How could I determine the solution of this system of equations?












1












$begingroup$


The problem I'm facing is this one (with the answer of the question underlined):



Problem



I reduced the matrix from this:



begin{bmatrix}1&2&-1&4\3&-1&5&2\4&1&(a^2-14)&a+2end{bmatrix}



to this:
begin{bmatrix}1&2&-1&4\0&1&-8/7&10/7\0&0&(a^2-18)&a-4end{bmatrix}



But after that I don't understand why does the system of equations has a solution when $a != 4$ or $a != -4$.



What I learned, applied to this problem, is that:




  • If $a^2 - 18 = 0 $, the system of equations could have infinite solutions or none solution.


Please, could somebody give me a hand on this?
Thanks.






linear-algebra systems-of-equations







share|cite|improve this question









$endgroup$












  • $begingroup$
    As a note: you can do neq to get $neq$.
    $endgroup$
    – Dave
    Dec 31 '18 at 0:16










  • $begingroup$
    If $a^2=18$, then the system is inconsistent. For it to have an infinite number of solutions, the last row of the reduced matrix must consist entirely of zeros, which can never happen.
    $endgroup$
    – amd
    Dec 31 '18 at 0:38
















1












$begingroup$


The problem I'm facing is this one (with the answer of the question underlined):



Problem



I reduced the matrix from this:



begin{bmatrix}1&2&-1&4\3&-1&5&2\4&1&(a^2-14)&a+2end{bmatrix}



to this:
begin{bmatrix}1&2&-1&4\0&1&-8/7&10/7\0&0&(a^2-18)&a-4end{bmatrix}



But after that I don't understand why does the system of equations has a solution when $a != 4$ or $a != -4$.



What I learned, applied to this problem, is that:




  • If $a^2 - 18 = 0 $, the system of equations could have infinite solutions or none solution.


Please, could somebody give me a hand on this?
Thanks.






linear-algebra systems-of-equations







share|cite|improve this question









$endgroup$












  • $begingroup$
    As a note: you can do neq to get $neq$.
    $endgroup$
    – Dave
    Dec 31 '18 at 0:16










  • $begingroup$
    If $a^2=18$, then the system is inconsistent. For it to have an infinite number of solutions, the last row of the reduced matrix must consist entirely of zeros, which can never happen.
    $endgroup$
    – amd
    Dec 31 '18 at 0:38














1












1








1





$begingroup$


The problem I'm facing is this one (with the answer of the question underlined):



Problem



I reduced the matrix from this:



begin{bmatrix}1&2&-1&4\3&-1&5&2\4&1&(a^2-14)&a+2end{bmatrix}



to this:
begin{bmatrix}1&2&-1&4\0&1&-8/7&10/7\0&0&(a^2-18)&a-4end{bmatrix}



But after that I don't understand why does the system of equations has a solution when $a != 4$ or $a != -4$.



What I learned, applied to this problem, is that:




  • If $a^2 - 18 = 0 $, the system of equations could have infinite solutions or none solution.


Please, could somebody give me a hand on this?
Thanks.






linear-algebra systems-of-equations







share|cite|improve this question









$endgroup$




The problem I'm facing is this one (with the answer of the question underlined):



Problem



I reduced the matrix from this:



begin{bmatrix}1&2&-1&4\3&-1&5&2\4&1&(a^2-14)&a+2end{bmatrix}



to this:
begin{bmatrix}1&2&-1&4\0&1&-8/7&10/7\0&0&(a^2-18)&a-4end{bmatrix}



But after that I don't understand why does the system of equations has a solution when $a != 4$ or $a != -4$.



What I learned, applied to this problem, is that:




  • If $a^2 - 18 = 0 $, the system of equations could have infinite solutions or none solution.


Please, could somebody give me a hand on this?
Thanks.






linear-algebra systems-of-equations




linear-algebra systems-of-equations






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 31 '18 at 0:15









Carlos Córdova S.Carlos Córdova S.

134




134












  • $begingroup$
    As a note: you can do neq to get $neq$.
    $endgroup$
    – Dave
    Dec 31 '18 at 0:16










  • $begingroup$
    If $a^2=18$, then the system is inconsistent. For it to have an infinite number of solutions, the last row of the reduced matrix must consist entirely of zeros, which can never happen.
    $endgroup$
    – amd
    Dec 31 '18 at 0:38


















  • $begingroup$
    As a note: you can do neq to get $neq$.
    $endgroup$
    – Dave
    Dec 31 '18 at 0:16










  • $begingroup$
    If $a^2=18$, then the system is inconsistent. For it to have an infinite number of solutions, the last row of the reduced matrix must consist entirely of zeros, which can never happen.
    $endgroup$
    – amd
    Dec 31 '18 at 0:38
















$begingroup$
As a note: you can do neq to get $neq$.
$endgroup$
– Dave
Dec 31 '18 at 0:16




$begingroup$
As a note: you can do neq to get $neq$.
$endgroup$
– Dave
Dec 31 '18 at 0:16












$begingroup$
If $a^2=18$, then the system is inconsistent. For it to have an infinite number of solutions, the last row of the reduced matrix must consist entirely of zeros, which can never happen.
$endgroup$
– amd
Dec 31 '18 at 0:38




$begingroup$
If $a^2=18$, then the system is inconsistent. For it to have an infinite number of solutions, the last row of the reduced matrix must consist entirely of zeros, which can never happen.
$endgroup$
– amd
Dec 31 '18 at 0:38










2 Answers
2






active

oldest

votes


















0












$begingroup$

Let us do it without matrix calculations. We have the linear equations
$$x+2 y-z=4 tag 1$$
$$3 x-y+5 z=2 tag 2$$
$$4 x+y+left(a^2-14right) z=2+a tag 3$$
Using $(1)$ and $(2)$ eliminate $x$ and $y$ as functions of $z$
$$x=frac{8}{7}-frac{9 z}{7} qquad text{and} qquad y=frac{8 z}{7}+frac{10}{7}$$ Plug these results in $(3)$ to get
$$left(a^2-18right) z=a-4$$
S0, the only problem is $a^2=18$ which makes $z$ undefined. If $a^2neq 18$, whatever could be $a$ there are solutions for $x,y,z$.



This is what you did show. Well done and $to +1$ for your post.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    I checked your matrix reduction and it looks good. Your approach looks good too. From here, you would have no solutions if and only if $a^2=18$ since your second matrix would become $$begin{bmatrix}1&2&-1&|&4 \ 0&1&-8/7&|&10/7 \ 0&0&0&|&pmsqrt{18}-4end{bmatrix}$$ which has no solutions (it could not have infinitely many solutions in this case). If $a=pm 4$ the system does have solutions.



    Perhaps there is an error in the given solution, because your calculations and method look fine to me (also, I'm not familiar with the language in which the problem is written, so I am not sure exactly what the problem is asking for).






    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%2f3057308%2fhow-could-i-determine-the-solution-of-this-system-of-equations%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      0












      $begingroup$

      Let us do it without matrix calculations. We have the linear equations
      $$x+2 y-z=4 tag 1$$
      $$3 x-y+5 z=2 tag 2$$
      $$4 x+y+left(a^2-14right) z=2+a tag 3$$
      Using $(1)$ and $(2)$ eliminate $x$ and $y$ as functions of $z$
      $$x=frac{8}{7}-frac{9 z}{7} qquad text{and} qquad y=frac{8 z}{7}+frac{10}{7}$$ Plug these results in $(3)$ to get
      $$left(a^2-18right) z=a-4$$
      S0, the only problem is $a^2=18$ which makes $z$ undefined. If $a^2neq 18$, whatever could be $a$ there are solutions for $x,y,z$.



      This is what you did show. Well done and $to +1$ for your post.






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        Let us do it without matrix calculations. We have the linear equations
        $$x+2 y-z=4 tag 1$$
        $$3 x-y+5 z=2 tag 2$$
        $$4 x+y+left(a^2-14right) z=2+a tag 3$$
        Using $(1)$ and $(2)$ eliminate $x$ and $y$ as functions of $z$
        $$x=frac{8}{7}-frac{9 z}{7} qquad text{and} qquad y=frac{8 z}{7}+frac{10}{7}$$ Plug these results in $(3)$ to get
        $$left(a^2-18right) z=a-4$$
        S0, the only problem is $a^2=18$ which makes $z$ undefined. If $a^2neq 18$, whatever could be $a$ there are solutions for $x,y,z$.



        This is what you did show. Well done and $to +1$ for your post.






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          Let us do it without matrix calculations. We have the linear equations
          $$x+2 y-z=4 tag 1$$
          $$3 x-y+5 z=2 tag 2$$
          $$4 x+y+left(a^2-14right) z=2+a tag 3$$
          Using $(1)$ and $(2)$ eliminate $x$ and $y$ as functions of $z$
          $$x=frac{8}{7}-frac{9 z}{7} qquad text{and} qquad y=frac{8 z}{7}+frac{10}{7}$$ Plug these results in $(3)$ to get
          $$left(a^2-18right) z=a-4$$
          S0, the only problem is $a^2=18$ which makes $z$ undefined. If $a^2neq 18$, whatever could be $a$ there are solutions for $x,y,z$.



          This is what you did show. Well done and $to +1$ for your post.






          share|cite|improve this answer









          $endgroup$



          Let us do it without matrix calculations. We have the linear equations
          $$x+2 y-z=4 tag 1$$
          $$3 x-y+5 z=2 tag 2$$
          $$4 x+y+left(a^2-14right) z=2+a tag 3$$
          Using $(1)$ and $(2)$ eliminate $x$ and $y$ as functions of $z$
          $$x=frac{8}{7}-frac{9 z}{7} qquad text{and} qquad y=frac{8 z}{7}+frac{10}{7}$$ Plug these results in $(3)$ to get
          $$left(a^2-18right) z=a-4$$
          S0, the only problem is $a^2=18$ which makes $z$ undefined. If $a^2neq 18$, whatever could be $a$ there are solutions for $x,y,z$.



          This is what you did show. Well done and $to +1$ for your post.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 31 '18 at 2:21









          Claude LeiboviciClaude Leibovici

          125k1158135




          125k1158135























              0












              $begingroup$

              I checked your matrix reduction and it looks good. Your approach looks good too. From here, you would have no solutions if and only if $a^2=18$ since your second matrix would become $$begin{bmatrix}1&2&-1&|&4 \ 0&1&-8/7&|&10/7 \ 0&0&0&|&pmsqrt{18}-4end{bmatrix}$$ which has no solutions (it could not have infinitely many solutions in this case). If $a=pm 4$ the system does have solutions.



              Perhaps there is an error in the given solution, because your calculations and method look fine to me (also, I'm not familiar with the language in which the problem is written, so I am not sure exactly what the problem is asking for).






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                I checked your matrix reduction and it looks good. Your approach looks good too. From here, you would have no solutions if and only if $a^2=18$ since your second matrix would become $$begin{bmatrix}1&2&-1&|&4 \ 0&1&-8/7&|&10/7 \ 0&0&0&|&pmsqrt{18}-4end{bmatrix}$$ which has no solutions (it could not have infinitely many solutions in this case). If $a=pm 4$ the system does have solutions.



                Perhaps there is an error in the given solution, because your calculations and method look fine to me (also, I'm not familiar with the language in which the problem is written, so I am not sure exactly what the problem is asking for).






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  I checked your matrix reduction and it looks good. Your approach looks good too. From here, you would have no solutions if and only if $a^2=18$ since your second matrix would become $$begin{bmatrix}1&2&-1&|&4 \ 0&1&-8/7&|&10/7 \ 0&0&0&|&pmsqrt{18}-4end{bmatrix}$$ which has no solutions (it could not have infinitely many solutions in this case). If $a=pm 4$ the system does have solutions.



                  Perhaps there is an error in the given solution, because your calculations and method look fine to me (also, I'm not familiar with the language in which the problem is written, so I am not sure exactly what the problem is asking for).






                  share|cite|improve this answer











                  $endgroup$



                  I checked your matrix reduction and it looks good. Your approach looks good too. From here, you would have no solutions if and only if $a^2=18$ since your second matrix would become $$begin{bmatrix}1&2&-1&|&4 \ 0&1&-8/7&|&10/7 \ 0&0&0&|&pmsqrt{18}-4end{bmatrix}$$ which has no solutions (it could not have infinitely many solutions in this case). If $a=pm 4$ the system does have solutions.



                  Perhaps there is an error in the given solution, because your calculations and method look fine to me (also, I'm not familiar with the language in which the problem is written, so I am not sure exactly what the problem is asking for).







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 31 '18 at 0:42

























                  answered Dec 31 '18 at 0:26









                  DaveDave

                  9,05311033




                  9,05311033






























                      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%2f3057308%2fhow-could-i-determine-the-solution-of-this-system-of-equations%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