Mixing
Is this matrix equation correct?
0
$begingroup$
In this equation I have the following: $B,I$ and $A$ and I have to find the X matrix.
$$3I + B -XA = -X$$
Then I calculated:
$$3I + B = XA -X$$
$$3I + B = X(A-I)$$
Is this correct?
$$3I + B + (A-I)^{-1} = X$$
matrices matrix-equations
edited Feb 1 at 13:47
Bernard
124k741117
asked Feb 1 at 13:41
AnBAnB
62
$endgroup$
add a comment |
0
$begingroup$
In this equation I have the following: $B,I$ and $A$ and I have to find the X matrix.
$$3I + B -XA = -X$$
Then I calculated:
$$3I + B = XA -X$$
$$3I + B = X(A-I)$$
Is this correct?
$$3I + B + (A-I)^{-1} = X$$
matrices matrix-equations
edited Feb 1 at 13:47
Bernard
124k741117
asked Feb 1 at 13:41
AnBAnB
62
$endgroup$
$begingroup$
$(A-I)^{-1}$ may not exist. Take $A=I$, $B=0$,, then there is no solution.
$endgroup$
– Dietrich Burde
Feb 1 at 13:45
$begingroup$
$$A$$ and $$B$$ are defined, $$I$$ is identity matrix. Why would not $$(A-I)^-1$$ exist?
$endgroup$
– AnB
Feb 1 at 13:49
$begingroup$
Why should $A-I$ be invertible? This is not always the case.
$endgroup$
– Dietrich Burde
Feb 1 at 14:46
$begingroup$
Check your own work: substitute the expression that you came up with for $X$ into the original equation and see what you get.
$endgroup$
– amd
Feb 1 at 18:31
add a comment |
0
0
0
$begingroup$
In this equation I have the following: $B,I$ and $A$ and I have to find the X matrix.
$$3I + B -XA = -X$$
Then I calculated:
$$3I + B = XA -X$$
$$3I + B = X(A-I)$$
Is this correct?
$$3I + B + (A-I)^{-1} = X$$
matrices matrix-equations
edited Feb 1 at 13:47
Bernard
124k741117
asked Feb 1 at 13:41
AnBAnB
62
$endgroup$
In this equation I have the following: $B,I$ and $A$ and I have to find the X matrix.
$$3I + B -XA = -X$$
Then I calculated:
$$3I + B = XA -X$$
$$3I + B = X(A-I)$$
Is this correct?
$$3I + B + (A-I)^{-1} = X$$
matrices matrix-equations
matrices matrix-equations
edited Feb 1 at 13:47
Bernard
124k741117
asked Feb 1 at 13:41
AnBAnB
62
edited Feb 1 at 13:47
Bernard
124k741117
asked Feb 1 at 13:41
AnBAnB
62
edited Feb 1 at 13:47
Bernard
124k741117
edited Feb 1 at 13:47
Bernard
124k741117
edited Feb 1 at 13:47
Bernard
124k741117
124k741117
asked Feb 1 at 13:41
AnBAnB
62
asked Feb 1 at 13:41
AnBAnB
62
asked Feb 1 at 13:41
AnBAnB
62
62
$begingroup$
$(A-I)^{-1}$ may not exist. Take $A=I$, $B=0$,, then there is no solution.
$endgroup$
– Dietrich Burde
Feb 1 at 13:45
$begingroup$
$$A$$ and $$B$$ are defined, $$I$$ is identity matrix. Why would not $$(A-I)^-1$$ exist?
$endgroup$
– AnB
Feb 1 at 13:49
$begingroup$
Why should $A-I$ be invertible? This is not always the case.
$endgroup$
– Dietrich Burde
Feb 1 at 14:46
$begingroup$
Check your own work: substitute the expression that you came up with for $X$ into the original equation and see what you get.
$endgroup$
– amd
Feb 1 at 18:31
add a comment |
$begingroup$
$(A-I)^{-1}$ may not exist. Take $A=I$, $B=0$,, then there is no solution.
$endgroup$
– Dietrich Burde
Feb 1 at 13:45
$begingroup$
$$A$$ and $$B$$ are defined, $$I$$ is identity matrix. Why would not $$(A-I)^-1$$ exist?
$endgroup$
– AnB
Feb 1 at 13:49
$begingroup$
Why should $A-I$ be invertible? This is not always the case.
$endgroup$
– Dietrich Burde
Feb 1 at 14:46
$begingroup$
Check your own work: substitute the expression that you came up with for $X$ into the original equation and see what you get.
$endgroup$
– amd
Feb 1 at 18:31
$begingroup$
$(A-I)^{-1}$ may not exist. Take $A=I$, $B=0$,, then there is no solution.
$endgroup$
– Dietrich Burde
Feb 1 at 13:45
$begingroup$
$(A-I)^{-1}$ may not exist. Take $A=I$, $B=0$,, then there is no solution.
$endgroup$
– Dietrich Burde
Feb 1 at 13:45
$begingroup$
$$A$$ and $$B$$ are defined, $$I$$ is identity matrix. Why would not $$(A-I)^-1$$ exist?
$endgroup$
– AnB
Feb 1 at 13:49
$begingroup$
$$A$$ and $$B$$ are defined, $$I$$ is identity matrix. Why would not $$(A-I)^-1$$ exist?
$endgroup$
– AnB
Feb 1 at 13:49
$begingroup$
Why should $A-I$ be invertible? This is not always the case.
$endgroup$
– Dietrich Burde
Feb 1 at 14:46
$begingroup$
Why should $A-I$ be invertible? This is not always the case.
$endgroup$
– Dietrich Burde
Feb 1 at 14:46
$begingroup$
Check your own work: substitute the expression that you came up with for $X$ into the original equation and see what you get.
$endgroup$
– amd
Feb 1 at 18:31
$begingroup$
Check your own work: substitute the expression that you came up with for $X$ into the original equation and see what you get.
$endgroup$
– amd
Feb 1 at 18:31
add a comment |
1 Answer
1
active
oldest
votes
0
$begingroup$
The last expression should replaced by $X=(3I+B) (A-I)^{-1}$, but note that, as pointed out by @Bernard, $(A-I)^{-1}$ may not exist (for instance, if $A=I$).
answered Feb 1 at 13:51
pendermathpendermath
56612
$endgroup$
$begingroup$
How do I know when to put $$(A-I)^-1$$ on the left side of $$(3I+B)$$ and when on the right side?
$endgroup$
– AnB
Feb 1 at 13:54
1
$begingroup$
Dear pendermath, I am Dietrich, not Bernhard, but otherwise the answer is great:)
$endgroup$
– Dietrich Burde
Feb 1 at 14:47
$begingroup$
@AnB The factor $(A-I) $ is on the right, so if you right multiply both sides of the equation by its inverse the two factors on the right hand side will cancel out.
$endgroup$
– pendermath
Feb 2 at 7:32
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
});
}
});
draft saved
draft discarded
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%2f3096236%2fis-this-matrix-equation-correct%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
0
$begingroup$
The last expression should replaced by $X=(3I+B) (A-I)^{-1}$, but note that, as pointed out by @Bernard, $(A-I)^{-1}$ may not exist (for instance, if $A=I$).
answered Feb 1 at 13:51
pendermathpendermath
56612
$endgroup$
$begingroup$
How do I know when to put $$(A-I)^-1$$ on the left side of $$(3I+B)$$ and when on the right side?
$endgroup$
– AnB
Feb 1 at 13:54
1
$begingroup$
Dear pendermath, I am Dietrich, not Bernhard, but otherwise the answer is great:)
$endgroup$
– Dietrich Burde
Feb 1 at 14:47
$begingroup$
@AnB The factor $(A-I) $ is on the right, so if you right multiply both sides of the equation by its inverse the two factors on the right hand side will cancel out.
$endgroup$
– pendermath
Feb 2 at 7:32
add a comment |
0
$begingroup$
The last expression should replaced by $X=(3I+B) (A-I)^{-1}$, but note that, as pointed out by @Bernard, $(A-I)^{-1}$ may not exist (for instance, if $A=I$).
answered Feb 1 at 13:51
pendermathpendermath
56612
$endgroup$
$begingroup$
How do I know when to put $$(A-I)^-1$$ on the left side of $$(3I+B)$$ and when on the right side?
$endgroup$
– AnB
Feb 1 at 13:54
1
$begingroup$
Dear pendermath, I am Dietrich, not Bernhard, but otherwise the answer is great:)
$endgroup$
– Dietrich Burde
Feb 1 at 14:47
$begingroup$
@AnB The factor $(A-I) $ is on the right, so if you right multiply both sides of the equation by its inverse the two factors on the right hand side will cancel out.
$endgroup$
– pendermath
Feb 2 at 7:32
add a comment |
0
0
0
$begingroup$
The last expression should replaced by $X=(3I+B) (A-I)^{-1}$, but note that, as pointed out by @Bernard, $(A-I)^{-1}$ may not exist (for instance, if $A=I$).
answered Feb 1 at 13:51
pendermathpendermath
56612
$endgroup$
The last expression should replaced by $X=(3I+B) (A-I)^{-1}$, but note that, as pointed out by @Bernard, $(A-I)^{-1}$ may not exist (for instance, if $A=I$).
answered Feb 1 at 13:51
pendermathpendermath
56612
answered Feb 1 at 13:51
pendermathpendermath
56612
answered Feb 1 at 13:51
pendermathpendermath
56612
answered Feb 1 at 13:51
pendermathpendermath
56612
56612
$begingroup$
How do I know when to put $$(A-I)^-1$$ on the left side of $$(3I+B)$$ and when on the right side?
$endgroup$
– AnB
Feb 1 at 13:54
1
$begingroup$
Dear pendermath, I am Dietrich, not Bernhard, but otherwise the answer is great:)
$endgroup$
– Dietrich Burde
Feb 1 at 14:47
$begingroup$
@AnB The factor $(A-I) $ is on the right, so if you right multiply both sides of the equation by its inverse the two factors on the right hand side will cancel out.
$endgroup$
– pendermath
Feb 2 at 7:32
add a comment |
$begingroup$
How do I know when to put $$(A-I)^-1$$ on the left side of $$(3I+B)$$ and when on the right side?
$endgroup$
– AnB
Feb 1 at 13:54
1
$begingroup$
Dear pendermath, I am Dietrich, not Bernhard, but otherwise the answer is great:)
$endgroup$
– Dietrich Burde
Feb 1 at 14:47
$begingroup$
@AnB The factor $(A-I) $ is on the right, so if you right multiply both sides of the equation by its inverse the two factors on the right hand side will cancel out.
$endgroup$
– pendermath
Feb 2 at 7:32
$begingroup$
How do I know when to put $$(A-I)^-1$$ on the left side of $$(3I+B)$$ and when on the right side?
$endgroup$
– AnB
Feb 1 at 13:54
$begingroup$
How do I know when to put $$(A-I)^-1$$ on the left side of $$(3I+B)$$ and when on the right side?
$endgroup$
– AnB
Feb 1 at 13:54
1
1
$begingroup$
Dear pendermath, I am Dietrich, not Bernhard, but otherwise the answer is great:)
$endgroup$
– Dietrich Burde
Feb 1 at 14:47
$begingroup$
Dear pendermath, I am Dietrich, not Bernhard, but otherwise the answer is great:)
$endgroup$
– Dietrich Burde
Feb 1 at 14:47
$begingroup$
@AnB The factor $(A-I) $ is on the right, so if you right multiply both sides of the equation by its inverse the two factors on the right hand side will cancel out.
$endgroup$
– pendermath
Feb 2 at 7:32
$begingroup$
@AnB The factor $(A-I) $ is on the right, so if you right multiply both sides of the equation by its inverse the two factors on the right hand side will cancel out.
$endgroup$
– pendermath
Feb 2 at 7:32
add a comment |
draft saved
draft discarded
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.
draft saved
draft discarded
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%2f3096236%2fis-this-matrix-equation-correct%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
$begingroup$
$(A-I)^{-1}$ may not exist. Take $A=I$, $B=0$,, then there is no solution.
$endgroup$
– Dietrich Burde
Feb 1 at 13:45
$begingroup$
$$A$$ and $$B$$ are defined, $$I$$ is identity matrix. Why would not $$(A-I)^-1$$ exist?
$endgroup$
– AnB
Feb 1 at 13:49
$begingroup$
Why should $A-I$ be invertible? This is not always the case.
$endgroup$
– Dietrich Burde
Feb 1 at 14:46
$begingroup$
Check your own work: substitute the expression that you came up with for $X$ into the original equation and see what you get.
$endgroup$
– amd
Feb 1 at 18:31