Matrices Equations: is it Okay to Transpose Both Sides?












1












$begingroup$


I have this matrices equation:
$$(AX+I)^T=2I$$



Is it possible to transpose both sides to get this?
$$
((AX+I)^T)^T=2(I)^T
\
AX+I=2I$$
And then
$$AX=I
\X=A^{-1}
$$



Thank you.






linear-algebra matrices matrix-equations







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    Its possible. In the last step, $A$ must be invertible.
    $endgroup$
    – Wuestenfux
    Dec 20 '18 at 11:46










  • $begingroup$
    It is the first property of matrix transposition.
    $endgroup$
    – farruhota
    Dec 20 '18 at 13:39
















1












$begingroup$


I have this matrices equation:
$$(AX+I)^T=2I$$



Is it possible to transpose both sides to get this?
$$
((AX+I)^T)^T=2(I)^T
\
AX+I=2I$$
And then
$$AX=I
\X=A^{-1}
$$



Thank you.






linear-algebra matrices matrix-equations







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    Its possible. In the last step, $A$ must be invertible.
    $endgroup$
    – Wuestenfux
    Dec 20 '18 at 11:46










  • $begingroup$
    It is the first property of matrix transposition.
    $endgroup$
    – farruhota
    Dec 20 '18 at 13:39














1












1








1





$begingroup$


I have this matrices equation:
$$(AX+I)^T=2I$$



Is it possible to transpose both sides to get this?
$$
((AX+I)^T)^T=2(I)^T
\
AX+I=2I$$
And then
$$AX=I
\X=A^{-1}
$$



Thank you.






linear-algebra matrices matrix-equations







share|cite|improve this question









$endgroup$




I have this matrices equation:
$$(AX+I)^T=2I$$



Is it possible to transpose both sides to get this?
$$
((AX+I)^T)^T=2(I)^T
\
AX+I=2I$$
And then
$$AX=I
\X=A^{-1}
$$



Thank you.






linear-algebra matrices matrix-equations




linear-algebra matrices matrix-equations






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 20 '18 at 11:44









NetanelNetanel

974




974








  • 2




    $begingroup$
    Its possible. In the last step, $A$ must be invertible.
    $endgroup$
    – Wuestenfux
    Dec 20 '18 at 11:46










  • $begingroup$
    It is the first property of matrix transposition.
    $endgroup$
    – farruhota
    Dec 20 '18 at 13:39














  • 2




    $begingroup$
    Its possible. In the last step, $A$ must be invertible.
    $endgroup$
    – Wuestenfux
    Dec 20 '18 at 11:46










  • $begingroup$
    It is the first property of matrix transposition.
    $endgroup$
    – farruhota
    Dec 20 '18 at 13:39








2




2




$begingroup$
Its possible. In the last step, $A$ must be invertible.
$endgroup$
– Wuestenfux
Dec 20 '18 at 11:46




$begingroup$
Its possible. In the last step, $A$ must be invertible.
$endgroup$
– Wuestenfux
Dec 20 '18 at 11:46












$begingroup$
It is the first property of matrix transposition.
$endgroup$
– farruhota
Dec 20 '18 at 13:39




$begingroup$
It is the first property of matrix transposition.
$endgroup$
– farruhota
Dec 20 '18 at 13:39










1 Answer
1






active

oldest

votes


















1












$begingroup$

All your steps are correct ! Your result $X=A^{-1}$ is also correct.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    $A$ must be invertible.
    $endgroup$
    – Thinking
    Dec 20 '18 at 11:46












  • $begingroup$
    If all matrices are square, then from $AX=I$ we get that $A$ is invertble !
    $endgroup$
    – Fred
    Dec 20 '18 at 11:50










  • $begingroup$
    It is necessary for $A$ to be invertible for the equation to have a solution indeed. But it is important to note that the equation has no solution if $A$ is not invertible.
    $endgroup$
    – nicomezi
    Dec 20 '18 at 12:05










  • $begingroup$
    What is your problem ? We have $(AX+I)^T=2I iff AX=I$. Hence, the equation $(AX+I)^T=2I $ implies that $A$ is invertible.
    $endgroup$
    – Fred
    Dec 20 '18 at 12:08










  • $begingroup$
    You are making the assumption that the equation has a solution in order to conclude that $A$ is invertible. That is all is being said here.
    $endgroup$
    – nicomezi
    Dec 20 '18 at 12:13











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


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3047451%2fmatrices-equations-is-it-okay-to-transpose-both-sides%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









1












$begingroup$

All your steps are correct ! Your result $X=A^{-1}$ is also correct.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    $A$ must be invertible.
    $endgroup$
    – Thinking
    Dec 20 '18 at 11:46












  • $begingroup$
    If all matrices are square, then from $AX=I$ we get that $A$ is invertble !
    $endgroup$
    – Fred
    Dec 20 '18 at 11:50










  • $begingroup$
    It is necessary for $A$ to be invertible for the equation to have a solution indeed. But it is important to note that the equation has no solution if $A$ is not invertible.
    $endgroup$
    – nicomezi
    Dec 20 '18 at 12:05










  • $begingroup$
    What is your problem ? We have $(AX+I)^T=2I iff AX=I$. Hence, the equation $(AX+I)^T=2I $ implies that $A$ is invertible.
    $endgroup$
    – Fred
    Dec 20 '18 at 12:08










  • $begingroup$
    You are making the assumption that the equation has a solution in order to conclude that $A$ is invertible. That is all is being said here.
    $endgroup$
    – nicomezi
    Dec 20 '18 at 12:13
















1












$begingroup$

All your steps are correct ! Your result $X=A^{-1}$ is also correct.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    $A$ must be invertible.
    $endgroup$
    – Thinking
    Dec 20 '18 at 11:46












  • $begingroup$
    If all matrices are square, then from $AX=I$ we get that $A$ is invertble !
    $endgroup$
    – Fred
    Dec 20 '18 at 11:50










  • $begingroup$
    It is necessary for $A$ to be invertible for the equation to have a solution indeed. But it is important to note that the equation has no solution if $A$ is not invertible.
    $endgroup$
    – nicomezi
    Dec 20 '18 at 12:05










  • $begingroup$
    What is your problem ? We have $(AX+I)^T=2I iff AX=I$. Hence, the equation $(AX+I)^T=2I $ implies that $A$ is invertible.
    $endgroup$
    – Fred
    Dec 20 '18 at 12:08










  • $begingroup$
    You are making the assumption that the equation has a solution in order to conclude that $A$ is invertible. That is all is being said here.
    $endgroup$
    – nicomezi
    Dec 20 '18 at 12:13














1












1








1





$begingroup$

All your steps are correct ! Your result $X=A^{-1}$ is also correct.






share|cite|improve this answer









$endgroup$



All your steps are correct ! Your result $X=A^{-1}$ is also correct.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 20 '18 at 11:46









FredFred

46.8k1848




46.8k1848








  • 1




    $begingroup$
    $A$ must be invertible.
    $endgroup$
    – Thinking
    Dec 20 '18 at 11:46












  • $begingroup$
    If all matrices are square, then from $AX=I$ we get that $A$ is invertble !
    $endgroup$
    – Fred
    Dec 20 '18 at 11:50










  • $begingroup$
    It is necessary for $A$ to be invertible for the equation to have a solution indeed. But it is important to note that the equation has no solution if $A$ is not invertible.
    $endgroup$
    – nicomezi
    Dec 20 '18 at 12:05










  • $begingroup$
    What is your problem ? We have $(AX+I)^T=2I iff AX=I$. Hence, the equation $(AX+I)^T=2I $ implies that $A$ is invertible.
    $endgroup$
    – Fred
    Dec 20 '18 at 12:08










  • $begingroup$
    You are making the assumption that the equation has a solution in order to conclude that $A$ is invertible. That is all is being said here.
    $endgroup$
    – nicomezi
    Dec 20 '18 at 12:13














  • 1




    $begingroup$
    $A$ must be invertible.
    $endgroup$
    – Thinking
    Dec 20 '18 at 11:46












  • $begingroup$
    If all matrices are square, then from $AX=I$ we get that $A$ is invertble !
    $endgroup$
    – Fred
    Dec 20 '18 at 11:50










  • $begingroup$
    It is necessary for $A$ to be invertible for the equation to have a solution indeed. But it is important to note that the equation has no solution if $A$ is not invertible.
    $endgroup$
    – nicomezi
    Dec 20 '18 at 12:05










  • $begingroup$
    What is your problem ? We have $(AX+I)^T=2I iff AX=I$. Hence, the equation $(AX+I)^T=2I $ implies that $A$ is invertible.
    $endgroup$
    – Fred
    Dec 20 '18 at 12:08










  • $begingroup$
    You are making the assumption that the equation has a solution in order to conclude that $A$ is invertible. That is all is being said here.
    $endgroup$
    – nicomezi
    Dec 20 '18 at 12:13








1




1




$begingroup$
$A$ must be invertible.
$endgroup$
– Thinking
Dec 20 '18 at 11:46






$begingroup$
$A$ must be invertible.
$endgroup$
– Thinking
Dec 20 '18 at 11:46














$begingroup$
If all matrices are square, then from $AX=I$ we get that $A$ is invertble !
$endgroup$
– Fred
Dec 20 '18 at 11:50




$begingroup$
If all matrices are square, then from $AX=I$ we get that $A$ is invertble !
$endgroup$
– Fred
Dec 20 '18 at 11:50












$begingroup$
It is necessary for $A$ to be invertible for the equation to have a solution indeed. But it is important to note that the equation has no solution if $A$ is not invertible.
$endgroup$
– nicomezi
Dec 20 '18 at 12:05




$begingroup$
It is necessary for $A$ to be invertible for the equation to have a solution indeed. But it is important to note that the equation has no solution if $A$ is not invertible.
$endgroup$
– nicomezi
Dec 20 '18 at 12:05












$begingroup$
What is your problem ? We have $(AX+I)^T=2I iff AX=I$. Hence, the equation $(AX+I)^T=2I $ implies that $A$ is invertible.
$endgroup$
– Fred
Dec 20 '18 at 12:08




$begingroup$
What is your problem ? We have $(AX+I)^T=2I iff AX=I$. Hence, the equation $(AX+I)^T=2I $ implies that $A$ is invertible.
$endgroup$
– Fred
Dec 20 '18 at 12:08












$begingroup$
You are making the assumption that the equation has a solution in order to conclude that $A$ is invertible. That is all is being said here.
$endgroup$
– nicomezi
Dec 20 '18 at 12:13




$begingroup$
You are making the assumption that the equation has a solution in order to conclude that $A$ is invertible. That is all is being said here.
$endgroup$
– nicomezi
Dec 20 '18 at 12:13


















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














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3047451%2fmatrices-equations-is-it-okay-to-transpose-both-sides%23new-answer', 'question_page');
}
);

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







-

Popular posts from this blog

Quarter-circle Tiles

Image
up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2
Read more

build a pushdown automaton that recognizes the reverse language of a given pushdown automaton?

Image
0 I think it's similiar to NFAs. I replace the finite states of the given automaton for start-states for my new automaton. I do it with $epsilon$ -transition from the start state to the actual finite states of the given automaton. The start state of the given automaton will be my accepting(finite) state of my new one. All transitions will be reversed of course to recognize the reversed word. But how do I have to change the stack of the pushdown automaton? That's tricky. formal-languages automata context-free-grammar share | cite | improve this question edited Nov 29 '18 at 14:20 joee
Read more

Mont Emei

Image
Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog
Read more