How to determine if the augmented matrix has no solution?
$begingroup$
"Convert the following system of linear equations into an augmented matrix and then solve it"
$$x_1 + x_2 - x_3 - x_4 = 7$$
$$2x_1 - x_2 + 3x_3 + x_4 = 1$$
$$x_1 - 5x_2 + 9x_3 - x_4 = 3$$
My approach:
So the augmented matrix, A, is
$$aug A = begin{bmatrix}1&1&-1&-1&7\2&-1&3&1&1\1&-5&9&-1&3end{bmatrix}$$
Then we perform elementary row operations to reduce the matrix into row echelon form:
$$B = begin{bmatrix}1&1&-1&-1&7\0&-3&5&3&-13\0&-6&10&0&-4end{bmatrix}$$
$$C = begin{bmatrix}1&1&-1&-1&7\0&-3&5&3&-13\0&0&0&-6&22end{bmatrix}$$
Matrix C is our system of linear equations in echelon form. From the last row of matrix C,
$$-6x_4 =22$$
$$x_4 = frac {-11}3$$
Now we let x_3 = t for t is any real number
Then we solve the two equations from the first two rows of matrix C,
$$x_2 = frac 13(-2 - 5t)$$
$$x_1 = frac 83t + 2frac 23 $$
SO this should be the solutions to the equations. However, according to the answer key the solution is the empty set. I inputted the regular matrix into a determinant calculator and indeed the determinant was zero. However, I managed to get a solution to the system of linear equations. Can someone tell me where I went wrong?
linear-algebra matrices systems-of-equations
edited Jan 14 at 21:31
Shubham Johri
5,427818
asked Jan 14 at 21:23
Wilson GuoWilson Guo
514
$endgroup$
|
show 1 more comment
$begingroup$
"Convert the following system of linear equations into an augmented matrix and then solve it"
$$x_1 + x_2 - x_3 - x_4 = 7$$
$$2x_1 - x_2 + 3x_3 + x_4 = 1$$
$$x_1 - 5x_2 + 9x_3 - x_4 = 3$$
My approach:
So the augmented matrix, A, is
$$aug A = begin{bmatrix}1&1&-1&-1&7\2&-1&3&1&1\1&-5&9&-1&3end{bmatrix}$$
Then we perform elementary row operations to reduce the matrix into row echelon form:
$$B = begin{bmatrix}1&1&-1&-1&7\0&-3&5&3&-13\0&-6&10&0&-4end{bmatrix}$$
$$C = begin{bmatrix}1&1&-1&-1&7\0&-3&5&3&-13\0&0&0&-6&22end{bmatrix}$$
Matrix C is our system of linear equations in echelon form. From the last row of matrix C,
$$-6x_4 =22$$
$$x_4 = frac {-11}3$$
Now we let x_3 = t for t is any real number
Then we solve the two equations from the first two rows of matrix C,
$$x_2 = frac 13(-2 - 5t)$$
$$x_1 = frac 83t + 2frac 23 $$
SO this should be the solutions to the equations. However, according to the answer key the solution is the empty set. I inputted the regular matrix into a determinant calculator and indeed the determinant was zero. However, I managed to get a solution to the system of linear equations. Can someone tell me where I went wrong?
linear-algebra matrices systems-of-equations
edited Jan 14 at 21:31
Shubham Johri
5,427818
asked Jan 14 at 21:23
Wilson GuoWilson Guo
514
$endgroup$
$begingroup$
Made a few small edits, you can revert if that was not your intention. Seemed like you made typos in $B_{35}$ and $C_{34}$. Also looks like you are missing a negative sign in $x_2$
$endgroup$
– Shubham Johri
Jan 14 at 21:33
1
$begingroup$
What is the determinant of a non-square matrix?
$endgroup$
– amd
Jan 14 at 21:33
1
$begingroup$
Answer keys have been known to be wrong. Have you double-checked that you’ve copied the problem correctly?
$endgroup$
– amd
Jan 14 at 21:35
$begingroup$
Your solution is correct and the system, as written, has infinitely many solutions. You get no solution if the coefficient of $x_4$ in the last equation is $5$ instead of $-1$.
$endgroup$
– egreg
Jan 14 at 21:40
1
$begingroup$
@ShubhamJohri I didn't check the final details, but the system indeed has infinitely many solutions contrary to what the answer key claims. Indeed the solution should have: $x_2=(2+5t)/3$, $x_1=(8-2t)/3$.
$endgroup$
– egreg
Jan 14 at 21:52
|
show 1 more comment
$begingroup$
"Convert the following system of linear equations into an augmented matrix and then solve it"
$$x_1 + x_2 - x_3 - x_4 = 7$$
$$2x_1 - x_2 + 3x_3 + x_4 = 1$$
$$x_1 - 5x_2 + 9x_3 - x_4 = 3$$
My approach:
So the augmented matrix, A, is
$$aug A = begin{bmatrix}1&1&-1&-1&7\2&-1&3&1&1\1&-5&9&-1&3end{bmatrix}$$
Then we perform elementary row operations to reduce the matrix into row echelon form:
$$B = begin{bmatrix}1&1&-1&-1&7\0&-3&5&3&-13\0&-6&10&0&-4end{bmatrix}$$
$$C = begin{bmatrix}1&1&-1&-1&7\0&-3&5&3&-13\0&0&0&-6&22end{bmatrix}$$
Matrix C is our system of linear equations in echelon form. From the last row of matrix C,
$$-6x_4 =22$$
$$x_4 = frac {-11}3$$
Now we let x_3 = t for t is any real number
Then we solve the two equations from the first two rows of matrix C,
$$x_2 = frac 13(-2 - 5t)$$
$$x_1 = frac 83t + 2frac 23 $$
SO this should be the solutions to the equations. However, according to the answer key the solution is the empty set. I inputted the regular matrix into a determinant calculator and indeed the determinant was zero. However, I managed to get a solution to the system of linear equations. Can someone tell me where I went wrong?
linear-algebra matrices systems-of-equations
edited Jan 14 at 21:31
Shubham Johri
5,427818
asked Jan 14 at 21:23
Wilson GuoWilson Guo
514
$endgroup$
"Convert the following system of linear equations into an augmented matrix and then solve it"
$$x_1 + x_2 - x_3 - x_4 = 7$$
$$2x_1 - x_2 + 3x_3 + x_4 = 1$$
$$x_1 - 5x_2 + 9x_3 - x_4 = 3$$
My approach:
So the augmented matrix, A, is
$$aug A = begin{bmatrix}1&1&-1&-1&7\2&-1&3&1&1\1&-5&9&-1&3end{bmatrix}$$
Then we perform elementary row operations to reduce the matrix into row echelon form:
$$B = begin{bmatrix}1&1&-1&-1&7\0&-3&5&3&-13\0&-6&10&0&-4end{bmatrix}$$
$$C = begin{bmatrix}1&1&-1&-1&7\0&-3&5&3&-13\0&0&0&-6&22end{bmatrix}$$
Matrix C is our system of linear equations in echelon form. From the last row of matrix C,
$$-6x_4 =22$$
$$x_4 = frac {-11}3$$
Now we let x_3 = t for t is any real number
Then we solve the two equations from the first two rows of matrix C,
$$x_2 = frac 13(-2 - 5t)$$
$$x_1 = frac 83t + 2frac 23 $$
SO this should be the solutions to the equations. However, according to the answer key the solution is the empty set. I inputted the regular matrix into a determinant calculator and indeed the determinant was zero. However, I managed to get a solution to the system of linear equations. Can someone tell me where I went wrong?
linear-algebra matrices systems-of-equations
linear-algebra matrices systems-of-equations
edited Jan 14 at 21:31
Shubham Johri
5,427818
asked Jan 14 at 21:23
Wilson GuoWilson Guo
514
edited Jan 14 at 21:31
Shubham Johri
5,427818
asked Jan 14 at 21:23
Wilson GuoWilson Guo
514
edited Jan 14 at 21:31
Shubham Johri
5,427818
edited Jan 14 at 21:31
Shubham Johri
5,427818
edited Jan 14 at 21:31
Shubham Johri
5,427818
5,427818
asked Jan 14 at 21:23
Wilson GuoWilson Guo
514
asked Jan 14 at 21:23
Wilson GuoWilson Guo
514
asked Jan 14 at 21:23
Wilson GuoWilson Guo
514
514
$begingroup$
Made a few small edits, you can revert if that was not your intention. Seemed like you made typos in $B_{35}$ and $C_{34}$. Also looks like you are missing a negative sign in $x_2$
$endgroup$
– Shubham Johri
Jan 14 at 21:33
1
$begingroup$
What is the determinant of a non-square matrix?
$endgroup$
– amd
Jan 14 at 21:33
1
$begingroup$
Answer keys have been known to be wrong. Have you double-checked that you’ve copied the problem correctly?
$endgroup$
– amd
Jan 14 at 21:35
$begingroup$
Your solution is correct and the system, as written, has infinitely many solutions. You get no solution if the coefficient of $x_4$ in the last equation is $5$ instead of $-1$.
$endgroup$
– egreg
Jan 14 at 21:40
1
$begingroup$
@ShubhamJohri I didn't check the final details, but the system indeed has infinitely many solutions contrary to what the answer key claims. Indeed the solution should have: $x_2=(2+5t)/3$, $x_1=(8-2t)/3$.
$endgroup$
– egreg
Jan 14 at 21:52
|
show 1 more comment
$begingroup$
Made a few small edits, you can revert if that was not your intention. Seemed like you made typos in $B_{35}$ and $C_{34}$. Also looks like you are missing a negative sign in $x_2$
$endgroup$
– Shubham Johri
Jan 14 at 21:33
1
$begingroup$
What is the determinant of a non-square matrix?
$endgroup$
– amd
Jan 14 at 21:33
1
$begingroup$
Answer keys have been known to be wrong. Have you double-checked that you’ve copied the problem correctly?
$endgroup$
– amd
Jan 14 at 21:35
$begingroup$
Your solution is correct and the system, as written, has infinitely many solutions. You get no solution if the coefficient of $x_4$ in the last equation is $5$ instead of $-1$.
$endgroup$
– egreg
Jan 14 at 21:40
1
$begingroup$
@ShubhamJohri I didn't check the final details, but the system indeed has infinitely many solutions contrary to what the answer key claims. Indeed the solution should have: $x_2=(2+5t)/3$, $x_1=(8-2t)/3$.
$endgroup$
– egreg
Jan 14 at 21:52
$begingroup$
Made a few small edits, you can revert if that was not your intention. Seemed like you made typos in $B_{35}$ and $C_{34}$. Also looks like you are missing a negative sign in $x_2$
$endgroup$
– Shubham Johri
Jan 14 at 21:33
$begingroup$
Made a few small edits, you can revert if that was not your intention. Seemed like you made typos in $B_{35}$ and $C_{34}$. Also looks like you are missing a negative sign in $x_2$
$endgroup$
– Shubham Johri
Jan 14 at 21:33
1
1
$begingroup$
What is the determinant of a non-square matrix?
$endgroup$
– amd
Jan 14 at 21:33
$begingroup$
What is the determinant of a non-square matrix?
$endgroup$
– amd
Jan 14 at 21:33
1
1
$begingroup$
Answer keys have been known to be wrong. Have you double-checked that you’ve copied the problem correctly?
$endgroup$
– amd
Jan 14 at 21:35
$begingroup$
Answer keys have been known to be wrong. Have you double-checked that you’ve copied the problem correctly?
$endgroup$
– amd
Jan 14 at 21:35
$begingroup$
Your solution is correct and the system, as written, has infinitely many solutions. You get no solution if the coefficient of $x_4$ in the last equation is $5$ instead of $-1$.
$endgroup$
– egreg
Jan 14 at 21:40
$begingroup$
Your solution is correct and the system, as written, has infinitely many solutions. You get no solution if the coefficient of $x_4$ in the last equation is $5$ instead of $-1$.
$endgroup$
– egreg
Jan 14 at 21:40
1
1
$begingroup$
@ShubhamJohri I didn't check the final details, but the system indeed has infinitely many solutions contrary to what the answer key claims. Indeed the solution should have: $x_2=(2+5t)/3$, $x_1=(8-2t)/3$.
$endgroup$
– egreg
Jan 14 at 21:52
$begingroup$
@ShubhamJohri I didn't check the final details, but the system indeed has infinitely many solutions contrary to what the answer key claims. Indeed the solution should have: $x_2=(2+5t)/3$, $x_1=(8-2t)/3$.
$endgroup$
– egreg
Jan 14 at 21:52
|
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%2f3073766%2fhow-to-determine-if-the-augmented-matrix-has-no-solution%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%2f3073766%2fhow-to-determine-if-the-augmented-matrix-has-no-solution%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
$begingroup$
Made a few small edits, you can revert if that was not your intention. Seemed like you made typos in $B_{35}$ and $C_{34}$. Also looks like you are missing a negative sign in $x_2$
$endgroup$
– Shubham Johri
Jan 14 at 21:33
1
$begingroup$
What is the determinant of a non-square matrix?
$endgroup$
– amd
Jan 14 at 21:33
1
$begingroup$
Answer keys have been known to be wrong. Have you double-checked that you’ve copied the problem correctly?
$endgroup$
– amd
Jan 14 at 21:35
$begingroup$
Your solution is correct and the system, as written, has infinitely many solutions. You get no solution if the coefficient of $x_4$ in the last equation is $5$ instead of $-1$.
$endgroup$
– egreg
Jan 14 at 21:40
1
$begingroup$
@ShubhamJohri I didn't check the final details, but the system indeed has infinitely many solutions contrary to what the answer key claims. Indeed the solution should have: $x_2=(2+5t)/3$, $x_1=(8-2t)/3$.
$endgroup$
– egreg
Jan 14 at 21:52