:14
I'm sorry. I want to solving the homogeneous system associated with the matrix which gives me the linear combinations of (1,0,0) and (-1,1,0)
– BM97
Dec 1 '18 at 14:21
The r.h.s. of your system should be a column vector.
– Bernard
Dec 1 '18 at 14:36
Yes, they are, i just couldn't write them in the right for
– BM97
Dec 1 '18 at 14:39