Using Cramer's Rule to solve a system of linear equation












0












$begingroup$


I have trouble solving the following question:
Question



How I approached it initially was as follow:



1) Since the questions asks for x3 only, I replaced the x3 column with the particular solution.
Step 1



2) Cramer's Rule asks of us to divide the det(Substituted Matrix in Step 1) by the det(Initial Matrix)
Step 2



3) To find the det of each 4x4 matrix I did C(1,1)(det(3x3 sub Matrix)). So 1(det(3x3 sub Matrix)). I evaluated the 3x3 sub Matrix as follow: 3x3 sub Matrix The red arrows are added and then subtracted to the addition of the blue arrows. Thus det(Substituted Matrix) = 3aei + 3bfg + 2cdh - 3ceg - 2afh - 3bdi.



4) I did step 3 for the initial Matrix as well and got det(Initial Matrix) = aei + bfg + cdh - ceg - afh - bdi



Now, I am unsure how to work around the determinants I just got. I should factor out something and then cancel the rest, but I don't see what. Did I miss something?






linear-algebra







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    I have trouble solving the following question:
    Question



    How I approached it initially was as follow:



    1) Since the questions asks for x3 only, I replaced the x3 column with the particular solution.
    Step 1



    2) Cramer's Rule asks of us to divide the det(Substituted Matrix in Step 1) by the det(Initial Matrix)
    Step 2



    3) To find the det of each 4x4 matrix I did C(1,1)(det(3x3 sub Matrix)). So 1(det(3x3 sub Matrix)). I evaluated the 3x3 sub Matrix as follow: 3x3 sub Matrix The red arrows are added and then subtracted to the addition of the blue arrows. Thus det(Substituted Matrix) = 3aei + 3bfg + 2cdh - 3ceg - 2afh - 3bdi.



    4) I did step 3 for the initial Matrix as well and got det(Initial Matrix) = aei + bfg + cdh - ceg - afh - bdi



    Now, I am unsure how to work around the determinants I just got. I should factor out something and then cancel the rest, but I don't see what. Did I miss something?






    linear-algebra







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      I have trouble solving the following question:
      Question



      How I approached it initially was as follow:



      1) Since the questions asks for x3 only, I replaced the x3 column with the particular solution.
      Step 1



      2) Cramer's Rule asks of us to divide the det(Substituted Matrix in Step 1) by the det(Initial Matrix)
      Step 2



      3) To find the det of each 4x4 matrix I did C(1,1)(det(3x3 sub Matrix)). So 1(det(3x3 sub Matrix)). I evaluated the 3x3 sub Matrix as follow: 3x3 sub Matrix The red arrows are added and then subtracted to the addition of the blue arrows. Thus det(Substituted Matrix) = 3aei + 3bfg + 2cdh - 3ceg - 2afh - 3bdi.



      4) I did step 3 for the initial Matrix as well and got det(Initial Matrix) = aei + bfg + cdh - ceg - afh - bdi



      Now, I am unsure how to work around the determinants I just got. I should factor out something and then cancel the rest, but I don't see what. Did I miss something?






      linear-algebra







      share|cite|improve this question









      $endgroup$




      I have trouble solving the following question:
      Question



      How I approached it initially was as follow:



      1) Since the questions asks for x3 only, I replaced the x3 column with the particular solution.
      Step 1



      2) Cramer's Rule asks of us to divide the det(Substituted Matrix in Step 1) by the det(Initial Matrix)
      Step 2



      3) To find the det of each 4x4 matrix I did C(1,1)(det(3x3 sub Matrix)). So 1(det(3x3 sub Matrix)). I evaluated the 3x3 sub Matrix as follow: 3x3 sub Matrix The red arrows are added and then subtracted to the addition of the blue arrows. Thus det(Substituted Matrix) = 3aei + 3bfg + 2cdh - 3ceg - 2afh - 3bdi.



      4) I did step 3 for the initial Matrix as well and got det(Initial Matrix) = aei + bfg + cdh - ceg - afh - bdi



      Now, I am unsure how to work around the determinants I just got. I should factor out something and then cancel the rest, but I don't see what. Did I miss something?






      linear-algebra




      linear-algebra






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 13 '18 at 1:00









      HeykanHeykan

      11




      11






















          1 Answer
          1






          active

          oldest

          votes


















          0












          $begingroup$

          It is clear that $begin{bmatrix}x_1\x_2\x_3\x_4 end{bmatrix} = begin{bmatrix}0\0\3\2 end{bmatrix}$ without Cramer's rule...



          What does Cramer say?



          $x_1 = frac {detbegin{bmatrix}0&0&0&0\3b+2c&a&b&c\3e+2f&d&e&f\3h+2i&g&h&i end{bmatrix}}{detbegin{bmatrix}1&0&0&0\0&a&b&c\0&d&e&f\0&g&h&i end{bmatrix}} = 0$



          We know this because one row is all 0's



          $x_2 = frac {detbegin{bmatrix}1&0&0&0\0&3b+2c&b&c\0&3e+2f&e&f\0&3h+2i&h&i end{bmatrix}}{detbegin{bmatrix}1&0&0&0\0&a&b&c\0&d&e&f\0&g&h&i end{bmatrix}} = 0$



          We know this because clearly the columns are not linearly independent.



          $x_3 = frac {detbegin{bmatrix}1&0&0&0\0&a&3b+2c&c\0&d&3e+2f&f\0&g&3h+2i&i end{bmatrix}}{detbegin{bmatrix}1&0&0&0\0&a&b&c\0&d&e&f\0&g&h&i end{bmatrix}} = 0$



          rather than calculating this out.



          $begin{bmatrix}1&0&0&0\0&a&3b+2c&c\0&d&3e+2f&f\0&g&3h+2i&i end{bmatrix} = begin{bmatrix}1&0&0&0\0&a&b&c\0&d&e&f\0&g&h&i end{bmatrix}begin{bmatrix}1&0&0&0\0&2&3&0\0&0&1&0\0&0&0&1 end{bmatrix}$



          The determinant of the product equals the product of the determinants. And



          $detbegin{bmatrix}1&0&0&0\0&1&0&0\0&0&3&0\0&0&2&1 end{bmatrix} = 2$



          And $x_4$ can be approached the same way.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            You should get $3$. The last matrix is off slightly.
            $endgroup$
            – Chris Custer
            Dec 13 '18 at 1:32












          • $begingroup$
            I think it's $begin{pmatrix}1&0&0&0\0&1&0&0\0&0&3&0\0&0&2&1end{pmatrix}$.
            $endgroup$
            – Chris Custer
            Dec 13 '18 at 1:45










          • $begingroup$
            @ChrisCuster thanks, I something came up and rushed the ending.
            $endgroup$
            – Doug M
            Dec 13 '18 at 16:38











          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%2f3037460%2fusing-cramers-rule-to-solve-a-system-of-linear-equation%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









          0












          $begingroup$

          It is clear that $begin{bmatrix}x_1\x_2\x_3\x_4 end{bmatrix} = begin{bmatrix}0\0\3\2 end{bmatrix}$ without Cramer's rule...



          What does Cramer say?



          $x_1 = frac {detbegin{bmatrix}0&0&0&0\3b+2c&a&b&c\3e+2f&d&e&f\3h+2i&g&h&i end{bmatrix}}{detbegin{bmatrix}1&0&0&0\0&a&b&c\0&d&e&f\0&g&h&i end{bmatrix}} = 0$



          We know this because one row is all 0's



          $x_2 = frac {detbegin{bmatrix}1&0&0&0\0&3b+2c&b&c\0&3e+2f&e&f\0&3h+2i&h&i end{bmatrix}}{detbegin{bmatrix}1&0&0&0\0&a&b&c\0&d&e&f\0&g&h&i end{bmatrix}} = 0$



          We know this because clearly the columns are not linearly independent.



          $x_3 = frac {detbegin{bmatrix}1&0&0&0\0&a&3b+2c&c\0&d&3e+2f&f\0&g&3h+2i&i end{bmatrix}}{detbegin{bmatrix}1&0&0&0\0&a&b&c\0&d&e&f\0&g&h&i end{bmatrix}} = 0$



          rather than calculating this out.



          $begin{bmatrix}1&0&0&0\0&a&3b+2c&c\0&d&3e+2f&f\0&g&3h+2i&i end{bmatrix} = begin{bmatrix}1&0&0&0\0&a&b&c\0&d&e&f\0&g&h&i end{bmatrix}begin{bmatrix}1&0&0&0\0&2&3&0\0&0&1&0\0&0&0&1 end{bmatrix}$



          The determinant of the product equals the product of the determinants. And



          $detbegin{bmatrix}1&0&0&0\0&1&0&0\0&0&3&0\0&0&2&1 end{bmatrix} = 2$



          And $x_4$ can be approached the same way.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            You should get $3$. The last matrix is off slightly.
            $endgroup$
            – Chris Custer
            Dec 13 '18 at 1:32












          • $begingroup$
            I think it's $begin{pmatrix}1&0&0&0\0&1&0&0\0&0&3&0\0&0&2&1end{pmatrix}$.
            $endgroup$
            – Chris Custer
            Dec 13 '18 at 1:45










          • $begingroup$
            @ChrisCuster thanks, I something came up and rushed the ending.
            $endgroup$
            – Doug M
            Dec 13 '18 at 16:38
















          0












          $begingroup$

          It is clear that $begin{bmatrix}x_1\x_2\x_3\x_4 end{bmatrix} = begin{bmatrix}0\0\3\2 end{bmatrix}$ without Cramer's rule...



          What does Cramer say?



          $x_1 = frac {detbegin{bmatrix}0&0&0&0\3b+2c&a&b&c\3e+2f&d&e&f\3h+2i&g&h&i end{bmatrix}}{detbegin{bmatrix}1&0&0&0\0&a&b&c\0&d&e&f\0&g&h&i end{bmatrix}} = 0$



          We know this because one row is all 0's



          $x_2 = frac {detbegin{bmatrix}1&0&0&0\0&3b+2c&b&c\0&3e+2f&e&f\0&3h+2i&h&i end{bmatrix}}{detbegin{bmatrix}1&0&0&0\0&a&b&c\0&d&e&f\0&g&h&i end{bmatrix}} = 0$



          We know this because clearly the columns are not linearly independent.



          $x_3 = frac {detbegin{bmatrix}1&0&0&0\0&a&3b+2c&c\0&d&3e+2f&f\0&g&3h+2i&i end{bmatrix}}{detbegin{bmatrix}1&0&0&0\0&a&b&c\0&d&e&f\0&g&h&i end{bmatrix}} = 0$



          rather than calculating this out.



          $begin{bmatrix}1&0&0&0\0&a&3b+2c&c\0&d&3e+2f&f\0&g&3h+2i&i end{bmatrix} = begin{bmatrix}1&0&0&0\0&a&b&c\0&d&e&f\0&g&h&i end{bmatrix}begin{bmatrix}1&0&0&0\0&2&3&0\0&0&1&0\0&0&0&1 end{bmatrix}$



          The determinant of the product equals the product of the determinants. And



          $detbegin{bmatrix}1&0&0&0\0&1&0&0\0&0&3&0\0&0&2&1 end{bmatrix} = 2$



          And $x_4$ can be approached the same way.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            You should get $3$. The last matrix is off slightly.
            $endgroup$
            – Chris Custer
            Dec 13 '18 at 1:32












          • $begingroup$
            I think it's $begin{pmatrix}1&0&0&0\0&1&0&0\0&0&3&0\0&0&2&1end{pmatrix}$.
            $endgroup$
            – Chris Custer
            Dec 13 '18 at 1:45










          • $begingroup$
            @ChrisCuster thanks, I something came up and rushed the ending.
            $endgroup$
            – Doug M
            Dec 13 '18 at 16:38














          0












          0








          0





          $begingroup$

          It is clear that $begin{bmatrix}x_1\x_2\x_3\x_4 end{bmatrix} = begin{bmatrix}0\0\3\2 end{bmatrix}$ without Cramer's rule...



          What does Cramer say?



          $x_1 = frac {detbegin{bmatrix}0&0&0&0\3b+2c&a&b&c\3e+2f&d&e&f\3h+2i&g&h&i end{bmatrix}}{detbegin{bmatrix}1&0&0&0\0&a&b&c\0&d&e&f\0&g&h&i end{bmatrix}} = 0$



          We know this because one row is all 0's



          $x_2 = frac {detbegin{bmatrix}1&0&0&0\0&3b+2c&b&c\0&3e+2f&e&f\0&3h+2i&h&i end{bmatrix}}{detbegin{bmatrix}1&0&0&0\0&a&b&c\0&d&e&f\0&g&h&i end{bmatrix}} = 0$



          We know this because clearly the columns are not linearly independent.



          $x_3 = frac {detbegin{bmatrix}1&0&0&0\0&a&3b+2c&c\0&d&3e+2f&f\0&g&3h+2i&i end{bmatrix}}{detbegin{bmatrix}1&0&0&0\0&a&b&c\0&d&e&f\0&g&h&i end{bmatrix}} = 0$



          rather than calculating this out.



          $begin{bmatrix}1&0&0&0\0&a&3b+2c&c\0&d&3e+2f&f\0&g&3h+2i&i end{bmatrix} = begin{bmatrix}1&0&0&0\0&a&b&c\0&d&e&f\0&g&h&i end{bmatrix}begin{bmatrix}1&0&0&0\0&2&3&0\0&0&1&0\0&0&0&1 end{bmatrix}$



          The determinant of the product equals the product of the determinants. And



          $detbegin{bmatrix}1&0&0&0\0&1&0&0\0&0&3&0\0&0&2&1 end{bmatrix} = 2$



          And $x_4$ can be approached the same way.






          share|cite|improve this answer











          $endgroup$



          It is clear that $begin{bmatrix}x_1\x_2\x_3\x_4 end{bmatrix} = begin{bmatrix}0\0\3\2 end{bmatrix}$ without Cramer's rule...



          What does Cramer say?



          $x_1 = frac {detbegin{bmatrix}0&0&0&0\3b+2c&a&b&c\3e+2f&d&e&f\3h+2i&g&h&i end{bmatrix}}{detbegin{bmatrix}1&0&0&0\0&a&b&c\0&d&e&f\0&g&h&i end{bmatrix}} = 0$



          We know this because one row is all 0's



          $x_2 = frac {detbegin{bmatrix}1&0&0&0\0&3b+2c&b&c\0&3e+2f&e&f\0&3h+2i&h&i end{bmatrix}}{detbegin{bmatrix}1&0&0&0\0&a&b&c\0&d&e&f\0&g&h&i end{bmatrix}} = 0$



          We know this because clearly the columns are not linearly independent.



          $x_3 = frac {detbegin{bmatrix}1&0&0&0\0&a&3b+2c&c\0&d&3e+2f&f\0&g&3h+2i&i end{bmatrix}}{detbegin{bmatrix}1&0&0&0\0&a&b&c\0&d&e&f\0&g&h&i end{bmatrix}} = 0$



          rather than calculating this out.



          $begin{bmatrix}1&0&0&0\0&a&3b+2c&c\0&d&3e+2f&f\0&g&3h+2i&i end{bmatrix} = begin{bmatrix}1&0&0&0\0&a&b&c\0&d&e&f\0&g&h&i end{bmatrix}begin{bmatrix}1&0&0&0\0&2&3&0\0&0&1&0\0&0&0&1 end{bmatrix}$



          The determinant of the product equals the product of the determinants. And



          $detbegin{bmatrix}1&0&0&0\0&1&0&0\0&0&3&0\0&0&2&1 end{bmatrix} = 2$



          And $x_4$ can be approached the same way.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 13 '18 at 16:37

























          answered Dec 13 '18 at 1:18









          Doug MDoug M

          44.8k31854




          44.8k31854












          • $begingroup$
            You should get $3$. The last matrix is off slightly.
            $endgroup$
            – Chris Custer
            Dec 13 '18 at 1:32












          • $begingroup$
            I think it's $begin{pmatrix}1&0&0&0\0&1&0&0\0&0&3&0\0&0&2&1end{pmatrix}$.
            $endgroup$
            – Chris Custer
            Dec 13 '18 at 1:45










          • $begingroup$
            @ChrisCuster thanks, I something came up and rushed the ending.
            $endgroup$
            – Doug M
            Dec 13 '18 at 16:38


















          • $begingroup$
            You should get $3$. The last matrix is off slightly.
            $endgroup$
            – Chris Custer
            Dec 13 '18 at 1:32












          • $begingroup$
            I think it's $begin{pmatrix}1&0&0&0\0&1&0&0\0&0&3&0\0&0&2&1end{pmatrix}$.
            $endgroup$
            – Chris Custer
            Dec 13 '18 at 1:45










          • $begingroup$
            @ChrisCuster thanks, I something came up and rushed the ending.
            $endgroup$
            – Doug M
            Dec 13 '18 at 16:38
















          $begingroup$
          You should get $3$. The last matrix is off slightly.
          $endgroup$
          – Chris Custer
          Dec 13 '18 at 1:32






          $begingroup$
          You should get $3$. The last matrix is off slightly.
          $endgroup$
          – Chris Custer
          Dec 13 '18 at 1:32














          $begingroup$
          I think it's $begin{pmatrix}1&0&0&0\0&1&0&0\0&0&3&0\0&0&2&1end{pmatrix}$.
          $endgroup$
          – Chris Custer
          Dec 13 '18 at 1:45




          $begingroup$
          I think it's $begin{pmatrix}1&0&0&0\0&1&0&0\0&0&3&0\0&0&2&1end{pmatrix}$.
          $endgroup$
          – Chris Custer
          Dec 13 '18 at 1:45












          $begingroup$
          @ChrisCuster thanks, I something came up and rushed the ending.
          $endgroup$
          – Doug M
          Dec 13 '18 at 16:38




          $begingroup$
          @ChrisCuster thanks, I something came up and rushed the ending.
          $endgroup$
          – Doug M
          Dec 13 '18 at 16:38


















          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%2f3037460%2fusing-cramers-rule-to-solve-a-system-of-linear-equation%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

          Wiesbaden

          Image
          Wappen Deutschlandkarte 50.0825 8.24 117 Koordinaten: 50° 5′  N , 8° 14′  O Basisdaten Bundesland: Hessen Regierungsbezirk: Darmstadt Höhe: 117 m ü. NHN Fläche: 203,93 km 2 Einwohner: 278.654 (31. Dez. 2017) [1] Bevölkerungsdichte: 1366 Einwohner je km 2 Postleitzahlen: 65183–65207, 55246, 55252 Vorlage:Infobox Gemeinde in Deutschland/Wartung/PLZ enthält Text Vorwahlen: 0611, 06122, 06127, 06134 Kfz-Kennzeichen: WI Gemeindeschlüssel: 06 4 14 000 LOCODE: DE WIB NUTS: DE714 Stadtgliederung: 26 Ortsbezirke Adresse der Stadtverwaltung: Schlossplatz 6 65183 Wiesbaden Webpräsenz: www.wiesbaden.de Oberbürgermeister: Sven Gerich (SPD) Lage der Landeshauptstadt Wiesbaden in Hessen und im Regierungsbezirk Darmstadt Offizielles Logo der Landeshauptstadt Wiesbaden Flagge der Landeshauptstadt Wiesbaden Wiesbaden is
          Read more

          Marschland

          Image
          Kilometerweite Marsch in den Vier- und Marschlanden am Elbdeich Die Wedeler Marschlandschaft in Schleswig-Holstein Der Hadelner Kanal zur Entwässerung der Marsch Die tief gelegene Marsch und ein Entwässerungskanal Das Schöpfwerk Otterndorf im Land Hadeln zur Entwässerung der Medem verfügte damals bereits über die größte Kreiselpumpe Europas. Das Schöpfwerk in Neuhaus für die Aue und des Neuhaus-Bülkauer Kanals Als Marsch(land) (v. niederdt., altsächs. mersc ) – auch Masch , Mersch oder Schwemmland genannt – bezeichnet man eine nacheiszeitlich entstandene geomorphologische Landform im Gebiet der nordwestdeutschen Küsten und Flüsse sowie vergleichbare Landformen weltweit. Inhaltsverzeichnis 1 Landschaft 2 Entstehung 3 Marschböden 4 Nutzung 5 Marschgebiete in Deutschland 6 Siehe auch 7 Literatur 8 Weblinks 9 Einzelnachweise Landschaft | Marschen sind generell flache Landstriche ohne natürliche Erhebungen. Si
          Read more

          Dieringhausen

          Image
          Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
          Read more