How to build a matrix out of one equation so that it can be solved with Gaussian elimination?












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$begingroup$


Find such
$ a, b, c, d ∈ ℝ $ that



$a(x^3 − x^2 + x − 1) + b(x^3 + x^2 + 3x−2) + c(x^2 + 3x + 1) +d(x^3 + 2x^2 − 2) + 7 = 0$



$∀x ∈ ℝ$ using Gauss' elimination.



How to get from the one equation to such matrix that Gaussian elimination can be used?






linear-algebra matrices gaussian-elimination







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$endgroup$

















    0












    $begingroup$


    Find such
    $ a, b, c, d ∈ ℝ $ that



    $a(x^3 − x^2 + x − 1) + b(x^3 + x^2 + 3x−2) + c(x^2 + 3x + 1) +d(x^3 + 2x^2 − 2) + 7 = 0$



    $∀x ∈ ℝ$ using Gauss' elimination.



    How to get from the one equation to such matrix that Gaussian elimination can be used?






    linear-algebra matrices gaussian-elimination







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Find such
      $ a, b, c, d ∈ ℝ $ that



      $a(x^3 − x^2 + x − 1) + b(x^3 + x^2 + 3x−2) + c(x^2 + 3x + 1) +d(x^3 + 2x^2 − 2) + 7 = 0$



      $∀x ∈ ℝ$ using Gauss' elimination.



      How to get from the one equation to such matrix that Gaussian elimination can be used?






      linear-algebra matrices gaussian-elimination







      share|cite|improve this question









      $endgroup$




      Find such
      $ a, b, c, d ∈ ℝ $ that



      $a(x^3 − x^2 + x − 1) + b(x^3 + x^2 + 3x−2) + c(x^2 + 3x + 1) +d(x^3 + 2x^2 − 2) + 7 = 0$



      $∀x ∈ ℝ$ using Gauss' elimination.



      How to get from the one equation to such matrix that Gaussian elimination can be used?






      linear-algebra matrices gaussian-elimination




      linear-algebra matrices gaussian-elimination






      share|cite|improve this question













      share|cite|improve this question











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      asked Oct 31 '18 at 23:50









      jasno1jasno1

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          $begingroup$

          Hint:



          Calculating the coefficients in each degree, you have to solve the system of linear equations:
          [begin{cases}begin{aligned}
          a +b+d&=0\
          -a+b+2d&=0\
          a+3b+2c&=0 \
          a+2b -c+2d&=7
          end{aligned} end{cases}
          Can you take it from there?






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            We have the following constraint




            • The coefficient of $x^3$ must be zero. ($a+b+d=0$)

            • The coefficient of $x^2$ must be zero. ($-a+b+c+2d=0$)

            • The coefficient of $x$ must be zero. ($a+3b+3c=0$)

            • The constant value must be zero. ($-a-2b+c-2d+7=0$)


            In matrix form:$$begin{bmatrix}1&1&0&1\-1&1&1&2\1&3&3&0\-1&-2&1&-2end{bmatrix}begin{bmatrix}a\b\c\dend{bmatrix}=begin{bmatrix}0\0\0\-7end{bmatrix}$$Therefore we need to perform operations on the following matrix$$begin{bmatrix}1&1&0&1&0\-1&1&1&2&0\1&3&3&0&0\-1&-2&1&-2&-7end{bmatrix}$$as below$$begin{bmatrix}1&1&0&1&0\0&2&1&3&0\0&2&3&-1&0\0&-1&1&-1&-7end{bmatrix}$$



            $$begin{bmatrix}2&0&-1&-1&0\0&2&1&3&0\0&0&2&-4&0\0&0&3&1&-14end{bmatrix}$$



            $$begin{bmatrix}4&0&0&-6&0\0&4&0&10&0\0&0&2&-4&0\0&0&0&14&-28end{bmatrix}$$or simply



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



            Let's continue what we were doing



            $$begin{bmatrix}2&0&0&0&-6\0&2&0&0&10\0&0&1&0&-4\0&0&0&1&-2end{bmatrix}$$which finally yields to



            $$begin{bmatrix}1&0&0&0&-3\0&1&0&0&5\0&0&1&0&-4\0&0&0&1&-2end{bmatrix}$$and at last$$begin{bmatrix}a\b\c\dend{bmatrix}=begin{bmatrix}-3\5\-4\-2end{bmatrix}$$






            share|cite|improve this answer









            $endgroup$













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              2 Answers
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              2 Answers
              2






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              0












              $begingroup$

              Hint:



              Calculating the coefficients in each degree, you have to solve the system of linear equations:
              [begin{cases}begin{aligned}
              a +b+d&=0\
              -a+b+2d&=0\
              a+3b+2c&=0 \
              a+2b -c+2d&=7
              end{aligned} end{cases}
              Can you take it from there?






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Hint:



                Calculating the coefficients in each degree, you have to solve the system of linear equations:
                [begin{cases}begin{aligned}
                a +b+d&=0\
                -a+b+2d&=0\
                a+3b+2c&=0 \
                a+2b -c+2d&=7
                end{aligned} end{cases}
                Can you take it from there?






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Hint:



                  Calculating the coefficients in each degree, you have to solve the system of linear equations:
                  [begin{cases}begin{aligned}
                  a +b+d&=0\
                  -a+b+2d&=0\
                  a+3b+2c&=0 \
                  a+2b -c+2d&=7
                  end{aligned} end{cases}
                  Can you take it from there?






                  share|cite|improve this answer









                  $endgroup$



                  Hint:



                  Calculating the coefficients in each degree, you have to solve the system of linear equations:
                  [begin{cases}begin{aligned}
                  a +b+d&=0\
                  -a+b+2d&=0\
                  a+3b+2c&=0 \
                  a+2b -c+2d&=7
                  end{aligned} end{cases}
                  Can you take it from there?







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 1 '18 at 0:06









                  BernardBernard

                  119k639112




                  119k639112























                      0












                      $begingroup$

                      We have the following constraint




                      • The coefficient of $x^3$ must be zero. ($a+b+d=0$)

                      • The coefficient of $x^2$ must be zero. ($-a+b+c+2d=0$)

                      • The coefficient of $x$ must be zero. ($a+3b+3c=0$)

                      • The constant value must be zero. ($-a-2b+c-2d+7=0$)


                      In matrix form:$$begin{bmatrix}1&1&0&1\-1&1&1&2\1&3&3&0\-1&-2&1&-2end{bmatrix}begin{bmatrix}a\b\c\dend{bmatrix}=begin{bmatrix}0\0\0\-7end{bmatrix}$$Therefore we need to perform operations on the following matrix$$begin{bmatrix}1&1&0&1&0\-1&1&1&2&0\1&3&3&0&0\-1&-2&1&-2&-7end{bmatrix}$$as below$$begin{bmatrix}1&1&0&1&0\0&2&1&3&0\0&2&3&-1&0\0&-1&1&-1&-7end{bmatrix}$$



                      $$begin{bmatrix}2&0&-1&-1&0\0&2&1&3&0\0&0&2&-4&0\0&0&3&1&-14end{bmatrix}$$



                      $$begin{bmatrix}4&0&0&-6&0\0&4&0&10&0\0&0&2&-4&0\0&0&0&14&-28end{bmatrix}$$or simply



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



                      Let's continue what we were doing



                      $$begin{bmatrix}2&0&0&0&-6\0&2&0&0&10\0&0&1&0&-4\0&0&0&1&-2end{bmatrix}$$which finally yields to



                      $$begin{bmatrix}1&0&0&0&-3\0&1&0&0&5\0&0&1&0&-4\0&0&0&1&-2end{bmatrix}$$and at last$$begin{bmatrix}a\b\c\dend{bmatrix}=begin{bmatrix}-3\5\-4\-2end{bmatrix}$$






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        We have the following constraint




                        • The coefficient of $x^3$ must be zero. ($a+b+d=0$)

                        • The coefficient of $x^2$ must be zero. ($-a+b+c+2d=0$)

                        • The coefficient of $x$ must be zero. ($a+3b+3c=0$)

                        • The constant value must be zero. ($-a-2b+c-2d+7=0$)


                        In matrix form:$$begin{bmatrix}1&1&0&1\-1&1&1&2\1&3&3&0\-1&-2&1&-2end{bmatrix}begin{bmatrix}a\b\c\dend{bmatrix}=begin{bmatrix}0\0\0\-7end{bmatrix}$$Therefore we need to perform operations on the following matrix$$begin{bmatrix}1&1&0&1&0\-1&1&1&2&0\1&3&3&0&0\-1&-2&1&-2&-7end{bmatrix}$$as below$$begin{bmatrix}1&1&0&1&0\0&2&1&3&0\0&2&3&-1&0\0&-1&1&-1&-7end{bmatrix}$$



                        $$begin{bmatrix}2&0&-1&-1&0\0&2&1&3&0\0&0&2&-4&0\0&0&3&1&-14end{bmatrix}$$



                        $$begin{bmatrix}4&0&0&-6&0\0&4&0&10&0\0&0&2&-4&0\0&0&0&14&-28end{bmatrix}$$or simply



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



                        Let's continue what we were doing



                        $$begin{bmatrix}2&0&0&0&-6\0&2&0&0&10\0&0&1&0&-4\0&0&0&1&-2end{bmatrix}$$which finally yields to



                        $$begin{bmatrix}1&0&0&0&-3\0&1&0&0&5\0&0&1&0&-4\0&0&0&1&-2end{bmatrix}$$and at last$$begin{bmatrix}a\b\c\dend{bmatrix}=begin{bmatrix}-3\5\-4\-2end{bmatrix}$$






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          We have the following constraint




                          • The coefficient of $x^3$ must be zero. ($a+b+d=0$)

                          • The coefficient of $x^2$ must be zero. ($-a+b+c+2d=0$)

                          • The coefficient of $x$ must be zero. ($a+3b+3c=0$)

                          • The constant value must be zero. ($-a-2b+c-2d+7=0$)


                          In matrix form:$$begin{bmatrix}1&1&0&1\-1&1&1&2\1&3&3&0\-1&-2&1&-2end{bmatrix}begin{bmatrix}a\b\c\dend{bmatrix}=begin{bmatrix}0\0\0\-7end{bmatrix}$$Therefore we need to perform operations on the following matrix$$begin{bmatrix}1&1&0&1&0\-1&1&1&2&0\1&3&3&0&0\-1&-2&1&-2&-7end{bmatrix}$$as below$$begin{bmatrix}1&1&0&1&0\0&2&1&3&0\0&2&3&-1&0\0&-1&1&-1&-7end{bmatrix}$$



                          $$begin{bmatrix}2&0&-1&-1&0\0&2&1&3&0\0&0&2&-4&0\0&0&3&1&-14end{bmatrix}$$



                          $$begin{bmatrix}4&0&0&-6&0\0&4&0&10&0\0&0&2&-4&0\0&0&0&14&-28end{bmatrix}$$or simply



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



                          Let's continue what we were doing



                          $$begin{bmatrix}2&0&0&0&-6\0&2&0&0&10\0&0&1&0&-4\0&0&0&1&-2end{bmatrix}$$which finally yields to



                          $$begin{bmatrix}1&0&0&0&-3\0&1&0&0&5\0&0&1&0&-4\0&0&0&1&-2end{bmatrix}$$and at last$$begin{bmatrix}a\b\c\dend{bmatrix}=begin{bmatrix}-3\5\-4\-2end{bmatrix}$$






                          share|cite|improve this answer









                          $endgroup$



                          We have the following constraint




                          • The coefficient of $x^3$ must be zero. ($a+b+d=0$)

                          • The coefficient of $x^2$ must be zero. ($-a+b+c+2d=0$)

                          • The coefficient of $x$ must be zero. ($a+3b+3c=0$)

                          • The constant value must be zero. ($-a-2b+c-2d+7=0$)


                          In matrix form:$$begin{bmatrix}1&1&0&1\-1&1&1&2\1&3&3&0\-1&-2&1&-2end{bmatrix}begin{bmatrix}a\b\c\dend{bmatrix}=begin{bmatrix}0\0\0\-7end{bmatrix}$$Therefore we need to perform operations on the following matrix$$begin{bmatrix}1&1&0&1&0\-1&1&1&2&0\1&3&3&0&0\-1&-2&1&-2&-7end{bmatrix}$$as below$$begin{bmatrix}1&1&0&1&0\0&2&1&3&0\0&2&3&-1&0\0&-1&1&-1&-7end{bmatrix}$$



                          $$begin{bmatrix}2&0&-1&-1&0\0&2&1&3&0\0&0&2&-4&0\0&0&3&1&-14end{bmatrix}$$



                          $$begin{bmatrix}4&0&0&-6&0\0&4&0&10&0\0&0&2&-4&0\0&0&0&14&-28end{bmatrix}$$or simply



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



                          Let's continue what we were doing



                          $$begin{bmatrix}2&0&0&0&-6\0&2&0&0&10\0&0&1&0&-4\0&0&0&1&-2end{bmatrix}$$which finally yields to



                          $$begin{bmatrix}1&0&0&0&-3\0&1&0&0&5\0&0&1&0&-4\0&0&0&1&-2end{bmatrix}$$and at last$$begin{bmatrix}a\b\c\dend{bmatrix}=begin{bmatrix}-3\5\-4\-2end{bmatrix}$$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Nov 24 '18 at 17:04









                          Mostafa AyazMostafa Ayaz

                          15.3k3939




                          15.3k3939






























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