Defined row operations for making a matrix ,containing varibel coefficients, into RREF
$begingroup$
I am trying to solve for the inverse for a given $3x3$ matrix. The matrix looks like this:
begin{pmatrix}1&0&1\ :a&0&1\ :1&a&0end{pmatrix}
I have tried solving it through carrying out row operations on the matrix above whilest in parallel carrying out the same operations on a $3x3$ Identity matrix.
I am worried that I might be preforming "undefined" or "non-approved" row operations on $A$ and that it, is the cause of the incorrect result. What i got so far looks like this:
begin{pmatrix}1&0&1\ :::::a&0&1\ :::::1&a&0end{pmatrix} $R_2to R_2 -aR_1$
$R_3to R_3-R_1$
begin{pmatrix}1&0&1\ ::::::0&0&1-a\ ::::::0&a&-1end{pmatrix}
Assuming $aneq 1$
Switch $R_2leftrightarrow R_3$
begin{pmatrix}1&0&1\ :::::0&a&-1\ :::::0&0&1-aend{pmatrix}
$R_3to frac{1}{1-a}R_3$begin{pmatrix}1&0&1\ ::::::0&a&-1\ ::::::0&0&1end{pmatrix}
$R_2to R_2+R_3$
$R_1to R_1-R_3$
begin{pmatrix}1&0&0\ :::::::0&a&0\ :::::::0&0&1end{pmatrix}
$R_2to R_2cdot(1/a)$
begin{pmatrix}1&0&0\ :::::::0&1&0\ :::::::0&0&1end{pmatrix}
This it what i am getting, if i do the same elementry row operations in the same sequence to the Identity matrix i get an incorrect result. What am i doing wrong?
(Sorry about the messy notation, first time poster)
Kind regards and many thanks!
linear-algebra matrices
edited Nov 30 '18 at 17:11
Thomas Shelby
2,199220
asked Nov 30 '18 at 16:43
Student123451Student123451
1
$endgroup$
add a comment |
$begingroup$
I am trying to solve for the inverse for a given $3x3$ matrix. The matrix looks like this:
begin{pmatrix}1&0&1\ :a&0&1\ :1&a&0end{pmatrix}
I have tried solving it through carrying out row operations on the matrix above whilest in parallel carrying out the same operations on a $3x3$ Identity matrix.
I am worried that I might be preforming "undefined" or "non-approved" row operations on $A$ and that it, is the cause of the incorrect result. What i got so far looks like this:
begin{pmatrix}1&0&1\ :::::a&0&1\ :::::1&a&0end{pmatrix} $R_2to R_2 -aR_1$
$R_3to R_3-R_1$
begin{pmatrix}1&0&1\ ::::::0&0&1-a\ ::::::0&a&-1end{pmatrix}
Assuming $aneq 1$
Switch $R_2leftrightarrow R_3$
begin{pmatrix}1&0&1\ :::::0&a&-1\ :::::0&0&1-aend{pmatrix}
$R_3to frac{1}{1-a}R_3$begin{pmatrix}1&0&1\ ::::::0&a&-1\ ::::::0&0&1end{pmatrix}
$R_2to R_2+R_3$
$R_1to R_1-R_3$
begin{pmatrix}1&0&0\ :::::::0&a&0\ :::::::0&0&1end{pmatrix}
$R_2to R_2cdot(1/a)$
begin{pmatrix}1&0&0\ :::::::0&1&0\ :::::::0&0&1end{pmatrix}
This it what i am getting, if i do the same elementry row operations in the same sequence to the Identity matrix i get an incorrect result. What am i doing wrong?
(Sorry about the messy notation, first time poster)
Kind regards and many thanks!
linear-algebra matrices
edited Nov 30 '18 at 17:11
Thomas Shelby
2,199220
asked Nov 30 '18 at 16:43
Student123451Student123451
1
$endgroup$
add a comment |
$begingroup$
I am trying to solve for the inverse for a given $3x3$ matrix. The matrix looks like this:
begin{pmatrix}1&0&1\ :a&0&1\ :1&a&0end{pmatrix}
I have tried solving it through carrying out row operations on the matrix above whilest in parallel carrying out the same operations on a $3x3$ Identity matrix.
I am worried that I might be preforming "undefined" or "non-approved" row operations on $A$ and that it, is the cause of the incorrect result. What i got so far looks like this:
begin{pmatrix}1&0&1\ :::::a&0&1\ :::::1&a&0end{pmatrix} $R_2to R_2 -aR_1$
$R_3to R_3-R_1$
begin{pmatrix}1&0&1\ ::::::0&0&1-a\ ::::::0&a&-1end{pmatrix}
Assuming $aneq 1$
Switch $R_2leftrightarrow R_3$
begin{pmatrix}1&0&1\ :::::0&a&-1\ :::::0&0&1-aend{pmatrix}
$R_3to frac{1}{1-a}R_3$begin{pmatrix}1&0&1\ ::::::0&a&-1\ ::::::0&0&1end{pmatrix}
$R_2to R_2+R_3$
$R_1to R_1-R_3$
begin{pmatrix}1&0&0\ :::::::0&a&0\ :::::::0&0&1end{pmatrix}
$R_2to R_2cdot(1/a)$
begin{pmatrix}1&0&0\ :::::::0&1&0\ :::::::0&0&1end{pmatrix}
This it what i am getting, if i do the same elementry row operations in the same sequence to the Identity matrix i get an incorrect result. What am i doing wrong?
(Sorry about the messy notation, first time poster)
Kind regards and many thanks!
linear-algebra matrices
edited Nov 30 '18 at 17:11
Thomas Shelby
2,199220
asked Nov 30 '18 at 16:43
Student123451Student123451
1
$endgroup$
I am trying to solve for the inverse for a given $3x3$ matrix. The matrix looks like this:
begin{pmatrix}1&0&1\ :a&0&1\ :1&a&0end{pmatrix}
I have tried solving it through carrying out row operations on the matrix above whilest in parallel carrying out the same operations on a $3x3$ Identity matrix.
I am worried that I might be preforming "undefined" or "non-approved" row operations on $A$ and that it, is the cause of the incorrect result. What i got so far looks like this:
begin{pmatrix}1&0&1\ :::::a&0&1\ :::::1&a&0end{pmatrix} $R_2to R_2 -aR_1$
$R_3to R_3-R_1$
begin{pmatrix}1&0&1\ ::::::0&0&1-a\ ::::::0&a&-1end{pmatrix}
Assuming $aneq 1$
Switch $R_2leftrightarrow R_3$
begin{pmatrix}1&0&1\ :::::0&a&-1\ :::::0&0&1-aend{pmatrix}
$R_3to frac{1}{1-a}R_3$begin{pmatrix}1&0&1\ ::::::0&a&-1\ ::::::0&0&1end{pmatrix}
$R_2to R_2+R_3$
$R_1to R_1-R_3$
begin{pmatrix}1&0&0\ :::::::0&a&0\ :::::::0&0&1end{pmatrix}
$R_2to R_2cdot(1/a)$
begin{pmatrix}1&0&0\ :::::::0&1&0\ :::::::0&0&1end{pmatrix}
This it what i am getting, if i do the same elementry row operations in the same sequence to the Identity matrix i get an incorrect result. What am i doing wrong?
(Sorry about the messy notation, first time poster)
Kind regards and many thanks!
linear-algebra matrices
linear-algebra matrices
edited Nov 30 '18 at 17:11
Thomas Shelby
2,199220
asked Nov 30 '18 at 16:43
Student123451Student123451
1
edited Nov 30 '18 at 17:11
Thomas Shelby
2,199220
asked Nov 30 '18 at 16:43
Student123451Student123451
1
edited Nov 30 '18 at 17:11
Thomas Shelby
2,199220
edited Nov 30 '18 at 17:11
Thomas Shelby
2,199220
edited Nov 30 '18 at 17:11
Thomas Shelby
2,199220
2,199220
asked Nov 30 '18 at 16:43
Student123451Student123451
1
asked Nov 30 '18 at 16:43
Student123451Student123451
1
asked Nov 30 '18 at 16:43
Student123451Student123451
1
1
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$a≠0,1$ or the matrix will be singular. The row transformations look okay. The inverse comes out to be $begin{pmatrix}frac1{1-a}&frac1{a-1}&0\ :frac1{a(a-1)}&frac1{a(1-a)}&frac1a\ :frac a{a-1}&frac1{1-a}&0end{pmatrix}$.
answered Nov 30 '18 at 17:17
Shubham JohriShubham Johri
4,666717
$endgroup$
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%2f3020302%2fdefined-row-operations-for-making-a-matrix-containing-varibel-coefficients-int%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
$begingroup$
$a≠0,1$ or the matrix will be singular. The row transformations look okay. The inverse comes out to be $begin{pmatrix}frac1{1-a}&frac1{a-1}&0\ :frac1{a(a-1)}&frac1{a(1-a)}&frac1a\ :frac a{a-1}&frac1{1-a}&0end{pmatrix}$.
answered Nov 30 '18 at 17:17
Shubham JohriShubham Johri
4,666717
$endgroup$
add a comment |
$begingroup$
$a≠0,1$ or the matrix will be singular. The row transformations look okay. The inverse comes out to be $begin{pmatrix}frac1{1-a}&frac1{a-1}&0\ :frac1{a(a-1)}&frac1{a(1-a)}&frac1a\ :frac a{a-1}&frac1{1-a}&0end{pmatrix}$.
answered Nov 30 '18 at 17:17
Shubham JohriShubham Johri
4,666717
$endgroup$
add a comment |
$begingroup$
$a≠0,1$ or the matrix will be singular. The row transformations look okay. The inverse comes out to be $begin{pmatrix}frac1{1-a}&frac1{a-1}&0\ :frac1{a(a-1)}&frac1{a(1-a)}&frac1a\ :frac a{a-1}&frac1{1-a}&0end{pmatrix}$.
answered Nov 30 '18 at 17:17
Shubham JohriShubham Johri
4,666717
$endgroup$
$a≠0,1$ or the matrix will be singular. The row transformations look okay. The inverse comes out to be $begin{pmatrix}frac1{1-a}&frac1{a-1}&0\ :frac1{a(a-1)}&frac1{a(1-a)}&frac1a\ :frac a{a-1}&frac1{1-a}&0end{pmatrix}$.
answered Nov 30 '18 at 17:17
Shubham JohriShubham Johri
4,666717
answered Nov 30 '18 at 17:17
Shubham JohriShubham Johri
4,666717
answered Nov 30 '18 at 17:17
Shubham JohriShubham Johri
4,666717
answered Nov 30 '18 at 17:17
Shubham JohriShubham Johri
4,666717
4,666717
add a comment |
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.
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%2f3020302%2fdefined-row-operations-for-making-a-matrix-containing-varibel-coefficients-int%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
