Matrices and Probability question












1












$begingroup$


Question:




Let Matrix $A$ be a non-singular Matrix and satisfies with Matrix $B$ such that $A^2B=A$ . Given $S={-1,0,1}$ and
$$ A= begin{bmatrix} a & b \ c & d \ end{bmatrix} $$
in which $a,b,c,din S $ . What is the probability that $det(A+B)=detA + detB $




I did manage to work out and get the equation $a^2+d^2=-2bc$ and calculate $frac{8}{81}$? But i'm not sure if its the correct answer.






linear-algebra probability matrices







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    Question:




    Let Matrix $A$ be a non-singular Matrix and satisfies with Matrix $B$ such that $A^2B=A$ . Given $S={-1,0,1}$ and
    $$ A= begin{bmatrix} a & b \ c & d \ end{bmatrix} $$
    in which $a,b,c,din S $ . What is the probability that $det(A+B)=detA + detB $




    I did manage to work out and get the equation $a^2+d^2=-2bc$ and calculate $frac{8}{81}$? But i'm not sure if its the correct answer.






    linear-algebra probability matrices







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      Question:




      Let Matrix $A$ be a non-singular Matrix and satisfies with Matrix $B$ such that $A^2B=A$ . Given $S={-1,0,1}$ and
      $$ A= begin{bmatrix} a & b \ c & d \ end{bmatrix} $$
      in which $a,b,c,din S $ . What is the probability that $det(A+B)=detA + detB $




      I did manage to work out and get the equation $a^2+d^2=-2bc$ and calculate $frac{8}{81}$? But i'm not sure if its the correct answer.






      linear-algebra probability matrices







      share|cite|improve this question











      $endgroup$




      Question:




      Let Matrix $A$ be a non-singular Matrix and satisfies with Matrix $B$ such that $A^2B=A$ . Given $S={-1,0,1}$ and
      $$ A= begin{bmatrix} a & b \ c & d \ end{bmatrix} $$
      in which $a,b,c,din S $ . What is the probability that $det(A+B)=detA + detB $




      I did manage to work out and get the equation $a^2+d^2=-2bc$ and calculate $frac{8}{81}$? But i'm not sure if its the correct answer.






      linear-algebra probability matrices




      linear-algebra probability matrices






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 1 at 9:15









      Maria Mazur

      48.7k1260122




      48.7k1260122










      asked Jan 1 at 7:17









      ThenewendingThenewending

      61




      61






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          Yes, you are correct!



          Since $A$ is non-singular:
          $$A^2B=A iff AB=I iff \
          B=A^{-1}=frac{1}{ad-bc}
          begin{pmatrix}
          d & -b\
          -c & a
          end{pmatrix};\
          A+B=begin{pmatrix}a+frac{d}{ad-bc}&b-frac{b}{ad-bc}\ c-frac{c}{ad-bc}&d+frac{a}{ad-bc}end{pmatrix}.$$
          So:
          $$det(A)+det(B)=ad-bc+frac{1}{(ad-bc)^2}(ad-bc)=frac{(ad-bc)^2+1}{ad-bc};\
          det(A+B)=ad+frac{a^2+d^2}{ad-bc}+frac{ad}{(ad-bc)^2}-bc+frac{2bc}{ad-bc}-frac{bc}{(ad-bc)^2};\
          det(A)+det(B)=det(A+B) Rightarrow \
          (ad-bc)^2+1=(ad-bc)^2+a^2+d^2+2bc+1 Rightarrow \
          a^2+d^2=-2bc.$$
          Note that $adne bc$ and the RHS must be positive. Then the favorable outcomes are:
          $$begin{array}{r|r|r|r}
          N&a&b&c&d\
          hline
          1&-1&-1&1&-1\
          2&1&-1&1&1\
          3&-1&1&-1&-1\
          4&1&1&-1&1
          end{array}$$
          There are total $3^4$ possible ways for $a,b,c,din {-1,0,1}$.



          Hence the required probability is:
          $$P=frac{n(text{favorable})}{n(text{total})}=frac4{81}.$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks so much! Btw in the outcome N=2,3,6,7 wouldn’t it give ad=bc?? In which it would make it a singular matrix?
            $endgroup$
            – Thenewending
            Jan 1 at 9:17












          • $begingroup$
            Yes, we must reject those cases, fixed.
            $endgroup$
            – farruhota
            Jan 1 at 9:20



















          0












          $begingroup$

          Use $$det(A)det(A+B)= det (A)(det(A)+det(B))$$



          so $$ det (A(A+B)) = det (A^2)+det (AB)$$



          so $$ det (A^2+I) = det (A^2)+1$$



          since $$ A^2= begin{bmatrix} a^2+bc & b(a+d) \ c(a+d) & d^2+bc \ end{bmatrix} = begin{bmatrix} x & y \ z & t \ end{bmatrix}$$



          we get $$(x+1)(t+1)-yz = xt-yz+1implies x+t=0$$



          so $$a^2+d^2+2bc=0$$






          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%2f3058263%2fmatrices-and-probability-question%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









            1












            $begingroup$

            Yes, you are correct!



            Since $A$ is non-singular:
            $$A^2B=A iff AB=I iff \
            B=A^{-1}=frac{1}{ad-bc}
            begin{pmatrix}
            d & -b\
            -c & a
            end{pmatrix};\
            A+B=begin{pmatrix}a+frac{d}{ad-bc}&b-frac{b}{ad-bc}\ c-frac{c}{ad-bc}&d+frac{a}{ad-bc}end{pmatrix}.$$
            So:
            $$det(A)+det(B)=ad-bc+frac{1}{(ad-bc)^2}(ad-bc)=frac{(ad-bc)^2+1}{ad-bc};\
            det(A+B)=ad+frac{a^2+d^2}{ad-bc}+frac{ad}{(ad-bc)^2}-bc+frac{2bc}{ad-bc}-frac{bc}{(ad-bc)^2};\
            det(A)+det(B)=det(A+B) Rightarrow \
            (ad-bc)^2+1=(ad-bc)^2+a^2+d^2+2bc+1 Rightarrow \
            a^2+d^2=-2bc.$$
            Note that $adne bc$ and the RHS must be positive. Then the favorable outcomes are:
            $$begin{array}{r|r|r|r}
            N&a&b&c&d\
            hline
            1&-1&-1&1&-1\
            2&1&-1&1&1\
            3&-1&1&-1&-1\
            4&1&1&-1&1
            end{array}$$
            There are total $3^4$ possible ways for $a,b,c,din {-1,0,1}$.



            Hence the required probability is:
            $$P=frac{n(text{favorable})}{n(text{total})}=frac4{81}.$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Thanks so much! Btw in the outcome N=2,3,6,7 wouldn’t it give ad=bc?? In which it would make it a singular matrix?
              $endgroup$
              – Thenewending
              Jan 1 at 9:17












            • $begingroup$
              Yes, we must reject those cases, fixed.
              $endgroup$
              – farruhota
              Jan 1 at 9:20
















            1












            $begingroup$

            Yes, you are correct!



            Since $A$ is non-singular:
            $$A^2B=A iff AB=I iff \
            B=A^{-1}=frac{1}{ad-bc}
            begin{pmatrix}
            d & -b\
            -c & a
            end{pmatrix};\
            A+B=begin{pmatrix}a+frac{d}{ad-bc}&b-frac{b}{ad-bc}\ c-frac{c}{ad-bc}&d+frac{a}{ad-bc}end{pmatrix}.$$
            So:
            $$det(A)+det(B)=ad-bc+frac{1}{(ad-bc)^2}(ad-bc)=frac{(ad-bc)^2+1}{ad-bc};\
            det(A+B)=ad+frac{a^2+d^2}{ad-bc}+frac{ad}{(ad-bc)^2}-bc+frac{2bc}{ad-bc}-frac{bc}{(ad-bc)^2};\
            det(A)+det(B)=det(A+B) Rightarrow \
            (ad-bc)^2+1=(ad-bc)^2+a^2+d^2+2bc+1 Rightarrow \
            a^2+d^2=-2bc.$$
            Note that $adne bc$ and the RHS must be positive. Then the favorable outcomes are:
            $$begin{array}{r|r|r|r}
            N&a&b&c&d\
            hline
            1&-1&-1&1&-1\
            2&1&-1&1&1\
            3&-1&1&-1&-1\
            4&1&1&-1&1
            end{array}$$
            There are total $3^4$ possible ways for $a,b,c,din {-1,0,1}$.



            Hence the required probability is:
            $$P=frac{n(text{favorable})}{n(text{total})}=frac4{81}.$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Thanks so much! Btw in the outcome N=2,3,6,7 wouldn’t it give ad=bc?? In which it would make it a singular matrix?
              $endgroup$
              – Thenewending
              Jan 1 at 9:17












            • $begingroup$
              Yes, we must reject those cases, fixed.
              $endgroup$
              – farruhota
              Jan 1 at 9:20














            1












            1








            1





            $begingroup$

            Yes, you are correct!



            Since $A$ is non-singular:
            $$A^2B=A iff AB=I iff \
            B=A^{-1}=frac{1}{ad-bc}
            begin{pmatrix}
            d & -b\
            -c & a
            end{pmatrix};\
            A+B=begin{pmatrix}a+frac{d}{ad-bc}&b-frac{b}{ad-bc}\ c-frac{c}{ad-bc}&d+frac{a}{ad-bc}end{pmatrix}.$$
            So:
            $$det(A)+det(B)=ad-bc+frac{1}{(ad-bc)^2}(ad-bc)=frac{(ad-bc)^2+1}{ad-bc};\
            det(A+B)=ad+frac{a^2+d^2}{ad-bc}+frac{ad}{(ad-bc)^2}-bc+frac{2bc}{ad-bc}-frac{bc}{(ad-bc)^2};\
            det(A)+det(B)=det(A+B) Rightarrow \
            (ad-bc)^2+1=(ad-bc)^2+a^2+d^2+2bc+1 Rightarrow \
            a^2+d^2=-2bc.$$
            Note that $adne bc$ and the RHS must be positive. Then the favorable outcomes are:
            $$begin{array}{r|r|r|r}
            N&a&b&c&d\
            hline
            1&-1&-1&1&-1\
            2&1&-1&1&1\
            3&-1&1&-1&-1\
            4&1&1&-1&1
            end{array}$$
            There are total $3^4$ possible ways for $a,b,c,din {-1,0,1}$.



            Hence the required probability is:
            $$P=frac{n(text{favorable})}{n(text{total})}=frac4{81}.$$






            share|cite|improve this answer











            $endgroup$



            Yes, you are correct!



            Since $A$ is non-singular:
            $$A^2B=A iff AB=I iff \
            B=A^{-1}=frac{1}{ad-bc}
            begin{pmatrix}
            d & -b\
            -c & a
            end{pmatrix};\
            A+B=begin{pmatrix}a+frac{d}{ad-bc}&b-frac{b}{ad-bc}\ c-frac{c}{ad-bc}&d+frac{a}{ad-bc}end{pmatrix}.$$
            So:
            $$det(A)+det(B)=ad-bc+frac{1}{(ad-bc)^2}(ad-bc)=frac{(ad-bc)^2+1}{ad-bc};\
            det(A+B)=ad+frac{a^2+d^2}{ad-bc}+frac{ad}{(ad-bc)^2}-bc+frac{2bc}{ad-bc}-frac{bc}{(ad-bc)^2};\
            det(A)+det(B)=det(A+B) Rightarrow \
            (ad-bc)^2+1=(ad-bc)^2+a^2+d^2+2bc+1 Rightarrow \
            a^2+d^2=-2bc.$$
            Note that $adne bc$ and the RHS must be positive. Then the favorable outcomes are:
            $$begin{array}{r|r|r|r}
            N&a&b&c&d\
            hline
            1&-1&-1&1&-1\
            2&1&-1&1&1\
            3&-1&1&-1&-1\
            4&1&1&-1&1
            end{array}$$
            There are total $3^4$ possible ways for $a,b,c,din {-1,0,1}$.



            Hence the required probability is:
            $$P=frac{n(text{favorable})}{n(text{total})}=frac4{81}.$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 1 at 9:20

























            answered Jan 1 at 9:00









            farruhotafarruhota

            21.6k2842




            21.6k2842












            • $begingroup$
              Thanks so much! Btw in the outcome N=2,3,6,7 wouldn’t it give ad=bc?? In which it would make it a singular matrix?
              $endgroup$
              – Thenewending
              Jan 1 at 9:17












            • $begingroup$
              Yes, we must reject those cases, fixed.
              $endgroup$
              – farruhota
              Jan 1 at 9:20


















            • $begingroup$
              Thanks so much! Btw in the outcome N=2,3,6,7 wouldn’t it give ad=bc?? In which it would make it a singular matrix?
              $endgroup$
              – Thenewending
              Jan 1 at 9:17












            • $begingroup$
              Yes, we must reject those cases, fixed.
              $endgroup$
              – farruhota
              Jan 1 at 9:20
















            $begingroup$
            Thanks so much! Btw in the outcome N=2,3,6,7 wouldn’t it give ad=bc?? In which it would make it a singular matrix?
            $endgroup$
            – Thenewending
            Jan 1 at 9:17






            $begingroup$
            Thanks so much! Btw in the outcome N=2,3,6,7 wouldn’t it give ad=bc?? In which it would make it a singular matrix?
            $endgroup$
            – Thenewending
            Jan 1 at 9:17














            $begingroup$
            Yes, we must reject those cases, fixed.
            $endgroup$
            – farruhota
            Jan 1 at 9:20




            $begingroup$
            Yes, we must reject those cases, fixed.
            $endgroup$
            – farruhota
            Jan 1 at 9:20











            0












            $begingroup$

            Use $$det(A)det(A+B)= det (A)(det(A)+det(B))$$



            so $$ det (A(A+B)) = det (A^2)+det (AB)$$



            so $$ det (A^2+I) = det (A^2)+1$$



            since $$ A^2= begin{bmatrix} a^2+bc & b(a+d) \ c(a+d) & d^2+bc \ end{bmatrix} = begin{bmatrix} x & y \ z & t \ end{bmatrix}$$



            we get $$(x+1)(t+1)-yz = xt-yz+1implies x+t=0$$



            so $$a^2+d^2+2bc=0$$






            share|cite|improve this answer











            $endgroup$


















              0












              $begingroup$

              Use $$det(A)det(A+B)= det (A)(det(A)+det(B))$$



              so $$ det (A(A+B)) = det (A^2)+det (AB)$$



              so $$ det (A^2+I) = det (A^2)+1$$



              since $$ A^2= begin{bmatrix} a^2+bc & b(a+d) \ c(a+d) & d^2+bc \ end{bmatrix} = begin{bmatrix} x & y \ z & t \ end{bmatrix}$$



              we get $$(x+1)(t+1)-yz = xt-yz+1implies x+t=0$$



              so $$a^2+d^2+2bc=0$$






              share|cite|improve this answer











              $endgroup$
















                0












                0








                0





                $begingroup$

                Use $$det(A)det(A+B)= det (A)(det(A)+det(B))$$



                so $$ det (A(A+B)) = det (A^2)+det (AB)$$



                so $$ det (A^2+I) = det (A^2)+1$$



                since $$ A^2= begin{bmatrix} a^2+bc & b(a+d) \ c(a+d) & d^2+bc \ end{bmatrix} = begin{bmatrix} x & y \ z & t \ end{bmatrix}$$



                we get $$(x+1)(t+1)-yz = xt-yz+1implies x+t=0$$



                so $$a^2+d^2+2bc=0$$






                share|cite|improve this answer











                $endgroup$



                Use $$det(A)det(A+B)= det (A)(det(A)+det(B))$$



                so $$ det (A(A+B)) = det (A^2)+det (AB)$$



                so $$ det (A^2+I) = det (A^2)+1$$



                since $$ A^2= begin{bmatrix} a^2+bc & b(a+d) \ c(a+d) & d^2+bc \ end{bmatrix} = begin{bmatrix} x & y \ z & t \ end{bmatrix}$$



                we get $$(x+1)(t+1)-yz = xt-yz+1implies x+t=0$$



                so $$a^2+d^2+2bc=0$$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 1 at 9:14

























                answered Jan 1 at 9:09









                Maria MazurMaria Mazur

                48.7k1260122




                48.7k1260122






























                    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%2f3058263%2fmatrices-and-probability-question%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

                    How do I know what Microsoft account the skydrive app is syncing to?

                    Image
                    up vote 4 down vote favorite 1 Some time ago, I set up skydrive for my wife on her mac. But now she wants to access the files online, I cannot remember what Microsoft account I setup and linked her skydive app to and nor can I see anything in the skydrive app settings to tell me either? How can I see the name of the Microsoft account the skydive app is currently linked to? I would like to know this for skydrive app for Mac and PC onedrive share | improve this question asked Feb 25 '13 at 4:52 Gary Barrett 260 2 5 12
                    Read more

                    When does type information flow backwards in C++?

                    Image
                    up vote 62 down vote favorite 19 I just watched Stephan T. Lavavej talk at CppCon 2018 on "Class Template Argument Deduction", where at some point he incidentally says: In C++ type information almost never flows backwards ... I had to say "almost" because there's one or two cases, possibly more but very few . Despite trying to figure out which cases he might be referring to, I couldn't come up with anything. Hence the question: In which cases the C++17 standard mandates that type information propagate backwards? c++ language-lawyer c++17 share | improve this question edited 7 hours ago scohe0
                    Read more

                    Grease: Live!

                    Image
                    A tradução deste artigo está abaixo da qualidade média aceitável. É possível que tenha sido feita por um tradutor automático ou por alguém que não conhece bem o português ou a língua original do texto. Caso queira colaborar com a Wikipédia, tente encontrar a página original e melhore este verbete conforme o guia de tradução. Grease: Live! Grease: Live!  (PRT/BRA)  Estados Unidos 2016 •   cor •   130 min. min   Direção Thomas Kail Alex Rudzinski Produção Marc Platt Adam Siegel Roteiro Allan Carr Robert Cary Warren Casey Jim Jacobs Jonathan Tolins Bronte Woodard Baseado em Grease (musical) Grease (filme) Elenco Julianne Hough Aaron Tveit Vanessa Hudgens Carlos Pena Jr. Keke Palmer Carly Rae Jepsen Kether Donohue Jordan Fisher David Del Rio Andrew Call Gênero Musical Música Tom Kitt Figurino William Ivey Long Companhia(s) produtora(s) Warner Bros. Studios Distribuição Fox Lançamento 31 de Janei
                    Read more