Mixing
Consider a system AX = B, where B ≠ 0. Let X and Y satisfy this system. Find all constants a and b such...
$begingroup$
I would like to know if the following approach is correct:
Given X and Y satisfy AX = B.
Hence, AX = B, AY = B.
X = Y.
Therefore, A(aX+bY) = A(aX+bX) = B. As AX = B, a + b must equal 1.
This approach seems very wrong to me at the X = Y step. Consider a system of equations:
3a + 2b + c = 1
a + 2b + c = 1
Where A = (a b c), X = (3 2 1)$^T$, Y = (1 2 3)$^T$, B = 1. Obviously, X ≠ Y.
However, the conclusion is correct (a + b = 1). Is there something wrong with this approach - it just accidentally leads to the correct answer - or my intuition is incorrect?
linear-algebra matrices systems-of-equations
asked Jan 30 at 8:39
TigerHixTigerHix
227
$endgroup$
add a comment |
$begingroup$
I would like to know if the following approach is correct:
Given X and Y satisfy AX = B.
Hence, AX = B, AY = B.
X = Y.
Therefore, A(aX+bY) = A(aX+bX) = B. As AX = B, a + b must equal 1.
This approach seems very wrong to me at the X = Y step. Consider a system of equations:
3a + 2b + c = 1
a + 2b + c = 1
Where A = (a b c), X = (3 2 1)$^T$, Y = (1 2 3)$^T$, B = 1. Obviously, X ≠ Y.
However, the conclusion is correct (a + b = 1). Is there something wrong with this approach - it just accidentally leads to the correct answer - or my intuition is incorrect?
linear-algebra matrices systems-of-equations
asked Jan 30 at 8:39
TigerHixTigerHix
227
$endgroup$
add a comment |
$begingroup$
I would like to know if the following approach is correct:
Given X and Y satisfy AX = B.
Hence, AX = B, AY = B.
X = Y.
Therefore, A(aX+bY) = A(aX+bX) = B. As AX = B, a + b must equal 1.
This approach seems very wrong to me at the X = Y step. Consider a system of equations:
3a + 2b + c = 1
a + 2b + c = 1
Where A = (a b c), X = (3 2 1)$^T$, Y = (1 2 3)$^T$, B = 1. Obviously, X ≠ Y.
However, the conclusion is correct (a + b = 1). Is there something wrong with this approach - it just accidentally leads to the correct answer - or my intuition is incorrect?
linear-algebra matrices systems-of-equations
asked Jan 30 at 8:39
TigerHixTigerHix
227
$endgroup$
I would like to know if the following approach is correct:
Given X and Y satisfy AX = B.
Hence, AX = B, AY = B.
X = Y.
Therefore, A(aX+bY) = A(aX+bX) = B. As AX = B, a + b must equal 1.
This approach seems very wrong to me at the X = Y step. Consider a system of equations:
3a + 2b + c = 1
a + 2b + c = 1
Where A = (a b c), X = (3 2 1)$^T$, Y = (1 2 3)$^T$, B = 1. Obviously, X ≠ Y.
However, the conclusion is correct (a + b = 1). Is there something wrong with this approach - it just accidentally leads to the correct answer - or my intuition is incorrect?
linear-algebra matrices systems-of-equations
linear-algebra matrices systems-of-equations
asked Jan 30 at 8:39
TigerHixTigerHix
227
asked Jan 30 at 8:39
TigerHixTigerHix
227
asked Jan 30 at 8:39
TigerHixTigerHix
227
asked Jan 30 at 8:39
TigerHixTigerHix
227
asked Jan 30 at 8:39
TigerHixTigerHix
227
227
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The approach is wrong. There is no reason to assume that $X=Y$. You made no justification of why the equation $X=Y$ should be true.
A better approach is this. Let $X,Y$ both satisfy the system, that is, let $AX=AY=B$. Also, let $aX+bY$ also satisfy the system. Now, because it satisfies the system, we know that $A(aX+bY)=B$. Further, we know that $A(aX+bY)=aAx+bAY$. Can you continue from here?
answered Jan 30 at 8:44
5xum5xum
91.8k394161
$endgroup$
$begingroup$
It makes much more sense now! I assume the steps are aAX + bAY = aAX + bAX (since AX = AY) = (a+b) AX = B. Compare with AX = B, we get a + b = 1. Thanks!
$endgroup$
– TigerHix
Jan 30 at 8:48
$begingroup$
However, why can't I derive X = Y by using AX = AY, A$^{-1}$AX = A$^{-1}$AY, X = Y?
$endgroup$
– TigerHix
Jan 30 at 8:48
1
$begingroup$
@TigerHix How do you know $A^{-1}$ exists?
$endgroup$
– 5xum
Jan 30 at 8:49
$begingroup$
I get it now. A is not a square matrix so $A^{-1}$ can't exist. That clears it up, thanks a lot!
$endgroup$
– TigerHix
Jan 30 at 8:51
2
$begingroup$
@TigerHix $A$ could be a square matrix. But you don't know if it is. Also, even if $A$ is a square matrix, you don't know if $A$ is invertible.
$endgroup$
– 5xum
Jan 30 at 8:59
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%2f3093262%2fconsider-a-system-ax-b-where-b-%25cc%25b8-0-let-x-and-y-satisfy-this-system-find-al%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
$begingroup$
The approach is wrong. There is no reason to assume that $X=Y$. You made no justification of why the equation $X=Y$ should be true.
A better approach is this. Let $X,Y$ both satisfy the system, that is, let $AX=AY=B$. Also, let $aX+bY$ also satisfy the system. Now, because it satisfies the system, we know that $A(aX+bY)=B$. Further, we know that $A(aX+bY)=aAx+bAY$. Can you continue from here?
answered Jan 30 at 8:44
5xum5xum
91.8k394161
$endgroup$
$begingroup$
It makes much more sense now! I assume the steps are aAX + bAY = aAX + bAX (since AX = AY) = (a+b) AX = B. Compare with AX = B, we get a + b = 1. Thanks!
$endgroup$
– TigerHix
Jan 30 at 8:48
$begingroup$
However, why can't I derive X = Y by using AX = AY, A$^{-1}$AX = A$^{-1}$AY, X = Y?
$endgroup$
– TigerHix
Jan 30 at 8:48
1
$begingroup$
@TigerHix How do you know $A^{-1}$ exists?
$endgroup$
– 5xum
Jan 30 at 8:49
$begingroup$
I get it now. A is not a square matrix so $A^{-1}$ can't exist. That clears it up, thanks a lot!
$endgroup$
– TigerHix
Jan 30 at 8:51
2
$begingroup$
@TigerHix $A$ could be a square matrix. But you don't know if it is. Also, even if $A$ is a square matrix, you don't know if $A$ is invertible.
$endgroup$
– 5xum
Jan 30 at 8:59
add a comment |
$begingroup$
The approach is wrong. There is no reason to assume that $X=Y$. You made no justification of why the equation $X=Y$ should be true.
A better approach is this. Let $X,Y$ both satisfy the system, that is, let $AX=AY=B$. Also, let $aX+bY$ also satisfy the system. Now, because it satisfies the system, we know that $A(aX+bY)=B$. Further, we know that $A(aX+bY)=aAx+bAY$. Can you continue from here?
answered Jan 30 at 8:44
5xum5xum
91.8k394161
$endgroup$
$begingroup$
It makes much more sense now! I assume the steps are aAX + bAY = aAX + bAX (since AX = AY) = (a+b) AX = B. Compare with AX = B, we get a + b = 1. Thanks!
$endgroup$
– TigerHix
Jan 30 at 8:48
$begingroup$
However, why can't I derive X = Y by using AX = AY, A$^{-1}$AX = A$^{-1}$AY, X = Y?
$endgroup$
– TigerHix
Jan 30 at 8:48
1
$begingroup$
@TigerHix How do you know $A^{-1}$ exists?
$endgroup$
– 5xum
Jan 30 at 8:49
$begingroup$
I get it now. A is not a square matrix so $A^{-1}$ can't exist. That clears it up, thanks a lot!
$endgroup$
– TigerHix
Jan 30 at 8:51
2
$begingroup$
@TigerHix $A$ could be a square matrix. But you don't know if it is. Also, even if $A$ is a square matrix, you don't know if $A$ is invertible.
$endgroup$
– 5xum
Jan 30 at 8:59
add a comment |
$begingroup$
The approach is wrong. There is no reason to assume that $X=Y$. You made no justification of why the equation $X=Y$ should be true.
A better approach is this. Let $X,Y$ both satisfy the system, that is, let $AX=AY=B$. Also, let $aX+bY$ also satisfy the system. Now, because it satisfies the system, we know that $A(aX+bY)=B$. Further, we know that $A(aX+bY)=aAx+bAY$. Can you continue from here?
answered Jan 30 at 8:44
5xum5xum
91.8k394161
$endgroup$
The approach is wrong. There is no reason to assume that $X=Y$. You made no justification of why the equation $X=Y$ should be true.
A better approach is this. Let $X,Y$ both satisfy the system, that is, let $AX=AY=B$. Also, let $aX+bY$ also satisfy the system. Now, because it satisfies the system, we know that $A(aX+bY)=B$. Further, we know that $A(aX+bY)=aAx+bAY$. Can you continue from here?
answered Jan 30 at 8:44
5xum5xum
91.8k394161
answered Jan 30 at 8:44
5xum5xum
91.8k394161
answered Jan 30 at 8:44
5xum5xum
91.8k394161
answered Jan 30 at 8:44
5xum5xum
91.8k394161
91.8k394161
$begingroup$
It makes much more sense now! I assume the steps are aAX + bAY = aAX + bAX (since AX = AY) = (a+b) AX = B. Compare with AX = B, we get a + b = 1. Thanks!
$endgroup$
– TigerHix
Jan 30 at 8:48
$begingroup$
However, why can't I derive X = Y by using AX = AY, A$^{-1}$AX = A$^{-1}$AY, X = Y?
$endgroup$
– TigerHix
Jan 30 at 8:48
1
$begingroup$
@TigerHix How do you know $A^{-1}$ exists?
$endgroup$
– 5xum
Jan 30 at 8:49
$begingroup$
I get it now. A is not a square matrix so $A^{-1}$ can't exist. That clears it up, thanks a lot!
$endgroup$
– TigerHix
Jan 30 at 8:51
2
$begingroup$
@TigerHix $A$ could be a square matrix. But you don't know if it is. Also, even if $A$ is a square matrix, you don't know if $A$ is invertible.
$endgroup$
– 5xum
Jan 30 at 8:59
add a comment |
$begingroup$
It makes much more sense now! I assume the steps are aAX + bAY = aAX + bAX (since AX = AY) = (a+b) AX = B. Compare with AX = B, we get a + b = 1. Thanks!
$endgroup$
– TigerHix
Jan 30 at 8:48
$begingroup$
However, why can't I derive X = Y by using AX = AY, A$^{-1}$AX = A$^{-1}$AY, X = Y?
$endgroup$
– TigerHix
Jan 30 at 8:48
1
$begingroup$
@TigerHix How do you know $A^{-1}$ exists?
$endgroup$
– 5xum
Jan 30 at 8:49
$begingroup$
I get it now. A is not a square matrix so $A^{-1}$ can't exist. That clears it up, thanks a lot!
$endgroup$
– TigerHix
Jan 30 at 8:51
2
$begingroup$
@TigerHix $A$ could be a square matrix. But you don't know if it is. Also, even if $A$ is a square matrix, you don't know if $A$ is invertible.
$endgroup$
– 5xum
Jan 30 at 8:59
$begingroup$
It makes much more sense now! I assume the steps are aAX + bAY = aAX + bAX (since AX = AY) = (a+b) AX = B. Compare with AX = B, we get a + b = 1. Thanks!
$endgroup$
– TigerHix
Jan 30 at 8:48
$begingroup$
It makes much more sense now! I assume the steps are aAX + bAY = aAX + bAX (since AX = AY) = (a+b) AX = B. Compare with AX = B, we get a + b = 1. Thanks!
$endgroup$
– TigerHix
Jan 30 at 8:48
$begingroup$
However, why can't I derive X = Y by using AX = AY, A$^{-1}$AX = A$^{-1}$AY, X = Y?
$endgroup$
– TigerHix
Jan 30 at 8:48
$begingroup$
However, why can't I derive X = Y by using AX = AY, A$^{-1}$AX = A$^{-1}$AY, X = Y?
$endgroup$
– TigerHix
Jan 30 at 8:48
1
1
$begingroup$
@TigerHix How do you know $A^{-1}$ exists?
$endgroup$
– 5xum
Jan 30 at 8:49
$begingroup$
@TigerHix How do you know $A^{-1}$ exists?
$endgroup$
– 5xum
Jan 30 at 8:49
$begingroup$
I get it now. A is not a square matrix so $A^{-1}$ can't exist. That clears it up, thanks a lot!
$endgroup$
– TigerHix
Jan 30 at 8:51
$begingroup$
I get it now. A is not a square matrix so $A^{-1}$ can't exist. That clears it up, thanks a lot!
$endgroup$
– TigerHix
Jan 30 at 8:51
2
2
$begingroup$
@TigerHix $A$ could be a square matrix. But you don't know if it is. Also, even if $A$ is a square matrix, you don't know if $A$ is invertible.
$endgroup$
– 5xum
Jan 30 at 8:59
$begingroup$
@TigerHix $A$ could be a square matrix. But you don't know if it is. Also, even if $A$ is a square matrix, you don't know if $A$ is invertible.
$endgroup$
– 5xum
Jan 30 at 8:59
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%2f3093262%2fconsider-a-system-ax-b-where-b-%25cc%25b8-0-let-x-and-y-satisfy-this-system-find-al%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
