Mixing
solutions to systems of linear equations and RREF
$begingroup$
following an example from here : https://en.wikipedia.org/wiki/Augmented_matrix
Given $A=pmatrix{1 ;;;1 ;;;2 \ 1 ;;;1 ;;;1 \ 2;;;2 ;;;2}, X=pmatrix{x\y\z}, B=pmatrix{3\1\2}$ forming the linear system:
$ AX=B$
Performing Gaussian elimination (by hand), starting with this augmented matrix:
$pmatrix{1 ;;;1 ;;;2 ;;;3 \ 1 ;;;1 ;;;1 ;;;1 \ 2;;;2 ;;;2 ;;;2}$
After doing the following Row operations:
L3=L3-2L2
L2=L2-L1
L2=-L2
I stopped at this point:
$pmatrix{1 ;;;1 ;;;2 ;;;3 \ 0 ;;;0 ;;;1 ;;;2 \ 0;;;0 ;;;0 ;;;0}$
it is clear that z=2
I decided to set x as a free variable and let x=t
then i get:
${x=t,y=1-t,z=2}$
E.g, let $t=10$,
then $X_{10}= pmatrix{10\ 1-10 \ 2} =pmatrix{10 \ -9 \ 2}$
and $AX_{10}=pmatrix{5\ 3 \ 6}$
The true RREF matrix is one step later with L1=L1-2L2:
$pmatrix{;;;;1 ;;;;;; 1 ;;;;; 0 ;;;; -1 \ 0 ;;;;; 0 ;;;;; 1 ;;;;; 2 \ 0 ;;;;; 0 ;;;;; 0 ;;;;; 0}$
then again if x is the free param. we have this time:
${x=t,y=-1-t,z=2}$
(Please, note that i get this time $y=-1-t$ instead of $y=1-t$ before!)
E.g, let $t=10$,
then $X_{{10}_{RREF}}= pmatrix{10\ -1-10 \ 2} =pmatrix{10 \ -11 \ 2}$
and this time 'it works' as $AX_{{10}_{RREF}} = pmatrix{3\1\2}$ which is indeed equal to $B$.
My question is: why does it make a difference to go to the FULL (or true?) RREF form? Why can't I stop 1 step before? It seems very strange, like if the system was 'wrong' at the point before L1=L1-2L2 ...
I am probably missing some fundamental point, but I hope that an answer to this will allow me to understand more about linear algebra (i know i could just say 'ok one must do the full RREF' but i m interested in the 'Why?')
linear-algebra systems-of-equations
asked Jan 28 at 13:42
MachupicchuMachupicchu
239
$endgroup$
|
show 1 more comment
$begingroup$
following an example from here : https://en.wikipedia.org/wiki/Augmented_matrix
Given $A=pmatrix{1 ;;;1 ;;;2 \ 1 ;;;1 ;;;1 \ 2;;;2 ;;;2}, X=pmatrix{x\y\z}, B=pmatrix{3\1\2}$ forming the linear system:
$ AX=B$
Performing Gaussian elimination (by hand), starting with this augmented matrix:
$pmatrix{1 ;;;1 ;;;2 ;;;3 \ 1 ;;;1 ;;;1 ;;;1 \ 2;;;2 ;;;2 ;;;2}$
After doing the following Row operations:
L3=L3-2L2
L2=L2-L1
L2=-L2
I stopped at this point:
$pmatrix{1 ;;;1 ;;;2 ;;;3 \ 0 ;;;0 ;;;1 ;;;2 \ 0;;;0 ;;;0 ;;;0}$
it is clear that z=2
I decided to set x as a free variable and let x=t
then i get:
${x=t,y=1-t,z=2}$
E.g, let $t=10$,
then $X_{10}= pmatrix{10\ 1-10 \ 2} =pmatrix{10 \ -9 \ 2}$
and $AX_{10}=pmatrix{5\ 3 \ 6}$
The true RREF matrix is one step later with L1=L1-2L2:
$pmatrix{;;;;1 ;;;;;; 1 ;;;;; 0 ;;;; -1 \ 0 ;;;;; 0 ;;;;; 1 ;;;;; 2 \ 0 ;;;;; 0 ;;;;; 0 ;;;;; 0}$
then again if x is the free param. we have this time:
${x=t,y=-1-t,z=2}$
(Please, note that i get this time $y=-1-t$ instead of $y=1-t$ before!)
E.g, let $t=10$,
then $X_{{10}_{RREF}}= pmatrix{10\ -1-10 \ 2} =pmatrix{10 \ -11 \ 2}$
and this time 'it works' as $AX_{{10}_{RREF}} = pmatrix{3\1\2}$ which is indeed equal to $B$.
My question is: why does it make a difference to go to the FULL (or true?) RREF form? Why can't I stop 1 step before? It seems very strange, like if the system was 'wrong' at the point before L1=L1-2L2 ...
I am probably missing some fundamental point, but I hope that an answer to this will allow me to understand more about linear algebra (i know i could just say 'ok one must do the full RREF' but i m interested in the 'Why?')
linear-algebra systems-of-equations
asked Jan 28 at 13:42
MachupicchuMachupicchu
239
$endgroup$
$begingroup$
You have $x+y+2z = 3$, which becomes $t+y + 4 = 3$, hence $y = -1-t$.
$endgroup$
– Catalin Zara
Jan 28 at 13:51
$begingroup$
Your first solution with $y=1-t$ does NOT solve the first row.
$endgroup$
– maxmilgram
Jan 28 at 13:53
$begingroup$
oh thanks so much!! I forgot the the 2 was the z factor! of course its 2*2... thanks the fact that z=2 and this factor was the confusion..
$endgroup$
– Machupicchu
Jan 28 at 13:53
$begingroup$
i suddenly thought there was something special with RREF but no... of course you can stop whenever you want and the system should always be in a coherent form .. right?
$endgroup$
– Machupicchu
Jan 28 at 13:55
$begingroup$
You say, correctly, that when you multiply your "$X_{10}$" by A you get $begin{bmatrix} 5 \ 3 \ 6end{bmatrix}$. But the problem was to find X such that AX was equal to $begin{bmatrix} 5 \ 3 \ 6end{bmatrix}$. That should have told you immediately that your "$X_{10}$" was wrong!
$endgroup$
– user247327
Jan 28 at 14:05
|
show 1 more comment
$begingroup$
following an example from here : https://en.wikipedia.org/wiki/Augmented_matrix
Given $A=pmatrix{1 ;;;1 ;;;2 \ 1 ;;;1 ;;;1 \ 2;;;2 ;;;2}, X=pmatrix{x\y\z}, B=pmatrix{3\1\2}$ forming the linear system:
$ AX=B$
Performing Gaussian elimination (by hand), starting with this augmented matrix:
$pmatrix{1 ;;;1 ;;;2 ;;;3 \ 1 ;;;1 ;;;1 ;;;1 \ 2;;;2 ;;;2 ;;;2}$
After doing the following Row operations:
L3=L3-2L2
L2=L2-L1
L2=-L2
I stopped at this point:
$pmatrix{1 ;;;1 ;;;2 ;;;3 \ 0 ;;;0 ;;;1 ;;;2 \ 0;;;0 ;;;0 ;;;0}$
it is clear that z=2
I decided to set x as a free variable and let x=t
then i get:
${x=t,y=1-t,z=2}$
E.g, let $t=10$,
then $X_{10}= pmatrix{10\ 1-10 \ 2} =pmatrix{10 \ -9 \ 2}$
and $AX_{10}=pmatrix{5\ 3 \ 6}$
The true RREF matrix is one step later with L1=L1-2L2:
$pmatrix{;;;;1 ;;;;;; 1 ;;;;; 0 ;;;; -1 \ 0 ;;;;; 0 ;;;;; 1 ;;;;; 2 \ 0 ;;;;; 0 ;;;;; 0 ;;;;; 0}$
then again if x is the free param. we have this time:
${x=t,y=-1-t,z=2}$
(Please, note that i get this time $y=-1-t$ instead of $y=1-t$ before!)
E.g, let $t=10$,
then $X_{{10}_{RREF}}= pmatrix{10\ -1-10 \ 2} =pmatrix{10 \ -11 \ 2}$
and this time 'it works' as $AX_{{10}_{RREF}} = pmatrix{3\1\2}$ which is indeed equal to $B$.
My question is: why does it make a difference to go to the FULL (or true?) RREF form? Why can't I stop 1 step before? It seems very strange, like if the system was 'wrong' at the point before L1=L1-2L2 ...
I am probably missing some fundamental point, but I hope that an answer to this will allow me to understand more about linear algebra (i know i could just say 'ok one must do the full RREF' but i m interested in the 'Why?')
linear-algebra systems-of-equations
asked Jan 28 at 13:42
MachupicchuMachupicchu
239
$endgroup$
following an example from here : https://en.wikipedia.org/wiki/Augmented_matrix
Given $A=pmatrix{1 ;;;1 ;;;2 \ 1 ;;;1 ;;;1 \ 2;;;2 ;;;2}, X=pmatrix{x\y\z}, B=pmatrix{3\1\2}$ forming the linear system:
$ AX=B$
Performing Gaussian elimination (by hand), starting with this augmented matrix:
$pmatrix{1 ;;;1 ;;;2 ;;;3 \ 1 ;;;1 ;;;1 ;;;1 \ 2;;;2 ;;;2 ;;;2}$
After doing the following Row operations:
L3=L3-2L2
L2=L2-L1
L2=-L2
I stopped at this point:
$pmatrix{1 ;;;1 ;;;2 ;;;3 \ 0 ;;;0 ;;;1 ;;;2 \ 0;;;0 ;;;0 ;;;0}$
it is clear that z=2
I decided to set x as a free variable and let x=t
then i get:
${x=t,y=1-t,z=2}$
E.g, let $t=10$,
then $X_{10}= pmatrix{10\ 1-10 \ 2} =pmatrix{10 \ -9 \ 2}$
and $AX_{10}=pmatrix{5\ 3 \ 6}$
The true RREF matrix is one step later with L1=L1-2L2:
$pmatrix{;;;;1 ;;;;;; 1 ;;;;; 0 ;;;; -1 \ 0 ;;;;; 0 ;;;;; 1 ;;;;; 2 \ 0 ;;;;; 0 ;;;;; 0 ;;;;; 0}$
then again if x is the free param. we have this time:
${x=t,y=-1-t,z=2}$
(Please, note that i get this time $y=-1-t$ instead of $y=1-t$ before!)
E.g, let $t=10$,
then $X_{{10}_{RREF}}= pmatrix{10\ -1-10 \ 2} =pmatrix{10 \ -11 \ 2}$
and this time 'it works' as $AX_{{10}_{RREF}} = pmatrix{3\1\2}$ which is indeed equal to $B$.
My question is: why does it make a difference to go to the FULL (or true?) RREF form? Why can't I stop 1 step before? It seems very strange, like if the system was 'wrong' at the point before L1=L1-2L2 ...
I am probably missing some fundamental point, but I hope that an answer to this will allow me to understand more about linear algebra (i know i could just say 'ok one must do the full RREF' but i m interested in the 'Why?')
linear-algebra systems-of-equations
linear-algebra systems-of-equations
asked Jan 28 at 13:42
MachupicchuMachupicchu
239
asked Jan 28 at 13:42
MachupicchuMachupicchu
239
edited Jan 28 at 13:47
Machupicchu
asked Jan 28 at 13:42
MachupicchuMachupicchu
239
asked Jan 28 at 13:42
MachupicchuMachupicchu
239
asked Jan 28 at 13:42
MachupicchuMachupicchu
239
239
$begingroup$
You have $x+y+2z = 3$, which becomes $t+y + 4 = 3$, hence $y = -1-t$.
$endgroup$
– Catalin Zara
Jan 28 at 13:51
$begingroup$
Your first solution with $y=1-t$ does NOT solve the first row.
$endgroup$
– maxmilgram
Jan 28 at 13:53
$begingroup$
oh thanks so much!! I forgot the the 2 was the z factor! of course its 2*2... thanks the fact that z=2 and this factor was the confusion..
$endgroup$
– Machupicchu
Jan 28 at 13:53
$begingroup$
i suddenly thought there was something special with RREF but no... of course you can stop whenever you want and the system should always be in a coherent form .. right?
$endgroup$
– Machupicchu
Jan 28 at 13:55
$begingroup$
You say, correctly, that when you multiply your "$X_{10}$" by A you get $begin{bmatrix} 5 \ 3 \ 6end{bmatrix}$. But the problem was to find X such that AX was equal to $begin{bmatrix} 5 \ 3 \ 6end{bmatrix}$. That should have told you immediately that your "$X_{10}$" was wrong!
$endgroup$
– user247327
Jan 28 at 14:05
|
show 1 more comment
$begingroup$
You have $x+y+2z = 3$, which becomes $t+y + 4 = 3$, hence $y = -1-t$.
$endgroup$
– Catalin Zara
Jan 28 at 13:51
$begingroup$
Your first solution with $y=1-t$ does NOT solve the first row.
$endgroup$
– maxmilgram
Jan 28 at 13:53
$begingroup$
oh thanks so much!! I forgot the the 2 was the z factor! of course its 2*2... thanks the fact that z=2 and this factor was the confusion..
$endgroup$
– Machupicchu
Jan 28 at 13:53
$begingroup$
i suddenly thought there was something special with RREF but no... of course you can stop whenever you want and the system should always be in a coherent form .. right?
$endgroup$
– Machupicchu
Jan 28 at 13:55
$begingroup$
You say, correctly, that when you multiply your "$X_{10}$" by A you get $begin{bmatrix} 5 \ 3 \ 6end{bmatrix}$. But the problem was to find X such that AX was equal to $begin{bmatrix} 5 \ 3 \ 6end{bmatrix}$. That should have told you immediately that your "$X_{10}$" was wrong!
$endgroup$
– user247327
Jan 28 at 14:05
$begingroup$
You have $x+y+2z = 3$, which becomes $t+y + 4 = 3$, hence $y = -1-t$.
$endgroup$
– Catalin Zara
Jan 28 at 13:51
$begingroup$
You have $x+y+2z = 3$, which becomes $t+y + 4 = 3$, hence $y = -1-t$.
$endgroup$
– Catalin Zara
Jan 28 at 13:51
$begingroup$
Your first solution with $y=1-t$ does NOT solve the first row.
$endgroup$
– maxmilgram
Jan 28 at 13:53
$begingroup$
Your first solution with $y=1-t$ does NOT solve the first row.
$endgroup$
– maxmilgram
Jan 28 at 13:53
$begingroup$
oh thanks so much!! I forgot the the 2 was the z factor! of course its 2*2... thanks the fact that z=2 and this factor was the confusion..
$endgroup$
– Machupicchu
Jan 28 at 13:53
$begingroup$
oh thanks so much!! I forgot the the 2 was the z factor! of course its 2*2... thanks the fact that z=2 and this factor was the confusion..
$endgroup$
– Machupicchu
Jan 28 at 13:53
$begingroup$
i suddenly thought there was something special with RREF but no... of course you can stop whenever you want and the system should always be in a coherent form .. right?
$endgroup$
– Machupicchu
Jan 28 at 13:55
$begingroup$
i suddenly thought there was something special with RREF but no... of course you can stop whenever you want and the system should always be in a coherent form .. right?
$endgroup$
– Machupicchu
Jan 28 at 13:55
$begingroup$
You say, correctly, that when you multiply your "$X_{10}$" by A you get $begin{bmatrix} 5 \ 3 \ 6end{bmatrix}$. But the problem was to find X such that AX was equal to $begin{bmatrix} 5 \ 3 \ 6end{bmatrix}$. That should have told you immediately that your "$X_{10}$" was wrong!
$endgroup$
– user247327
Jan 28 at 14:05
$begingroup$
You say, correctly, that when you multiply your "$X_{10}$" by A you get $begin{bmatrix} 5 \ 3 \ 6end{bmatrix}$. But the problem was to find X such that AX was equal to $begin{bmatrix} 5 \ 3 \ 6end{bmatrix}$. That should have told you immediately that your "$X_{10}$" was wrong!
$endgroup$
– user247327
Jan 28 at 14:05
|
show 1 more 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%2f3090866%2fsolutions-to-systems-of-linear-equations-and-rref%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%2f3090866%2fsolutions-to-systems-of-linear-equations-and-rref%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
$begingroup$
You have $x+y+2z = 3$, which becomes $t+y + 4 = 3$, hence $y = -1-t$.
$endgroup$
– Catalin Zara
Jan 28 at 13:51
$begingroup$
Your first solution with $y=1-t$ does NOT solve the first row.
$endgroup$
– maxmilgram
Jan 28 at 13:53
$begingroup$
oh thanks so much!! I forgot the the 2 was the z factor! of course its 2*2... thanks the fact that z=2 and this factor was the confusion..
$endgroup$
– Machupicchu
Jan 28 at 13:53
$begingroup$
i suddenly thought there was something special with RREF but no... of course you can stop whenever you want and the system should always be in a coherent form .. right?
$endgroup$
– Machupicchu
Jan 28 at 13:55
$begingroup$
You say, correctly, that when you multiply your "$X_{10}$" by A you get $begin{bmatrix} 5 \ 3 \ 6end{bmatrix}$. But the problem was to find X such that AX was equal to $begin{bmatrix} 5 \ 3 \ 6end{bmatrix}$. That should have told you immediately that your "$X_{10}$" was wrong!
$endgroup$
– user247327
Jan 28 at 14:05