Invertible matrix properties of a matrix












1












$begingroup$


I have here the following question:




Let $X$ be the $5 times 5$ matrix "full of ones":



$X = begin{pmatrix}1 & 1 & 1 & 1 & 1 \ 1 & 1 & 1 & 1 & 1 \ 1 & 1 & 1 & 1 & 1 \ 1 & 1 & 1 & 1 & 1\ 1 & 1 & 1 & 1 & 1end{pmatrix}$



$(a)$ Is $X$ invertible? Explain.



$(b)$ Find a number $c$ such that $X^2=cX$.



$(c)$ Compute $(X-aI_5)(X+(a-c)I_5)$, where $a$ is a real number, $c$ is the constant from part $(b)$, $aneq 0$, and $aneq c$. If $M=(X-aI_5)$, what is $M^{-1}$? (You may express your answer in terms of $X,I_5,a$, and $c.$)




I already did part $(a)$ and $(b)$.



$(a)$ No. It's not invertible since there are two or more identical rows or columns so the determinant would be $0$ and hence it is not invertible.



$(b)$ $X^2 = begin{pmatrix}5 & 5 & 5 & 5 & 5 \ 5 & 5 & 5 & 5 & 5 \ 5 & 5 & 5 & 5 & 5 \ 5 & 5 & 5 & 5 & 5\ 5 & 5 & 5 & 5 & 5 end{pmatrix}$



Thus, $c=5$.



$(c)$ I almost got part $(c)$ too. It's the last part I can't figure out.



$=(X-aI_5)(X+(a-c)I_5)$



$=-a(a-c)I_5$



If $M=(X-aI_5)$, that would mean if I multiply both sides by $(X+(a-c)I_5)$, I get that:



$$M(X+(a-c)I_5)=(X-aI_5)(X+(a-c)I_5)$$
$$M(X+(a-c)I_5)=-a(a-c)I_5$$



I can literally see the answer in front of me, but it's not quite there. How do I proceed from here?



For reference, the answer should be:



$M^{-1}=frac{X+(a-c)I_5}{-a(a-c)}$






linear-algebra matrices inverse







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    HINT:$MM^{-1}=M^{-1}M=I $.
    $endgroup$
    – Thomas Shelby
    Dec 15 '18 at 8:18










  • $begingroup$
    Yes I know that but how does that help? My problem is that I need to divide what are essentially matrices, and I don't think that's allowed.
    $endgroup$
    – Future Math person
    Dec 15 '18 at 8:21






  • 1




    $begingroup$
    Note that $-a(a-c) in Bbb R $.
    $endgroup$
    – Thomas Shelby
    Dec 15 '18 at 8:26






  • 1




    $begingroup$
    Never mind I get it! Thanks a lot!!
    $endgroup$
    – Future Math person
    Dec 15 '18 at 8:29






  • 1




    $begingroup$
    Just divide the whole equation by the scalar and compare.
    $endgroup$
    – Thomas Shelby
    Dec 15 '18 at 8:29
















1












$begingroup$


I have here the following question:




Let $X$ be the $5 times 5$ matrix "full of ones":



$X = begin{pmatrix}1 & 1 & 1 & 1 & 1 \ 1 & 1 & 1 & 1 & 1 \ 1 & 1 & 1 & 1 & 1 \ 1 & 1 & 1 & 1 & 1\ 1 & 1 & 1 & 1 & 1end{pmatrix}$



$(a)$ Is $X$ invertible? Explain.



$(b)$ Find a number $c$ such that $X^2=cX$.



$(c)$ Compute $(X-aI_5)(X+(a-c)I_5)$, where $a$ is a real number, $c$ is the constant from part $(b)$, $aneq 0$, and $aneq c$. If $M=(X-aI_5)$, what is $M^{-1}$? (You may express your answer in terms of $X,I_5,a$, and $c.$)




I already did part $(a)$ and $(b)$.



$(a)$ No. It's not invertible since there are two or more identical rows or columns so the determinant would be $0$ and hence it is not invertible.



$(b)$ $X^2 = begin{pmatrix}5 & 5 & 5 & 5 & 5 \ 5 & 5 & 5 & 5 & 5 \ 5 & 5 & 5 & 5 & 5 \ 5 & 5 & 5 & 5 & 5\ 5 & 5 & 5 & 5 & 5 end{pmatrix}$



Thus, $c=5$.



$(c)$ I almost got part $(c)$ too. It's the last part I can't figure out.



$=(X-aI_5)(X+(a-c)I_5)$



$=-a(a-c)I_5$



If $M=(X-aI_5)$, that would mean if I multiply both sides by $(X+(a-c)I_5)$, I get that:



$$M(X+(a-c)I_5)=(X-aI_5)(X+(a-c)I_5)$$
$$M(X+(a-c)I_5)=-a(a-c)I_5$$



I can literally see the answer in front of me, but it's not quite there. How do I proceed from here?



For reference, the answer should be:



$M^{-1}=frac{X+(a-c)I_5}{-a(a-c)}$






linear-algebra matrices inverse







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    HINT:$MM^{-1}=M^{-1}M=I $.
    $endgroup$
    – Thomas Shelby
    Dec 15 '18 at 8:18










  • $begingroup$
    Yes I know that but how does that help? My problem is that I need to divide what are essentially matrices, and I don't think that's allowed.
    $endgroup$
    – Future Math person
    Dec 15 '18 at 8:21






  • 1




    $begingroup$
    Note that $-a(a-c) in Bbb R $.
    $endgroup$
    – Thomas Shelby
    Dec 15 '18 at 8:26






  • 1




    $begingroup$
    Never mind I get it! Thanks a lot!!
    $endgroup$
    – Future Math person
    Dec 15 '18 at 8:29






  • 1




    $begingroup$
    Just divide the whole equation by the scalar and compare.
    $endgroup$
    – Thomas Shelby
    Dec 15 '18 at 8:29














1












1








1


1



$begingroup$


I have here the following question:




Let $X$ be the $5 times 5$ matrix "full of ones":



$X = begin{pmatrix}1 & 1 & 1 & 1 & 1 \ 1 & 1 & 1 & 1 & 1 \ 1 & 1 & 1 & 1 & 1 \ 1 & 1 & 1 & 1 & 1\ 1 & 1 & 1 & 1 & 1end{pmatrix}$



$(a)$ Is $X$ invertible? Explain.



$(b)$ Find a number $c$ such that $X^2=cX$.



$(c)$ Compute $(X-aI_5)(X+(a-c)I_5)$, where $a$ is a real number, $c$ is the constant from part $(b)$, $aneq 0$, and $aneq c$. If $M=(X-aI_5)$, what is $M^{-1}$? (You may express your answer in terms of $X,I_5,a$, and $c.$)




I already did part $(a)$ and $(b)$.



$(a)$ No. It's not invertible since there are two or more identical rows or columns so the determinant would be $0$ and hence it is not invertible.



$(b)$ $X^2 = begin{pmatrix}5 & 5 & 5 & 5 & 5 \ 5 & 5 & 5 & 5 & 5 \ 5 & 5 & 5 & 5 & 5 \ 5 & 5 & 5 & 5 & 5\ 5 & 5 & 5 & 5 & 5 end{pmatrix}$



Thus, $c=5$.



$(c)$ I almost got part $(c)$ too. It's the last part I can't figure out.



$=(X-aI_5)(X+(a-c)I_5)$



$=-a(a-c)I_5$



If $M=(X-aI_5)$, that would mean if I multiply both sides by $(X+(a-c)I_5)$, I get that:



$$M(X+(a-c)I_5)=(X-aI_5)(X+(a-c)I_5)$$
$$M(X+(a-c)I_5)=-a(a-c)I_5$$



I can literally see the answer in front of me, but it's not quite there. How do I proceed from here?



For reference, the answer should be:



$M^{-1}=frac{X+(a-c)I_5}{-a(a-c)}$






linear-algebra matrices inverse







share|cite|improve this question











$endgroup$




I have here the following question:




Let $X$ be the $5 times 5$ matrix "full of ones":



$X = begin{pmatrix}1 & 1 & 1 & 1 & 1 \ 1 & 1 & 1 & 1 & 1 \ 1 & 1 & 1 & 1 & 1 \ 1 & 1 & 1 & 1 & 1\ 1 & 1 & 1 & 1 & 1end{pmatrix}$



$(a)$ Is $X$ invertible? Explain.



$(b)$ Find a number $c$ such that $X^2=cX$.



$(c)$ Compute $(X-aI_5)(X+(a-c)I_5)$, where $a$ is a real number, $c$ is the constant from part $(b)$, $aneq 0$, and $aneq c$. If $M=(X-aI_5)$, what is $M^{-1}$? (You may express your answer in terms of $X,I_5,a$, and $c.$)




I already did part $(a)$ and $(b)$.



$(a)$ No. It's not invertible since there are two or more identical rows or columns so the determinant would be $0$ and hence it is not invertible.



$(b)$ $X^2 = begin{pmatrix}5 & 5 & 5 & 5 & 5 \ 5 & 5 & 5 & 5 & 5 \ 5 & 5 & 5 & 5 & 5 \ 5 & 5 & 5 & 5 & 5\ 5 & 5 & 5 & 5 & 5 end{pmatrix}$



Thus, $c=5$.



$(c)$ I almost got part $(c)$ too. It's the last part I can't figure out.



$=(X-aI_5)(X+(a-c)I_5)$



$=-a(a-c)I_5$



If $M=(X-aI_5)$, that would mean if I multiply both sides by $(X+(a-c)I_5)$, I get that:



$$M(X+(a-c)I_5)=(X-aI_5)(X+(a-c)I_5)$$
$$M(X+(a-c)I_5)=-a(a-c)I_5$$



I can literally see the answer in front of me, but it's not quite there. How do I proceed from here?



For reference, the answer should be:



$M^{-1}=frac{X+(a-c)I_5}{-a(a-c)}$






linear-algebra matrices inverse




linear-algebra matrices inverse






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 15 '18 at 8:26







Future Math person

















asked Dec 15 '18 at 8:14









Future Math personFuture Math person

977817




977817








  • 1




    $begingroup$
    HINT:$MM^{-1}=M^{-1}M=I $.
    $endgroup$
    – Thomas Shelby
    Dec 15 '18 at 8:18










  • $begingroup$
    Yes I know that but how does that help? My problem is that I need to divide what are essentially matrices, and I don't think that's allowed.
    $endgroup$
    – Future Math person
    Dec 15 '18 at 8:21






  • 1




    $begingroup$
    Note that $-a(a-c) in Bbb R $.
    $endgroup$
    – Thomas Shelby
    Dec 15 '18 at 8:26






  • 1




    $begingroup$
    Never mind I get it! Thanks a lot!!
    $endgroup$
    – Future Math person
    Dec 15 '18 at 8:29






  • 1




    $begingroup$
    Just divide the whole equation by the scalar and compare.
    $endgroup$
    – Thomas Shelby
    Dec 15 '18 at 8:29














  • 1




    $begingroup$
    HINT:$MM^{-1}=M^{-1}M=I $.
    $endgroup$
    – Thomas Shelby
    Dec 15 '18 at 8:18










  • $begingroup$
    Yes I know that but how does that help? My problem is that I need to divide what are essentially matrices, and I don't think that's allowed.
    $endgroup$
    – Future Math person
    Dec 15 '18 at 8:21






  • 1




    $begingroup$
    Note that $-a(a-c) in Bbb R $.
    $endgroup$
    – Thomas Shelby
    Dec 15 '18 at 8:26






  • 1




    $begingroup$
    Never mind I get it! Thanks a lot!!
    $endgroup$
    – Future Math person
    Dec 15 '18 at 8:29






  • 1




    $begingroup$
    Just divide the whole equation by the scalar and compare.
    $endgroup$
    – Thomas Shelby
    Dec 15 '18 at 8:29








1




1




$begingroup$
HINT:$MM^{-1}=M^{-1}M=I $.
$endgroup$
– Thomas Shelby
Dec 15 '18 at 8:18




$begingroup$
HINT:$MM^{-1}=M^{-1}M=I $.
$endgroup$
– Thomas Shelby
Dec 15 '18 at 8:18












$begingroup$
Yes I know that but how does that help? My problem is that I need to divide what are essentially matrices, and I don't think that's allowed.
$endgroup$
– Future Math person
Dec 15 '18 at 8:21




$begingroup$
Yes I know that but how does that help? My problem is that I need to divide what are essentially matrices, and I don't think that's allowed.
$endgroup$
– Future Math person
Dec 15 '18 at 8:21




1




1




$begingroup$
Note that $-a(a-c) in Bbb R $.
$endgroup$
– Thomas Shelby
Dec 15 '18 at 8:26




$begingroup$
Note that $-a(a-c) in Bbb R $.
$endgroup$
– Thomas Shelby
Dec 15 '18 at 8:26




1




1




$begingroup$
Never mind I get it! Thanks a lot!!
$endgroup$
– Future Math person
Dec 15 '18 at 8:29




$begingroup$
Never mind I get it! Thanks a lot!!
$endgroup$
– Future Math person
Dec 15 '18 at 8:29




1




1




$begingroup$
Just divide the whole equation by the scalar and compare.
$endgroup$
– Thomas Shelby
Dec 15 '18 at 8:29




$begingroup$
Just divide the whole equation by the scalar and compare.
$endgroup$
– Thomas Shelby
Dec 15 '18 at 8:29










1 Answer
1






active

oldest

votes


















2












$begingroup$

Hint. You have $M(X+(a-c)I_5)=-a(a-c)I_5$. Pre-multiply by $M^{-1}$.






share|cite|improve this answer









$endgroup$













    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%2f3040265%2finvertible-matrix-properties-of-a-matrix%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









    2












    $begingroup$

    Hint. You have $M(X+(a-c)I_5)=-a(a-c)I_5$. Pre-multiply by $M^{-1}$.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Hint. You have $M(X+(a-c)I_5)=-a(a-c)I_5$. Pre-multiply by $M^{-1}$.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Hint. You have $M(X+(a-c)I_5)=-a(a-c)I_5$. Pre-multiply by $M^{-1}$.






        share|cite|improve this answer









        $endgroup$



        Hint. You have $M(X+(a-c)I_5)=-a(a-c)I_5$. Pre-multiply by $M^{-1}$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 15 '18 at 8:28









        Shubham JohriShubham Johri

        5,204718




        5,204718






























            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%2f3040265%2finvertible-matrix-properties-of-a-matrix%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

            Plaza Victoria

            Image
            Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l
            Read more

            In PowerPoint, is there a keyboard shortcut for bulleted / numbered list?

            Image
            .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 2 OneNote has two useful keyboard shortcuts for bulleted and numbered lists, namely Ctrl + . and Ctrl + - . Is there something similar for PowerPoint where the work with lists is also very common? Haven't found anything here: Use keyboard shortcuts in PowerPoint. keyboard-shortcuts microsoft-powerpoint share | improve this question edited Aug 14 '14 at 14:49 Linger 2,826 10 28 40
            Read more

            How to put 3 figures in Latex with 2 figures side by side and 1 below these side by side images but in...

            Image
            1 0 I want to put 3 images in LateX such as 2 figures are side by side horizotally and 3rd figure below these 2 side by side figures but in middle. I have following code: begin{figure}[H] centering begin{minipage}{.5textwidth} centering includegraphics[width=.50linewidth]{Text/Images/Genelec_8010_AP.jpg} captionof{figure}{Genelec 8010 AP} label{fig:Genelec 8010} end{minipage}% begin{minipage}{.5textwidth} centering includegraphics[width=.69linewidth]{Text/Images/Genelec_8020_CPM.jpg} captionof{figure}{Genelec 8020 CPM} label{fig:Genelec 8020} end{minipage}% end{figure} %%Here is my 3rd picture code begin{figure} centering includegraphics[width=5cm]{Text/Images/Genelec_8030_BPM.jpg} caption{Genelec 8030 BPM} label{fig:Genelec 8030} end{figure} I want to ask ho
            Read more