Is echelon form a requirement to show a matrix has no solutions?












2














Do I need to get a matrix into echelon form to prove that it has no solutions, or do I only need a pivot in the last column. For example, would the following row operation show that there are no solutions to the linear system represented that by the augmented matrix below?



$begin{bmatrix}1&-3&0&5\-1&1&5&2\-1&1&5&3end{bmatrix}$ $Rightarrow$ $begin{bmatrix}1&-3&0&5\-1&1&5&2\0&0&0&1end{bmatrix}$






linear-algebra matrices







share|cite|improve this question







New contributor




Gaussian Elimination is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • What entry did you pivot on?
    – coffeemath
    7 hours ago










  • I'm not sure what definition of pivot you are using. By pivot, I mean the "leading entry in a row".
    – Gaussian Elimination
    7 hours ago
















2














Do I need to get a matrix into echelon form to prove that it has no solutions, or do I only need a pivot in the last column. For example, would the following row operation show that there are no solutions to the linear system represented that by the augmented matrix below?



$begin{bmatrix}1&-3&0&5\-1&1&5&2\-1&1&5&3end{bmatrix}$ $Rightarrow$ $begin{bmatrix}1&-3&0&5\-1&1&5&2\0&0&0&1end{bmatrix}$






linear-algebra matrices







share|cite|improve this question







New contributor




Gaussian Elimination is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • What entry did you pivot on?
    – coffeemath
    7 hours ago










  • I'm not sure what definition of pivot you are using. By pivot, I mean the "leading entry in a row".
    – Gaussian Elimination
    7 hours ago














2












2








2


1





Do I need to get a matrix into echelon form to prove that it has no solutions, or do I only need a pivot in the last column. For example, would the following row operation show that there are no solutions to the linear system represented that by the augmented matrix below?



$begin{bmatrix}1&-3&0&5\-1&1&5&2\-1&1&5&3end{bmatrix}$ $Rightarrow$ $begin{bmatrix}1&-3&0&5\-1&1&5&2\0&0&0&1end{bmatrix}$






linear-algebra matrices







share|cite|improve this question







New contributor




Gaussian Elimination is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











Do I need to get a matrix into echelon form to prove that it has no solutions, or do I only need a pivot in the last column. For example, would the following row operation show that there are no solutions to the linear system represented that by the augmented matrix below?



$begin{bmatrix}1&-3&0&5\-1&1&5&2\-1&1&5&3end{bmatrix}$ $Rightarrow$ $begin{bmatrix}1&-3&0&5\-1&1&5&2\0&0&0&1end{bmatrix}$






linear-algebra matrices




linear-algebra matrices






share|cite|improve this question







New contributor




Gaussian Elimination is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question







New contributor




Gaussian Elimination is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question






New contributor




Gaussian Elimination is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 7 hours ago









Gaussian Elimination

112




112




New contributor




Gaussian Elimination is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Gaussian Elimination is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Gaussian Elimination is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • What entry did you pivot on?
    – coffeemath
    7 hours ago










  • I'm not sure what definition of pivot you are using. By pivot, I mean the "leading entry in a row".
    – Gaussian Elimination
    7 hours ago


















  • What entry did you pivot on?
    – coffeemath
    7 hours ago










  • I'm not sure what definition of pivot you are using. By pivot, I mean the "leading entry in a row".
    – Gaussian Elimination
    7 hours ago
















What entry did you pivot on?
– coffeemath
7 hours ago




What entry did you pivot on?
– coffeemath
7 hours ago












I'm not sure what definition of pivot you are using. By pivot, I mean the "leading entry in a row".
– Gaussian Elimination
7 hours ago




I'm not sure what definition of pivot you are using. By pivot, I mean the "leading entry in a row".
– Gaussian Elimination
7 hours ago










3 Answers
3






active

oldest

votes


















3














Your transformation suffices; if you translate your matrix back into a system of linear equations, the last row gives the equation $0 cdot x + 0 cdot y + 0 cdot z = 1$.



This is indeed the beauty of the representation of systems of linear equations as matrices: you perform simple operations on the rows of the matrix but preserve the solution set of the original system. At the end, you just read off the solutions.






share|cite|improve this answer





























    2














    No, it's not required. Once you've performed row operations (as you have) to a point where you reach an inconsistency, you can conclude the system has no solutions. Going all the way to echelon form, or rref, won't change that.






    share|cite|improve this answer





























      2














      You only have to prove the matrix of the homogeneous part and the augmented matrix do not have the same rank. In the present case, it is enough: the matrix of the homogeneous part has rank $2$ and the augmented matrix has rank $3$.






      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
        });


        }
        });






        Gaussian Elimination is a new contributor. Be nice, and check out our Code of Conduct.










        draft saved

        draft discarded


















        StackExchange.ready(
        function () {
        StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3052473%2fis-echelon-form-a-requirement-to-show-a-matrix-has-no-solutions%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









        3














        Your transformation suffices; if you translate your matrix back into a system of linear equations, the last row gives the equation $0 cdot x + 0 cdot y + 0 cdot z = 1$.



        This is indeed the beauty of the representation of systems of linear equations as matrices: you perform simple operations on the rows of the matrix but preserve the solution set of the original system. At the end, you just read off the solutions.






        share|cite|improve this answer


























          3














          Your transformation suffices; if you translate your matrix back into a system of linear equations, the last row gives the equation $0 cdot x + 0 cdot y + 0 cdot z = 1$.



          This is indeed the beauty of the representation of systems of linear equations as matrices: you perform simple operations on the rows of the matrix but preserve the solution set of the original system. At the end, you just read off the solutions.






          share|cite|improve this answer
























            3












            3








            3






            Your transformation suffices; if you translate your matrix back into a system of linear equations, the last row gives the equation $0 cdot x + 0 cdot y + 0 cdot z = 1$.



            This is indeed the beauty of the representation of systems of linear equations as matrices: you perform simple operations on the rows of the matrix but preserve the solution set of the original system. At the end, you just read off the solutions.






            share|cite|improve this answer












            Your transformation suffices; if you translate your matrix back into a system of linear equations, the last row gives the equation $0 cdot x + 0 cdot y + 0 cdot z = 1$.



            This is indeed the beauty of the representation of systems of linear equations as matrices: you perform simple operations on the rows of the matrix but preserve the solution set of the original system. At the end, you just read off the solutions.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 7 hours ago









            MacRance

            956




            956























                2














                No, it's not required. Once you've performed row operations (as you have) to a point where you reach an inconsistency, you can conclude the system has no solutions. Going all the way to echelon form, or rref, won't change that.






                share|cite|improve this answer


























                  2














                  No, it's not required. Once you've performed row operations (as you have) to a point where you reach an inconsistency, you can conclude the system has no solutions. Going all the way to echelon form, or rref, won't change that.






                  share|cite|improve this answer
























                    2












                    2








                    2






                    No, it's not required. Once you've performed row operations (as you have) to a point where you reach an inconsistency, you can conclude the system has no solutions. Going all the way to echelon form, or rref, won't change that.






                    share|cite|improve this answer












                    No, it's not required. Once you've performed row operations (as you have) to a point where you reach an inconsistency, you can conclude the system has no solutions. Going all the way to echelon form, or rref, won't change that.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 7 hours ago









                    pwerth

                    1,609411




                    1,609411























                        2














                        You only have to prove the matrix of the homogeneous part and the augmented matrix do not have the same rank. In the present case, it is enough: the matrix of the homogeneous part has rank $2$ and the augmented matrix has rank $3$.






                        share|cite|improve this answer


























                          2














                          You only have to prove the matrix of the homogeneous part and the augmented matrix do not have the same rank. In the present case, it is enough: the matrix of the homogeneous part has rank $2$ and the augmented matrix has rank $3$.






                          share|cite|improve this answer
























                            2












                            2








                            2






                            You only have to prove the matrix of the homogeneous part and the augmented matrix do not have the same rank. In the present case, it is enough: the matrix of the homogeneous part has rank $2$ and the augmented matrix has rank $3$.






                            share|cite|improve this answer












                            You only have to prove the matrix of the homogeneous part and the augmented matrix do not have the same rank. In the present case, it is enough: the matrix of the homogeneous part has rank $2$ and the augmented matrix has rank $3$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 7 hours ago









                            Bernard

                            118k638112




                            118k638112






















                                Gaussian Elimination is a new contributor. Be nice, and check out our Code of Conduct.










                                draft saved

                                draft discarded


















                                Gaussian Elimination is a new contributor. Be nice, and check out our Code of Conduct.













                                Gaussian Elimination is a new contributor. Be nice, and check out our Code of Conduct.












                                Gaussian Elimination is a new contributor. Be nice, and check out our Code of Conduct.
















                                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%2f3052473%2fis-echelon-form-a-requirement-to-show-a-matrix-has-no-solutions%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

                                サソリ

                                Image
                                この記事は検証可能な参考文献や出典が全く示されていないか、不十分です。 出典を追加して記事の信頼性向上にご協力ください。 （ 2019年1月 ） 「 サソリ 」のその他の用法については「サソリ (曖昧さ回避)」をご覧ください。 「 スコーピオン 」と「 スコルピオン 」はこの項目へ転送されています。その他の用法については「スコーピオン」をご覧ください。 サソリ目 生息年代: 430–0 Ma PreЄ Є O S D C P T J K Pg N シルル紀前期-現世 ダイオウサソリ（ Pandinus Imperator ） 分類 界 : 動物界 Animalia 門 : 節足動物門 Arthropoda 亜門 : 鋏角亜門 Chelicerata 綱 : クモガタ綱 Arachnida 目 : サソリ目 Scorpiones 学名 Scorpiones C. L. Koch, 1837 英名 Scorpion 上科 Pseudochactoidea キョクトウサソリ上科 Buthoidea Chaeriloidea Chactoidea Iuroidea Scorpionoidea サソリ （ サソリ類 、 蠍 、 蝎 、 scorpion ）は、鋏角亜門・クモガタ綱の サソリ目 （ Scorpiones ）に属する節足動物の総称である。体の前端に鋏型の触肢、後端に毒針を有する捕食者である。 1700種以上を含め、最古の化石記録は4億3千万年前のシルル紀まで遡る。有毒生物として広く知られるが、人間の命にかかわる毒を持つものはごく一部の種類に過ぎない。 目次 1 形態 1.1 前体 1.2 後体 1.2.1 中体 1.2.2 終体 2 雌雄 3 生態 3.1 天敵 3.2 生殖 4 分布 5 毒 6 蛍光現象 7 分類
                                Read more

                                広島県道265号伴広島線

                                Image
                                一般県道 広島県道265号伴広島線 路線延長 km 起点 広島県広島市安佐南区伴東7丁目 終点 広島県広島市中区榎町 接続する 主な道路 （記法） 広島県道38号広島豊平線 広島県道71号広島湯来線 山陽自動車道五日市IC 国道183号（ 国道261号重複） ■テンプレート（■ノート ■使い方） ■PJ道路 広島県道265号伴広島線 （ひろしまけんどう265ごう ともひろしません）は広島県広島市安佐南区から広島県広島市中区を結ぶ一般県道である。 目次 1 路線データ 2 通過する自治体 3 見どころ 4 接続道路 4.1 本線 4.2 旧道 5 別名 6 関連項目 路線データ 本線 起点：広島市安佐南区伴東7丁目・大原下橋南詰交差点（広島県道38号広島豊平線（本線・旧道）交点） 終点：広島市中区榎町・十日市交差点（国道183号（国道261号重用）交点） 総延長： 旧道 起点：広島市安佐南区伴東4丁目・新細坂交差点（広島県道38号広島豊平線交点） 終点：広島市安佐南区伴中央4丁目・寺組橋(南)交差点（広島県道265号伴広島線（本線）交点） 総延長： 通過する自治体 広島県 広島市（安佐南区、佐伯区、西区、中区） 見どころ 広島高速交通広島新交通1号線（アストラムライン） 広島広域公園 広島広域公園陸上競技場（広島ビッグアーチ） 己斐峠 己斐橋（太田川放水路） 天満橋（天満川） 接続道路 本線 広島県道38号広島豊平線（本線・旧道）（広島市安佐南区伴東7丁目・大原下橋南詰交差点〔起点〕） 広島県道71号広島湯来線（本線）（広島市安佐南区伴中央2丁目・伴交番入口交差点 - 広島市安佐南区伴中央4丁目・寺組橋(南)交差点間で重用） 広島県道71号広島湯来線（旧道）（広島県道265号伴広島線（旧道）重用）（広島市安佐南区伴中央4丁目・寺組橋(南)交差点） 広島県道71号広島湯来線（広島市安佐南区伴中央4丁目・寺組橋(南)交差点 - 広島市佐伯区五日市町石内・五月が丘(南)交差点間で重用）
                                Read more

                                Setup Asymptote in Texstudio

                                Image
                                0 Good time of the day! I am trying to compile the following file: documentclass[12pt]{article} usepackage[inline]{asymptote} begin{document} Here is a venn diagram produced with Asymptote, drawn to width 4cm: defA{A} defB{V{B}} begin{figure} begin{center} begin{asy} size(4cm,0); pen colour1=red; pen colour2=green; pair z0=(0,0); pair z1=(-1,0); pair z2=(1,0); real r=1.5; path c1=circle(z1,r); path c2=circle(z2,r); fill(c1,colour1); fill(c2,colour2); picture intersection=new picture; fill(intersection,c1,colour1+colour2); clip(intersection,c2); add(intersection); draw(c1); draw(c2); //draw("$A$",box,z1)
                                Read more