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












4














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




















  • 1




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






  • 1




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
















4














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




















  • 1




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






  • 1




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














4












4








4


2





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















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

75.6k897189




75.6k897189










asked Jan 6 at 0:16









DD90DD90

2638




2638








  • 1




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






  • 1




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














  • 1




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






  • 1




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








1




1




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




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




1




1




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




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










3 Answers
3






active

oldest

votes


















5














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





















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










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










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






  • 1




    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...
    – Mindlack
    Jan 6 at 0:52










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



















2














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































    0














    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





















      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














      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





















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










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










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






      • 1




        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...
        – Mindlack
        Jan 6 at 0:52










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
















      5














      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





















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










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










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






      • 1




        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...
        – Mindlack
        Jan 6 at 0:52










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














      5












      5








      5






      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












      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.

      102k6111207




      102k6111207












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










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










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






      • 1




        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...
        – Mindlack
        Jan 6 at 0:52










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


















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










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










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






      • 1




        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...
        – Mindlack
        Jan 6 at 0:52










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
















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




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












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




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












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




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




      1




      1




      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...
      – Mindlack
      Jan 6 at 0:52




      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...
      – Mindlack
      Jan 6 at 0:52












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




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











      2














      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




























        2














        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


























          2












          2








          2






          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














          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

          29.4k21050




          29.4k21050























              0














              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


























                0














                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
























                  0












                  0








                  0






                  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












                  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

                  878211




                  878211






























                      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.





                      Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                      Please pay close attention to the following guidance:


                      • 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.


                      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

                      Mario Kart Wii

                      Image
                      Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro...
                      Read more

                      The Binding of Isaac: Rebirth/Afterbirth

                      Image
                      Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! The Binding of Isaac: Rebirth/Afterbirth From The Cutting Room Floor < The Binding of Isaac: Rebirth Jump to: navigation, search This is a sub-page of The Binding of Isaac: Rebirth . Afterbirth is the first expansion pack for Rebirth and features a variety of improvements. Contents 1 Unused Graphics 1.1 Old Man Keeper 1.2 Skinless Hush 1.3 Miscellaneous Unused Hush Sprites 1.4 Generosity achievement 1.5 Afterbirth+ Room Editor 1.6 Bearded Ultra Greed 1.7 Early Ultra Greed 1.8 Unused Brownie Expression 1.9 Frail's Spinning Head 1.10 Mom's Dead Hand Alt 2 Unused Audio 2.1 Unused Dead Womb Layer 2.2 Unused Hush Battle Outro 2.3 The Stain Grunt Unused Graphics Old Man Keeper Image file: gfxuimain me...
                      Read more

                      What does “Dominus providebit” mean?

                      Image
                      5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1 ...
                      Read more