Let $A= begin{pmatrix} 8&2 \ -8&-2 end{pmatrix}$. Find the entry in the first row and second column...












4












$begingroup$


I have tried diagonalizing the matrix and obtained:
$A=PDP^{-1}$.

Where:
$P=begin{pmatrix} 1&1 \ -4&-1 end{pmatrix}$
$D=begin{pmatrix} 0&0 \ 0&6 end{pmatrix}$
$P^{-1}=frac{1}{3}begin{pmatrix} -1&-1 \ 4&1 end{pmatrix}$.



So that $A^{2014}=PD^{2014}P^{-1}$

But i just want to know whether there is an alternate method.






linear-algebra eigenvalues-eigenvectors diagonalization







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    This is the standard elegant way to do it.I don't know another method other than brute force.
    $endgroup$
    – Ethan Bolker
    Jan 6 at 0:19






  • 1




    $begingroup$
    I think that what you did is very well, other ways will come to the decomposition anyways
    $endgroup$
    – José Alejandro Aburto Araneda
    Jan 6 at 0:19
















4












$begingroup$


I have tried diagonalizing the matrix and obtained:
$A=PDP^{-1}$.

Where:
$P=begin{pmatrix} 1&1 \ -4&-1 end{pmatrix}$
$D=begin{pmatrix} 0&0 \ 0&6 end{pmatrix}$
$P^{-1}=frac{1}{3}begin{pmatrix} -1&-1 \ 4&1 end{pmatrix}$.



So that $A^{2014}=PD^{2014}P^{-1}$

But i just want to know whether there is an alternate method.






linear-algebra eigenvalues-eigenvectors diagonalization







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    This is the standard elegant way to do it.I don't know another method other than brute force.
    $endgroup$
    – Ethan Bolker
    Jan 6 at 0:19






  • 1




    $begingroup$
    I think that what you did is very well, other ways will come to the decomposition anyways
    $endgroup$
    – José Alejandro Aburto Araneda
    Jan 6 at 0:19














4












4








4


2



$begingroup$


I have tried diagonalizing the matrix and obtained:
$A=PDP^{-1}$.

Where:
$P=begin{pmatrix} 1&1 \ -4&-1 end{pmatrix}$
$D=begin{pmatrix} 0&0 \ 0&6 end{pmatrix}$
$P^{-1}=frac{1}{3}begin{pmatrix} -1&-1 \ 4&1 end{pmatrix}$.



So that $A^{2014}=PD^{2014}P^{-1}$

But i just want to know whether there is an alternate method.






linear-algebra eigenvalues-eigenvectors diagonalization







share|cite|improve this question











$endgroup$




I have tried diagonalizing the matrix and obtained:
$A=PDP^{-1}$.

Where:
$P=begin{pmatrix} 1&1 \ -4&-1 end{pmatrix}$
$D=begin{pmatrix} 0&0 \ 0&6 end{pmatrix}$
$P^{-1}=frac{1}{3}begin{pmatrix} -1&-1 \ 4&1 end{pmatrix}$.



So that $A^{2014}=PD^{2014}P^{-1}$

But i just want to know whether there is an alternate method.






linear-algebra eigenvalues-eigenvectors diagonalization




linear-algebra eigenvalues-eigenvectors diagonalization






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 6 at 0:18









vadim123

76.5k897191




76.5k897191










asked Jan 6 at 0:16









DD90DD90

2598




2598








  • 1




    $begingroup$
    This is the standard elegant way to do it.I don't know another method other than brute force.
    $endgroup$
    – Ethan Bolker
    Jan 6 at 0:19






  • 1




    $begingroup$
    I think that what you did is very well, other ways will come to the decomposition anyways
    $endgroup$
    – José Alejandro Aburto Araneda
    Jan 6 at 0:19














  • 1




    $begingroup$
    This is the standard elegant way to do it.I don't know another method other than brute force.
    $endgroup$
    – Ethan Bolker
    Jan 6 at 0:19






  • 1




    $begingroup$
    I think that what you did is very well, other ways will come to the decomposition anyways
    $endgroup$
    – José Alejandro Aburto Araneda
    Jan 6 at 0:19








1




1




$begingroup$
This is the standard elegant way to do it.I don't know another method other than brute force.
$endgroup$
– Ethan Bolker
Jan 6 at 0:19




$begingroup$
This is the standard elegant way to do it.I don't know another method other than brute force.
$endgroup$
– Ethan Bolker
Jan 6 at 0:19




1




1




$begingroup$
I think that what you did is very well, other ways will come to the decomposition anyways
$endgroup$
– José Alejandro Aburto Araneda
Jan 6 at 0:19




$begingroup$
I think that what you did is very well, other ways will come to the decomposition anyways
$endgroup$
– José Alejandro Aburto Araneda
Jan 6 at 0:19










3 Answers
3






active

oldest

votes


















5












$begingroup$

Hint Show that
$$A^2=6A$$



Note: You can show the stronger statement:
$$begin{pmatrix} 8&2 \ -8&-2 end{pmatrix}begin{pmatrix} a&b \ -a&-b end{pmatrix}=6begin{pmatrix} a&b \ -a&-b end{pmatrix}$$
but this is overkill.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This is a cleaner and easier argument. Of course, it relies on the fact that the matrix $A$ has rank $1$.
    $endgroup$
    – Mindlack
    Jan 6 at 0:25










  • $begingroup$
    Very nice indeed!
    $endgroup$
    – Mike
    Jan 6 at 0:33










  • $begingroup$
    Can you please explain it a little further how to proceed with this
    $endgroup$
    – DD90
    Jan 6 at 0:35






  • 1




    $begingroup$
    If $A^2=6A$, then $A^3=A^2A=6AA=6*6A=6^2A$. Moreover, $A^4=A^3A=6^2AA=6^2*6A=6^3A$, and so on...
    $endgroup$
    – Mindlack
    Jan 6 at 0:52










  • $begingroup$
    @DD90 $$A^3=A^2 cdot A=6A cdot A =6 A^2 = 6^2A \A^4=A^3 cdot A=6^2A cdot A =6^2 A^2 = 6^3A \...$$
    $endgroup$
    – N. S.
    Jan 6 at 3:34



















2












$begingroup$

By the Cayley-Hamilton theorem, $A^{2014}=aI+bA$ for some unknown coefficients $a$ and $b$. This equation is also satisfied by $A$’s eigenvalues, which generates the system of equations $a=0$, $a+6b=6^{2014}$, from which $b=6^{2013}$. The required entry of $A^{2014}$ is therefore $2cdot6^{2013}$.



Note, too, that the eigenvalues of $A$ can be found by inspection. Its rows are obviously linearly dependent, so one of its eigenvalues is $0$. The other eigenvalue is then equal to $A$’s trace.






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    Since $A$ is clearly rank-1 (the two columns are multiples), you can directly decompose it into the outer product of two vectors:
    $$A=begin{bmatrix}1\-1end{bmatrix}begin{bmatrix}8&2end{bmatrix}$$



    Which now immediately gives you
    $$A^{2014}=left(begin{bmatrix}1\-1end{bmatrix}begin{bmatrix}8&2end{bmatrix}right)^{2014}=begin{bmatrix}1\-1end{bmatrix}left(begin{bmatrix}8&2end{bmatrix}begin{bmatrix}1\-1end{bmatrix}right)^{2013}begin{bmatrix}8&2end{bmatrix}=begin{bmatrix}1\-1end{bmatrix}6^{2013}begin{bmatrix}8&2end{bmatrix}=6^{2013}A$$






    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%2f3063354%2flet-a-beginpmatrix-82-8-2-endpmatrix-find-the-entry-in-the-first%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      5












      $begingroup$

      Hint Show that
      $$A^2=6A$$



      Note: You can show the stronger statement:
      $$begin{pmatrix} 8&2 \ -8&-2 end{pmatrix}begin{pmatrix} a&b \ -a&-b end{pmatrix}=6begin{pmatrix} a&b \ -a&-b end{pmatrix}$$
      but this is overkill.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        This is a cleaner and easier argument. Of course, it relies on the fact that the matrix $A$ has rank $1$.
        $endgroup$
        – Mindlack
        Jan 6 at 0:25










      • $begingroup$
        Very nice indeed!
        $endgroup$
        – Mike
        Jan 6 at 0:33










      • $begingroup$
        Can you please explain it a little further how to proceed with this
        $endgroup$
        – DD90
        Jan 6 at 0:35






      • 1




        $begingroup$
        If $A^2=6A$, then $A^3=A^2A=6AA=6*6A=6^2A$. Moreover, $A^4=A^3A=6^2AA=6^2*6A=6^3A$, and so on...
        $endgroup$
        – Mindlack
        Jan 6 at 0:52










      • $begingroup$
        @DD90 $$A^3=A^2 cdot A=6A cdot A =6 A^2 = 6^2A \A^4=A^3 cdot A=6^2A cdot A =6^2 A^2 = 6^3A \...$$
        $endgroup$
        – N. S.
        Jan 6 at 3:34
















      5












      $begingroup$

      Hint Show that
      $$A^2=6A$$



      Note: You can show the stronger statement:
      $$begin{pmatrix} 8&2 \ -8&-2 end{pmatrix}begin{pmatrix} a&b \ -a&-b end{pmatrix}=6begin{pmatrix} a&b \ -a&-b end{pmatrix}$$
      but this is overkill.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        This is a cleaner and easier argument. Of course, it relies on the fact that the matrix $A$ has rank $1$.
        $endgroup$
        – Mindlack
        Jan 6 at 0:25










      • $begingroup$
        Very nice indeed!
        $endgroup$
        – Mike
        Jan 6 at 0:33










      • $begingroup$
        Can you please explain it a little further how to proceed with this
        $endgroup$
        – DD90
        Jan 6 at 0:35






      • 1




        $begingroup$
        If $A^2=6A$, then $A^3=A^2A=6AA=6*6A=6^2A$. Moreover, $A^4=A^3A=6^2AA=6^2*6A=6^3A$, and so on...
        $endgroup$
        – Mindlack
        Jan 6 at 0:52










      • $begingroup$
        @DD90 $$A^3=A^2 cdot A=6A cdot A =6 A^2 = 6^2A \A^4=A^3 cdot A=6^2A cdot A =6^2 A^2 = 6^3A \...$$
        $endgroup$
        – N. S.
        Jan 6 at 3:34














      5












      5








      5





      $begingroup$

      Hint Show that
      $$A^2=6A$$



      Note: You can show the stronger statement:
      $$begin{pmatrix} 8&2 \ -8&-2 end{pmatrix}begin{pmatrix} a&b \ -a&-b end{pmatrix}=6begin{pmatrix} a&b \ -a&-b end{pmatrix}$$
      but this is overkill.






      share|cite|improve this answer









      $endgroup$



      Hint Show that
      $$A^2=6A$$



      Note: You can show the stronger statement:
      $$begin{pmatrix} 8&2 \ -8&-2 end{pmatrix}begin{pmatrix} a&b \ -a&-b end{pmatrix}=6begin{pmatrix} a&b \ -a&-b end{pmatrix}$$
      but this is overkill.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Jan 6 at 0:23









      N. S.N. S.

      105k7115210




      105k7115210












      • $begingroup$
        This is a cleaner and easier argument. Of course, it relies on the fact that the matrix $A$ has rank $1$.
        $endgroup$
        – Mindlack
        Jan 6 at 0:25










      • $begingroup$
        Very nice indeed!
        $endgroup$
        – Mike
        Jan 6 at 0:33










      • $begingroup$
        Can you please explain it a little further how to proceed with this
        $endgroup$
        – DD90
        Jan 6 at 0:35






      • 1




        $begingroup$
        If $A^2=6A$, then $A^3=A^2A=6AA=6*6A=6^2A$. Moreover, $A^4=A^3A=6^2AA=6^2*6A=6^3A$, and so on...
        $endgroup$
        – Mindlack
        Jan 6 at 0:52










      • $begingroup$
        @DD90 $$A^3=A^2 cdot A=6A cdot A =6 A^2 = 6^2A \A^4=A^3 cdot A=6^2A cdot A =6^2 A^2 = 6^3A \...$$
        $endgroup$
        – N. S.
        Jan 6 at 3:34


















      • $begingroup$
        This is a cleaner and easier argument. Of course, it relies on the fact that the matrix $A$ has rank $1$.
        $endgroup$
        – Mindlack
        Jan 6 at 0:25










      • $begingroup$
        Very nice indeed!
        $endgroup$
        – Mike
        Jan 6 at 0:33










      • $begingroup$
        Can you please explain it a little further how to proceed with this
        $endgroup$
        – DD90
        Jan 6 at 0:35






      • 1




        $begingroup$
        If $A^2=6A$, then $A^3=A^2A=6AA=6*6A=6^2A$. Moreover, $A^4=A^3A=6^2AA=6^2*6A=6^3A$, and so on...
        $endgroup$
        – Mindlack
        Jan 6 at 0:52










      • $begingroup$
        @DD90 $$A^3=A^2 cdot A=6A cdot A =6 A^2 = 6^2A \A^4=A^3 cdot A=6^2A cdot A =6^2 A^2 = 6^3A \...$$
        $endgroup$
        – N. S.
        Jan 6 at 3:34
















      $begingroup$
      This is a cleaner and easier argument. Of course, it relies on the fact that the matrix $A$ has rank $1$.
      $endgroup$
      – Mindlack
      Jan 6 at 0:25




      $begingroup$
      This is a cleaner and easier argument. Of course, it relies on the fact that the matrix $A$ has rank $1$.
      $endgroup$
      – Mindlack
      Jan 6 at 0:25












      $begingroup$
      Very nice indeed!
      $endgroup$
      – Mike
      Jan 6 at 0:33




      $begingroup$
      Very nice indeed!
      $endgroup$
      – Mike
      Jan 6 at 0:33












      $begingroup$
      Can you please explain it a little further how to proceed with this
      $endgroup$
      – DD90
      Jan 6 at 0:35




      $begingroup$
      Can you please explain it a little further how to proceed with this
      $endgroup$
      – DD90
      Jan 6 at 0:35




      1




      1




      $begingroup$
      If $A^2=6A$, then $A^3=A^2A=6AA=6*6A=6^2A$. Moreover, $A^4=A^3A=6^2AA=6^2*6A=6^3A$, and so on...
      $endgroup$
      – Mindlack
      Jan 6 at 0:52




      $begingroup$
      If $A^2=6A$, then $A^3=A^2A=6AA=6*6A=6^2A$. Moreover, $A^4=A^3A=6^2AA=6^2*6A=6^3A$, and so on...
      $endgroup$
      – Mindlack
      Jan 6 at 0:52












      $begingroup$
      @DD90 $$A^3=A^2 cdot A=6A cdot A =6 A^2 = 6^2A \A^4=A^3 cdot A=6^2A cdot A =6^2 A^2 = 6^3A \...$$
      $endgroup$
      – N. S.
      Jan 6 at 3:34




      $begingroup$
      @DD90 $$A^3=A^2 cdot A=6A cdot A =6 A^2 = 6^2A \A^4=A^3 cdot A=6^2A cdot A =6^2 A^2 = 6^3A \...$$
      $endgroup$
      – N. S.
      Jan 6 at 3:34











      2












      $begingroup$

      By the Cayley-Hamilton theorem, $A^{2014}=aI+bA$ for some unknown coefficients $a$ and $b$. This equation is also satisfied by $A$’s eigenvalues, which generates the system of equations $a=0$, $a+6b=6^{2014}$, from which $b=6^{2013}$. The required entry of $A^{2014}$ is therefore $2cdot6^{2013}$.



      Note, too, that the eigenvalues of $A$ can be found by inspection. Its rows are obviously linearly dependent, so one of its eigenvalues is $0$. The other eigenvalue is then equal to $A$’s trace.






      share|cite|improve this answer











      $endgroup$


















        2












        $begingroup$

        By the Cayley-Hamilton theorem, $A^{2014}=aI+bA$ for some unknown coefficients $a$ and $b$. This equation is also satisfied by $A$’s eigenvalues, which generates the system of equations $a=0$, $a+6b=6^{2014}$, from which $b=6^{2013}$. The required entry of $A^{2014}$ is therefore $2cdot6^{2013}$.



        Note, too, that the eigenvalues of $A$ can be found by inspection. Its rows are obviously linearly dependent, so one of its eigenvalues is $0$. The other eigenvalue is then equal to $A$’s trace.






        share|cite|improve this answer











        $endgroup$
















          2












          2








          2





          $begingroup$

          By the Cayley-Hamilton theorem, $A^{2014}=aI+bA$ for some unknown coefficients $a$ and $b$. This equation is also satisfied by $A$’s eigenvalues, which generates the system of equations $a=0$, $a+6b=6^{2014}$, from which $b=6^{2013}$. The required entry of $A^{2014}$ is therefore $2cdot6^{2013}$.



          Note, too, that the eigenvalues of $A$ can be found by inspection. Its rows are obviously linearly dependent, so one of its eigenvalues is $0$. The other eigenvalue is then equal to $A$’s trace.






          share|cite|improve this answer











          $endgroup$



          By the Cayley-Hamilton theorem, $A^{2014}=aI+bA$ for some unknown coefficients $a$ and $b$. This equation is also satisfied by $A$’s eigenvalues, which generates the system of equations $a=0$, $a+6b=6^{2014}$, from which $b=6^{2013}$. The required entry of $A^{2014}$ is therefore $2cdot6^{2013}$.



          Note, too, that the eigenvalues of $A$ can be found by inspection. Its rows are obviously linearly dependent, so one of its eigenvalues is $0$. The other eigenvalue is then equal to $A$’s trace.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 6 at 1:08

























          answered Jan 6 at 1:02









          amdamd

          31.6k21052




          31.6k21052























              0












              $begingroup$

              Since $A$ is clearly rank-1 (the two columns are multiples), you can directly decompose it into the outer product of two vectors:
              $$A=begin{bmatrix}1\-1end{bmatrix}begin{bmatrix}8&2end{bmatrix}$$



              Which now immediately gives you
              $$A^{2014}=left(begin{bmatrix}1\-1end{bmatrix}begin{bmatrix}8&2end{bmatrix}right)^{2014}=begin{bmatrix}1\-1end{bmatrix}left(begin{bmatrix}8&2end{bmatrix}begin{bmatrix}1\-1end{bmatrix}right)^{2013}begin{bmatrix}8&2end{bmatrix}=begin{bmatrix}1\-1end{bmatrix}6^{2013}begin{bmatrix}8&2end{bmatrix}=6^{2013}A$$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Since $A$ is clearly rank-1 (the two columns are multiples), you can directly decompose it into the outer product of two vectors:
                $$A=begin{bmatrix}1\-1end{bmatrix}begin{bmatrix}8&2end{bmatrix}$$



                Which now immediately gives you
                $$A^{2014}=left(begin{bmatrix}1\-1end{bmatrix}begin{bmatrix}8&2end{bmatrix}right)^{2014}=begin{bmatrix}1\-1end{bmatrix}left(begin{bmatrix}8&2end{bmatrix}begin{bmatrix}1\-1end{bmatrix}right)^{2013}begin{bmatrix}8&2end{bmatrix}=begin{bmatrix}1\-1end{bmatrix}6^{2013}begin{bmatrix}8&2end{bmatrix}=6^{2013}A$$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Since $A$ is clearly rank-1 (the two columns are multiples), you can directly decompose it into the outer product of two vectors:
                  $$A=begin{bmatrix}1\-1end{bmatrix}begin{bmatrix}8&2end{bmatrix}$$



                  Which now immediately gives you
                  $$A^{2014}=left(begin{bmatrix}1\-1end{bmatrix}begin{bmatrix}8&2end{bmatrix}right)^{2014}=begin{bmatrix}1\-1end{bmatrix}left(begin{bmatrix}8&2end{bmatrix}begin{bmatrix}1\-1end{bmatrix}right)^{2013}begin{bmatrix}8&2end{bmatrix}=begin{bmatrix}1\-1end{bmatrix}6^{2013}begin{bmatrix}8&2end{bmatrix}=6^{2013}A$$






                  share|cite|improve this answer









                  $endgroup$



                  Since $A$ is clearly rank-1 (the two columns are multiples), you can directly decompose it into the outer product of two vectors:
                  $$A=begin{bmatrix}1\-1end{bmatrix}begin{bmatrix}8&2end{bmatrix}$$



                  Which now immediately gives you
                  $$A^{2014}=left(begin{bmatrix}1\-1end{bmatrix}begin{bmatrix}8&2end{bmatrix}right)^{2014}=begin{bmatrix}1\-1end{bmatrix}left(begin{bmatrix}8&2end{bmatrix}begin{bmatrix}1\-1end{bmatrix}right)^{2013}begin{bmatrix}8&2end{bmatrix}=begin{bmatrix}1\-1end{bmatrix}6^{2013}begin{bmatrix}8&2end{bmatrix}=6^{2013}A$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 6 at 3:26









                  obscuransobscurans

                  1,152311




                  1,152311






























                      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%2f3063354%2flet-a-beginpmatrix-82-8-2-endpmatrix-find-the-entry-in-the-first%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

                      Probability when a professor distributes a quiz and homework assignment to a class of n students.

                      Image
                      0 0 $begingroup$ Need help with this problem. Suppose our lazy professor collects a quiz and a homework assignment from a class of n students one day, then distributes both the quizzes and the homework assignments back to the class in a random fashion for grading. Each student receives one quiz and one homework assignment to grade. (a) What is the probability that every student receives someone else's quiz to grade, and someone else's homework to grade? (b) What is the probability that no student receives both their own quiz and their own homework assignment to grade? In this case, some students may receive their own quiz, and others may receive their own homework assignment. (c) Compute the limiting probability as n approaches infinity in each case. ...
                      Read more

                      Aardman Animations

                      Image
                      Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com ...
                      Read more

                      Are they similar matrix

                      Image
                      0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
                      Read more