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
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          active

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          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











          $endgroup$













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            2 Answers
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            2 Answers
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            active

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            active

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            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






























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