Matrix determinant effect

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
4
down vote

favorite












Determinant of the matrix $$A= beginbmatrixa&b&c\d&e&f\g&h&iendbmatrix$$
is $$det A=4.$$ So what is the determinant of beginbmatrix3a&3b&3c\-d&-e&-f\g-a&h-b&i-cendbmatrix



I found that the row operations that were done were $3R1, -1.R2,$ and the last one doesn't matter.



So is the determinant $3times 4times(-1)= -12$
or we have to do the inverse $4times 1/3 times 1/(-1)?$






linear-algebra determinant







share|cite|improve this question



















  • 3




    Yes the answer is -12. Your approach is correct.
    – SchrodingersCat
    1 hour ago














up vote
4
down vote

favorite












Determinant of the matrix $$A= beginbmatrixa&b&c\d&e&f\g&h&iendbmatrix$$
is $$det A=4.$$ So what is the determinant of beginbmatrix3a&3b&3c\-d&-e&-f\g-a&h-b&i-cendbmatrix



I found that the row operations that were done were $3R1, -1.R2,$ and the last one doesn't matter.



So is the determinant $3times 4times(-1)= -12$
or we have to do the inverse $4times 1/3 times 1/(-1)?$






linear-algebra determinant







share|cite|improve this question



















  • 3




    Yes the answer is -12. Your approach is correct.
    – SchrodingersCat
    1 hour ago












up vote
4
down vote

favorite









up vote
4
down vote

favorite











Determinant of the matrix $$A= beginbmatrixa&b&c\d&e&f\g&h&iendbmatrix$$
is $$det A=4.$$ So what is the determinant of beginbmatrix3a&3b&3c\-d&-e&-f\g-a&h-b&i-cendbmatrix



I found that the row operations that were done were $3R1, -1.R2,$ and the last one doesn't matter.



So is the determinant $3times 4times(-1)= -12$
or we have to do the inverse $4times 1/3 times 1/(-1)?$






linear-algebra determinant







share|cite|improve this question















Determinant of the matrix $$A= beginbmatrixa&b&c\d&e&f\g&h&iendbmatrix$$
is $$det A=4.$$ So what is the determinant of beginbmatrix3a&3b&3c\-d&-e&-f\g-a&h-b&i-cendbmatrix



I found that the row operations that were done were $3R1, -1.R2,$ and the last one doesn't matter.



So is the determinant $3times 4times(-1)= -12$
or we have to do the inverse $4times 1/3 times 1/(-1)?$






linear-algebra determinant




linear-algebra determinant






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 1 hour ago









user376343

2,1011715




2,1011715










asked 1 hour ago









Bakarr

345




345







  • 3




    Yes the answer is -12. Your approach is correct.
    – SchrodingersCat
    1 hour ago












  • 3




    Yes the answer is -12. Your approach is correct.
    – SchrodingersCat
    1 hour ago







3




3




Yes the answer is -12. Your approach is correct.
– SchrodingersCat
1 hour ago




Yes the answer is -12. Your approach is correct.
– SchrodingersCat
1 hour ago










2 Answers
2






active

oldest

votes

















up vote
4
down vote



accepted










Yes that's correct indeed by the Properties of the determinant we have that



$$detbeginbmatrixa&b&c\d&e&f\g&h&iendbmatrix=detbeginbmatrixa&b&c\d&e&f\g-a&h-b&i-cendbmatrix=\=frac13 detbeginbmatrix3a&3b&3c\d&e&f\g-a&h-b&i-cendbmatrix=-frac13 detbeginbmatrix3a&3b&3c\-d&-e&-f\g-a&h-b&i-cendbmatrix$$






share|cite|improve this answer





























    up vote
    4
    down vote













    Sometimes you can't tell if matrix $B$ can be generated from matrix $B$ or not. In such a case, one may expand each determinant and compare the result of each. However, this approach requires careful attention to signs!



    Let $X$ be the determinant of the first matrix and $Y$ be the determinant of the 2nd matrix, then we have:



    $ X = a(ei - fh) - b(di - fg) + c(dh - eg) $



    $ X = aei - afh - bdi + bfg + cdh - ceg $



    $ Y = 3afh - 3eai + 3ecg + 3bdi - 3bfg - 3cdh $



    $ fracY-3 = -afh + eai -ecg -bdi + bfg + cdh $



    You could re-arrange the terms of any of the equations to see that:



    $X=fracY-3$



    Since X=4,



    $Y=-12$






    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',
      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%2f2991712%2fmatrix-determinant-effect%23new-answer', 'question_page');

      );

      Post as a guest

















      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      4
      down vote



      accepted










      Yes that's correct indeed by the Properties of the determinant we have that



      $$detbeginbmatrixa&b&c\d&e&f\g&h&iendbmatrix=detbeginbmatrixa&b&c\d&e&f\g-a&h-b&i-cendbmatrix=\=frac13 detbeginbmatrix3a&3b&3c\d&e&f\g-a&h-b&i-cendbmatrix=-frac13 detbeginbmatrix3a&3b&3c\-d&-e&-f\g-a&h-b&i-cendbmatrix$$






      share|cite|improve this answer


























        up vote
        4
        down vote



        accepted










        Yes that's correct indeed by the Properties of the determinant we have that



        $$detbeginbmatrixa&b&c\d&e&f\g&h&iendbmatrix=detbeginbmatrixa&b&c\d&e&f\g-a&h-b&i-cendbmatrix=\=frac13 detbeginbmatrix3a&3b&3c\d&e&f\g-a&h-b&i-cendbmatrix=-frac13 detbeginbmatrix3a&3b&3c\-d&-e&-f\g-a&h-b&i-cendbmatrix$$






        share|cite|improve this answer
























          up vote
          4
          down vote



          accepted







          up vote
          4
          down vote



          accepted






          Yes that's correct indeed by the Properties of the determinant we have that



          $$detbeginbmatrixa&b&c\d&e&f\g&h&iendbmatrix=detbeginbmatrixa&b&c\d&e&f\g-a&h-b&i-cendbmatrix=\=frac13 detbeginbmatrix3a&3b&3c\d&e&f\g-a&h-b&i-cendbmatrix=-frac13 detbeginbmatrix3a&3b&3c\-d&-e&-f\g-a&h-b&i-cendbmatrix$$






          share|cite|improve this answer














          Yes that's correct indeed by the Properties of the determinant we have that



          $$detbeginbmatrixa&b&c\d&e&f\g&h&iendbmatrix=detbeginbmatrixa&b&c\d&e&f\g-a&h-b&i-cendbmatrix=\=frac13 detbeginbmatrix3a&3b&3c\d&e&f\g-a&h-b&i-cendbmatrix=-frac13 detbeginbmatrix3a&3b&3c\-d&-e&-f\g-a&h-b&i-cendbmatrix$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 1 hour ago

























          answered 1 hour ago









          gimusi

          82.9k74091




          82.9k74091




















              up vote
              4
              down vote













              Sometimes you can't tell if matrix $B$ can be generated from matrix $B$ or not. In such a case, one may expand each determinant and compare the result of each. However, this approach requires careful attention to signs!



              Let $X$ be the determinant of the first matrix and $Y$ be the determinant of the 2nd matrix, then we have:



              $ X = a(ei - fh) - b(di - fg) + c(dh - eg) $



              $ X = aei - afh - bdi + bfg + cdh - ceg $



              $ Y = 3afh - 3eai + 3ecg + 3bdi - 3bfg - 3cdh $



              $ fracY-3 = -afh + eai -ecg -bdi + bfg + cdh $



              You could re-arrange the terms of any of the equations to see that:



              $X=fracY-3$



              Since X=4,



              $Y=-12$






              share|cite|improve this answer


























                up vote
                4
                down vote













                Sometimes you can't tell if matrix $B$ can be generated from matrix $B$ or not. In such a case, one may expand each determinant and compare the result of each. However, this approach requires careful attention to signs!



                Let $X$ be the determinant of the first matrix and $Y$ be the determinant of the 2nd matrix, then we have:



                $ X = a(ei - fh) - b(di - fg) + c(dh - eg) $



                $ X = aei - afh - bdi + bfg + cdh - ceg $



                $ Y = 3afh - 3eai + 3ecg + 3bdi - 3bfg - 3cdh $



                $ fracY-3 = -afh + eai -ecg -bdi + bfg + cdh $



                You could re-arrange the terms of any of the equations to see that:



                $X=fracY-3$



                Since X=4,



                $Y=-12$






                share|cite|improve this answer
























                  up vote
                  4
                  down vote










                  up vote
                  4
                  down vote









                  Sometimes you can't tell if matrix $B$ can be generated from matrix $B$ or not. In such a case, one may expand each determinant and compare the result of each. However, this approach requires careful attention to signs!



                  Let $X$ be the determinant of the first matrix and $Y$ be the determinant of the 2nd matrix, then we have:



                  $ X = a(ei - fh) - b(di - fg) + c(dh - eg) $



                  $ X = aei - afh - bdi + bfg + cdh - ceg $



                  $ Y = 3afh - 3eai + 3ecg + 3bdi - 3bfg - 3cdh $



                  $ fracY-3 = -afh + eai -ecg -bdi + bfg + cdh $



                  You could re-arrange the terms of any of the equations to see that:



                  $X=fracY-3$



                  Since X=4,



                  $Y=-12$






                  share|cite|improve this answer














                  Sometimes you can't tell if matrix $B$ can be generated from matrix $B$ or not. In such a case, one may expand each determinant and compare the result of each. However, this approach requires careful attention to signs!



                  Let $X$ be the determinant of the first matrix and $Y$ be the determinant of the 2nd matrix, then we have:



                  $ X = a(ei - fh) - b(di - fg) + c(dh - eg) $



                  $ X = aei - afh - bdi + bfg + cdh - ceg $



                  $ Y = 3afh - 3eai + 3ecg + 3bdi - 3bfg - 3cdh $



                  $ fracY-3 = -afh + eai -ecg -bdi + bfg + cdh $



                  You could re-arrange the terms of any of the equations to see that:



                  $X=fracY-3$



                  Since X=4,



                  $Y=-12$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited 33 mins ago

























                  answered 52 mins ago









                  NoChance

                  3,54621221




                  3,54621221



























                       

                      draft saved


                      draft discarded















































                       


                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function ()
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2991712%2fmatrix-determinant-effect%23new-answer', 'question_page');

                      );

                      Post as a guest
































































                      Comments

                      Post a Comment

                      Popular posts from this blog

                      List of Gilmore Girls characters

                      Image
                      The season three cast—(Back row): Lane, Michel, Paris, Emily, Richard, Sookie, Miss Patty, Kirk; Front row: Jess, Luke, Lorelai, Rory, Dean This is a list of characters for the comedy-drama television series Gilmore Girls . Contents 1 Main characters 1.1 Lorelai Gilmore 1.2 Rory Gilmore 1.3 Sookie St. James 1.4 Lane Kim 1.5 Michel Gerard 1.6 Luke Danes 1.7 Emily Gilmore 1.8 Richard Gilmore 1.9 Paris Geller 1.10 Dean Forester 1.11 Jess Mariano 1.12 Kirk Gleason 1.13 Jason Stiles 1.14 Logan Huntzberger 2 Major recurring characters 2.1 Christopher Hayden 2.2 Jackson Belleville 2.3 Mrs. Kim 2.4 Tristin DuGray 2.5 Max Medina 2.6 Taylor Doose 2.7 Dave Rygalski 2.8 Marty 3 Recurring characters 3.1 Stars Hollow 3.2 Chilton 3.3 Yale 3.4 Other 4 Notable guest stars 5 Unseen characters 6 Notes 7 References Main characters Actor Character Appearances Season 1 Season 2 Season 3 Season 4 Season 5 Season 6 Season 7 A Year in the Life Lauren Graham Lorelai Gilmore...
                      Read more

                      What does second last employer means? [closed]

                      Image
                      Clash Royale CLAN TAG #URR8PPP .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty margin-bottom:0; up vote -3 down vote favorite Say, for example I have worked in the following companies, Company A => 2010 - 2011 Company B => 2011 - 2012 Company C => 2012 - Till Date Company D => Have offer Now assume Company D is asking to submit documents from my second last employer. So from the list of companies given, which company would be qualified as second last employer? software-industry employer share | improve this question asked Aug 10 '14 at 5:16 Rajaprabhu Aravindasamy 559 1 6 13 closed as off-topic by Jim G., jmort253 ♦ Aug 11 '14 at 1:58 This question appears to be off-topic. The users who voted to close gave this specific reason: "Questions seeking advice on company-specific regulations, agreements, or policies should be directed to your manager or HR department. Questions that address...
                      Read more

                      One-line joke

                      A one-liner is a joke that is delivered in a single line. A good one-liner is said to be pithy - concise and meaningful. [1] Comedians and actors use this comedic method as part of their act, e.g. Jimmy Carr, Tommy Cooper, Rodney Dangerfield, Norm Macdonald, Ken Dodd, Stewart Francis, Zach Galifianakis, Mitch Hedberg, Anthony Jeselnik, Milton Jones, Shappi Khorsandi, Jay London, Mark Linn-Baker, Demetri Martin, Groucho Marx, Dan Mintz, Emo Philips, Tim Vine, Steven Wright and Henny Youngman and Arnold Schwarzenegger. [ citation needed ] Many fictional characters are also known to deliver one-liners, including James Bond, who usually includes pithy and laconic quips after disposing of a villain. [ citation needed ] Examples "A baby seal walks into a club." "A dyslexic man walks into a bra." (Anonymous) "There are three types of people, those who can count and those who can't." "Venison's dear, isn't it?" (Jimmy Carr) "An escala...
                      Read more