Determinant of matrix $Ainmathbb{R}^{n,n}$ [closed]












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$begingroup$


How to prove that for matrix $Ainmathbb{R}^{n,n}$ we have
$$det A = det
begin{vmatrix}
x & x& x& ...&x&x\
1-x& 1&1 & ...&1& 1\
0& 1-x& 1& ...&1&1\
0& 0&1-x& ...&1&1\
&&ldots&& \
0& 0& 0& ...&1-x&1\
end{vmatrix} =x^n
$$






matrices determinant







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closed as off-topic by Martin R, Saad, Dando18, A. Pongrácz, dantopa Dec 8 '18 at 17:22


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Martin R, Saad, Dando18, A. Pongrácz

If this question can be reworded to fit the rules in the help center, please edit the question.


















    0












    $begingroup$


    How to prove that for matrix $Ainmathbb{R}^{n,n}$ we have
    $$det A = det
    begin{vmatrix}
    x & x& x& ...&x&x\
    1-x& 1&1 & ...&1& 1\
    0& 1-x& 1& ...&1&1\
    0& 0&1-x& ...&1&1\
    &&ldots&& \
    0& 0& 0& ...&1-x&1\
    end{vmatrix} =x^n
    $$






    matrices determinant







    share|cite|improve this question









    $endgroup$



    closed as off-topic by Martin R, Saad, Dando18, A. Pongrácz, dantopa Dec 8 '18 at 17:22


    This question appears to be off-topic. The users who voted to close gave this specific reason:


    • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Martin R, Saad, Dando18, A. Pongrácz

    If this question can be reworded to fit the rules in the help center, please edit the question.
















      0












      0








      0


      0



      $begingroup$


      How to prove that for matrix $Ainmathbb{R}^{n,n}$ we have
      $$det A = det
      begin{vmatrix}
      x & x& x& ...&x&x\
      1-x& 1&1 & ...&1& 1\
      0& 1-x& 1& ...&1&1\
      0& 0&1-x& ...&1&1\
      &&ldots&& \
      0& 0& 0& ...&1-x&1\
      end{vmatrix} =x^n
      $$






      matrices determinant







      share|cite|improve this question









      $endgroup$




      How to prove that for matrix $Ainmathbb{R}^{n,n}$ we have
      $$det A = det
      begin{vmatrix}
      x & x& x& ...&x&x\
      1-x& 1&1 & ...&1& 1\
      0& 1-x& 1& ...&1&1\
      0& 0&1-x& ...&1&1\
      &&ldots&& \
      0& 0& 0& ...&1-x&1\
      end{vmatrix} =x^n
      $$






      matrices determinant




      matrices determinant






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




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      asked Dec 7 '18 at 7:54









      avan1235avan1235

      1946




      1946




      closed as off-topic by Martin R, Saad, Dando18, A. Pongrácz, dantopa Dec 8 '18 at 17:22


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Martin R, Saad, Dando18, A. Pongrácz

      If this question can be reworded to fit the rules in the help center, please edit the question.




      closed as off-topic by Martin R, Saad, Dando18, A. Pongrácz, dantopa Dec 8 '18 at 17:22


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Martin R, Saad, Dando18, A. Pongrácz

      If this question can be reworded to fit the rules in the help center, please edit the question.






















          2 Answers
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          $begingroup$

          Subtract every column other than the last by the last column. Then move the last column to the first (this gives you a factor of $(-1)^{n-1}$ in the determinant). The result is a lower triangular matrix whose main diagonal is $(x,-x,-x,ldots,-x)$. Thus the determinant of the original matrix is $(-1)^{n-1}x(-x)^{n-1}=x^n$.






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            Let the determinant of $Ainmathbb R^{ntimes n}$ be a polynomial in $x$ denoted as $P_n(x); P_1(x)=det[x]=x$. Expand along the first column,



            $det A=P_n(x)=
            begin{vmatrix}
            x& x & x & ldots & x& x \
            1-x & 1 & 1&ldots &1& 1 \
            0 & 1-x & 1 & ldots &1& 1 \
            0 & 0 & 1-x & ldots &1& 1 \
            vdots&vdots&vdots&&vdots&vdots\
            0 & 0 & 0 & ldots&1-x&1 notag
            end{vmatrix}_{n}\=xbegin{vmatrix}
            1& 1 &ldots&1 \
            1-x & 1&ldots &1\
            0 & 1-x& ldots &1\
            0 & 0& ldots &1\
            vdots&vdots&&vdots\
            0 & 0& ldots&1-xnotag
            end{vmatrix}_{n-1}+(x-1)begin{vmatrix}
            x& x&ldots & x\
            1-x & 1&ldots &1\
            0 & 1-x& ldots &1\
            0 & 0& ldots &1\
            vdots&vdots&&vdots\
            0 & 0 & ldots&1-xnotag
            end{vmatrix}_{n-1}$



            Note that the first determinant is just $frac {P_{n-1}(x)}x$ and the second one $P_{n-1}(x)$.



            $implies P_n(x)=xcdotfrac{P_{n-1}(x)}x+(x-1)P_{n-1}(x)\ =xP_{n-1}(x)\ =x^2P_{n-2}(x)\ =x^3P_{n-3}(x)\ =x^{n-1}P_{n-(n-1)}(x)\ =x^n$






            share|cite|improve this answer











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              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              3












              $begingroup$

              Subtract every column other than the last by the last column. Then move the last column to the first (this gives you a factor of $(-1)^{n-1}$ in the determinant). The result is a lower triangular matrix whose main diagonal is $(x,-x,-x,ldots,-x)$. Thus the determinant of the original matrix is $(-1)^{n-1}x(-x)^{n-1}=x^n$.






              share|cite|improve this answer









              $endgroup$


















                3












                $begingroup$

                Subtract every column other than the last by the last column. Then move the last column to the first (this gives you a factor of $(-1)^{n-1}$ in the determinant). The result is a lower triangular matrix whose main diagonal is $(x,-x,-x,ldots,-x)$. Thus the determinant of the original matrix is $(-1)^{n-1}x(-x)^{n-1}=x^n$.






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  Subtract every column other than the last by the last column. Then move the last column to the first (this gives you a factor of $(-1)^{n-1}$ in the determinant). The result is a lower triangular matrix whose main diagonal is $(x,-x,-x,ldots,-x)$. Thus the determinant of the original matrix is $(-1)^{n-1}x(-x)^{n-1}=x^n$.






                  share|cite|improve this answer









                  $endgroup$



                  Subtract every column other than the last by the last column. Then move the last column to the first (this gives you a factor of $(-1)^{n-1}$ in the determinant). The result is a lower triangular matrix whose main diagonal is $(x,-x,-x,ldots,-x)$. Thus the determinant of the original matrix is $(-1)^{n-1}x(-x)^{n-1}=x^n$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 7 '18 at 9:36









                  user1551user1551

                  72.1k566127




                  72.1k566127























                      1












                      $begingroup$

                      Let the determinant of $Ainmathbb R^{ntimes n}$ be a polynomial in $x$ denoted as $P_n(x); P_1(x)=det[x]=x$. Expand along the first column,



                      $det A=P_n(x)=
                      begin{vmatrix}
                      x& x & x & ldots & x& x \
                      1-x & 1 & 1&ldots &1& 1 \
                      0 & 1-x & 1 & ldots &1& 1 \
                      0 & 0 & 1-x & ldots &1& 1 \
                      vdots&vdots&vdots&&vdots&vdots\
                      0 & 0 & 0 & ldots&1-x&1 notag
                      end{vmatrix}_{n}\=xbegin{vmatrix}
                      1& 1 &ldots&1 \
                      1-x & 1&ldots &1\
                      0 & 1-x& ldots &1\
                      0 & 0& ldots &1\
                      vdots&vdots&&vdots\
                      0 & 0& ldots&1-xnotag
                      end{vmatrix}_{n-1}+(x-1)begin{vmatrix}
                      x& x&ldots & x\
                      1-x & 1&ldots &1\
                      0 & 1-x& ldots &1\
                      0 & 0& ldots &1\
                      vdots&vdots&&vdots\
                      0 & 0 & ldots&1-xnotag
                      end{vmatrix}_{n-1}$



                      Note that the first determinant is just $frac {P_{n-1}(x)}x$ and the second one $P_{n-1}(x)$.



                      $implies P_n(x)=xcdotfrac{P_{n-1}(x)}x+(x-1)P_{n-1}(x)\ =xP_{n-1}(x)\ =x^2P_{n-2}(x)\ =x^3P_{n-3}(x)\ =x^{n-1}P_{n-(n-1)}(x)\ =x^n$






                      share|cite|improve this answer











                      $endgroup$


















                        1












                        $begingroup$

                        Let the determinant of $Ainmathbb R^{ntimes n}$ be a polynomial in $x$ denoted as $P_n(x); P_1(x)=det[x]=x$. Expand along the first column,



                        $det A=P_n(x)=
                        begin{vmatrix}
                        x& x & x & ldots & x& x \
                        1-x & 1 & 1&ldots &1& 1 \
                        0 & 1-x & 1 & ldots &1& 1 \
                        0 & 0 & 1-x & ldots &1& 1 \
                        vdots&vdots&vdots&&vdots&vdots\
                        0 & 0 & 0 & ldots&1-x&1 notag
                        end{vmatrix}_{n}\=xbegin{vmatrix}
                        1& 1 &ldots&1 \
                        1-x & 1&ldots &1\
                        0 & 1-x& ldots &1\
                        0 & 0& ldots &1\
                        vdots&vdots&&vdots\
                        0 & 0& ldots&1-xnotag
                        end{vmatrix}_{n-1}+(x-1)begin{vmatrix}
                        x& x&ldots & x\
                        1-x & 1&ldots &1\
                        0 & 1-x& ldots &1\
                        0 & 0& ldots &1\
                        vdots&vdots&&vdots\
                        0 & 0 & ldots&1-xnotag
                        end{vmatrix}_{n-1}$



                        Note that the first determinant is just $frac {P_{n-1}(x)}x$ and the second one $P_{n-1}(x)$.



                        $implies P_n(x)=xcdotfrac{P_{n-1}(x)}x+(x-1)P_{n-1}(x)\ =xP_{n-1}(x)\ =x^2P_{n-2}(x)\ =x^3P_{n-3}(x)\ =x^{n-1}P_{n-(n-1)}(x)\ =x^n$






                        share|cite|improve this answer











                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          Let the determinant of $Ainmathbb R^{ntimes n}$ be a polynomial in $x$ denoted as $P_n(x); P_1(x)=det[x]=x$. Expand along the first column,



                          $det A=P_n(x)=
                          begin{vmatrix}
                          x& x & x & ldots & x& x \
                          1-x & 1 & 1&ldots &1& 1 \
                          0 & 1-x & 1 & ldots &1& 1 \
                          0 & 0 & 1-x & ldots &1& 1 \
                          vdots&vdots&vdots&&vdots&vdots\
                          0 & 0 & 0 & ldots&1-x&1 notag
                          end{vmatrix}_{n}\=xbegin{vmatrix}
                          1& 1 &ldots&1 \
                          1-x & 1&ldots &1\
                          0 & 1-x& ldots &1\
                          0 & 0& ldots &1\
                          vdots&vdots&&vdots\
                          0 & 0& ldots&1-xnotag
                          end{vmatrix}_{n-1}+(x-1)begin{vmatrix}
                          x& x&ldots & x\
                          1-x & 1&ldots &1\
                          0 & 1-x& ldots &1\
                          0 & 0& ldots &1\
                          vdots&vdots&&vdots\
                          0 & 0 & ldots&1-xnotag
                          end{vmatrix}_{n-1}$



                          Note that the first determinant is just $frac {P_{n-1}(x)}x$ and the second one $P_{n-1}(x)$.



                          $implies P_n(x)=xcdotfrac{P_{n-1}(x)}x+(x-1)P_{n-1}(x)\ =xP_{n-1}(x)\ =x^2P_{n-2}(x)\ =x^3P_{n-3}(x)\ =x^{n-1}P_{n-(n-1)}(x)\ =x^n$






                          share|cite|improve this answer











                          $endgroup$



                          Let the determinant of $Ainmathbb R^{ntimes n}$ be a polynomial in $x$ denoted as $P_n(x); P_1(x)=det[x]=x$. Expand along the first column,



                          $det A=P_n(x)=
                          begin{vmatrix}
                          x& x & x & ldots & x& x \
                          1-x & 1 & 1&ldots &1& 1 \
                          0 & 1-x & 1 & ldots &1& 1 \
                          0 & 0 & 1-x & ldots &1& 1 \
                          vdots&vdots&vdots&&vdots&vdots\
                          0 & 0 & 0 & ldots&1-x&1 notag
                          end{vmatrix}_{n}\=xbegin{vmatrix}
                          1& 1 &ldots&1 \
                          1-x & 1&ldots &1\
                          0 & 1-x& ldots &1\
                          0 & 0& ldots &1\
                          vdots&vdots&&vdots\
                          0 & 0& ldots&1-xnotag
                          end{vmatrix}_{n-1}+(x-1)begin{vmatrix}
                          x& x&ldots & x\
                          1-x & 1&ldots &1\
                          0 & 1-x& ldots &1\
                          0 & 0& ldots &1\
                          vdots&vdots&&vdots\
                          0 & 0 & ldots&1-xnotag
                          end{vmatrix}_{n-1}$



                          Note that the first determinant is just $frac {P_{n-1}(x)}x$ and the second one $P_{n-1}(x)$.



                          $implies P_n(x)=xcdotfrac{P_{n-1}(x)}x+(x-1)P_{n-1}(x)\ =xP_{n-1}(x)\ =x^2P_{n-2}(x)\ =x^3P_{n-3}(x)\ =x^{n-1}P_{n-(n-1)}(x)\ =x^n$







                          share|cite|improve this answer














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                          edited Dec 7 '18 at 7:59

























                          answered Dec 7 '18 at 7:56









                          Shubham JohriShubham Johri

                          4,689717




                          4,689717















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