Solving $2x-y=5$ and $x+y=1$ by Gaussian elimination [on hold]











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I am reading a book called "Linear algebra and its applications" by Gilbert Strang.



Chapter 1 is called “Matrices and Gaussian elimination”, the example in 1.1 I understand, Strang goes through the steps used.



Can someone please explain the steps to solving the two linear equations below using Gaussian elimination?



It's quite obvious that $x = 2$ and $y = 3$ but I'm struggling the understand the steps using Gaussian elimination to work it out.



begin{cases}
2x - y = 5 \[4px]
x + y = 1
end{cases}










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put on hold as off-topic by amWhy, user21820, Did, José Carlos Santos, Jyrki Lahtonen yesterday


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  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, user21820, Did, José Carlos Santos, Jyrki Lahtonen

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  • 3




    Since $2+3=5neq1$, I don't see how you find it “obvious” that $x=2$ and $y=3$.
    – José Carlos Santos
    Dec 2 at 10:40










  • Sorry my mistake I was meant to write 2x - y = 1 and x+ y = 5
    – Paul
    2 days ago















up vote
-2
down vote

favorite












I am reading a book called "Linear algebra and its applications" by Gilbert Strang.



Chapter 1 is called “Matrices and Gaussian elimination”, the example in 1.1 I understand, Strang goes through the steps used.



Can someone please explain the steps to solving the two linear equations below using Gaussian elimination?



It's quite obvious that $x = 2$ and $y = 3$ but I'm struggling the understand the steps using Gaussian elimination to work it out.



begin{cases}
2x - y = 5 \[4px]
x + y = 1
end{cases}










share|cite|improve this question















put on hold as off-topic by amWhy, user21820, Did, José Carlos Santos, Jyrki Lahtonen yesterday


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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, user21820, Did, José Carlos Santos, Jyrki Lahtonen

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









  • 3




    Since $2+3=5neq1$, I don't see how you find it “obvious” that $x=2$ and $y=3$.
    – José Carlos Santos
    Dec 2 at 10:40










  • Sorry my mistake I was meant to write 2x - y = 1 and x+ y = 5
    – Paul
    2 days ago













up vote
-2
down vote

favorite









up vote
-2
down vote

favorite











I am reading a book called "Linear algebra and its applications" by Gilbert Strang.



Chapter 1 is called “Matrices and Gaussian elimination”, the example in 1.1 I understand, Strang goes through the steps used.



Can someone please explain the steps to solving the two linear equations below using Gaussian elimination?



It's quite obvious that $x = 2$ and $y = 3$ but I'm struggling the understand the steps using Gaussian elimination to work it out.



begin{cases}
2x - y = 5 \[4px]
x + y = 1
end{cases}










share|cite|improve this question















I am reading a book called "Linear algebra and its applications" by Gilbert Strang.



Chapter 1 is called “Matrices and Gaussian elimination”, the example in 1.1 I understand, Strang goes through the steps used.



Can someone please explain the steps to solving the two linear equations below using Gaussian elimination?



It's quite obvious that $x = 2$ and $y = 3$ but I'm struggling the understand the steps using Gaussian elimination to work it out.



begin{cases}
2x - y = 5 \[4px]
x + y = 1
end{cases}







linear-algebra algebra-precalculus






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edited Dec 2 at 14:48









user21820

38.1k541150




38.1k541150










asked Dec 2 at 10:29









Paul

187




187




put on hold as off-topic by amWhy, user21820, Did, José Carlos Santos, Jyrki Lahtonen yesterday


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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, user21820, Did, José Carlos Santos, Jyrki Lahtonen

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




put on hold as off-topic by amWhy, user21820, Did, José Carlos Santos, Jyrki Lahtonen yesterday


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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, user21820, Did, José Carlos Santos, Jyrki Lahtonen

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








  • 3




    Since $2+3=5neq1$, I don't see how you find it “obvious” that $x=2$ and $y=3$.
    – José Carlos Santos
    Dec 2 at 10:40










  • Sorry my mistake I was meant to write 2x - y = 1 and x+ y = 5
    – Paul
    2 days ago














  • 3




    Since $2+3=5neq1$, I don't see how you find it “obvious” that $x=2$ and $y=3$.
    – José Carlos Santos
    Dec 2 at 10:40










  • Sorry my mistake I was meant to write 2x - y = 1 and x+ y = 5
    – Paul
    2 days ago








3




3




Since $2+3=5neq1$, I don't see how you find it “obvious” that $x=2$ and $y=3$.
– José Carlos Santos
Dec 2 at 10:40




Since $2+3=5neq1$, I don't see how you find it “obvious” that $x=2$ and $y=3$.
– José Carlos Santos
Dec 2 at 10:40












Sorry my mistake I was meant to write 2x - y = 1 and x+ y = 5
– Paul
2 days ago




Sorry my mistake I was meant to write 2x - y = 1 and x+ y = 5
– Paul
2 days ago










2 Answers
2






active

oldest

votes

















up vote
3
down vote



accepted










There is something wrong with the given solution, indeed we have that




  • $2x - y = 5$

  • $x + y = 1 iff 2x+2y=2$


then subtracting the first one from the second we obtain the equivalent system




  • $2x - y = 5$

  • $0+3y=-3 implies y=-1$


and then




  • $x=2$

  • $y=-1$






share|cite|improve this answer




























    up vote
    3
    down vote













    It depends on what kind of Gaussian elimination you use.



    If you use Gauss-Crout with also backward elimination, the steps are
    begin{align}
    left[begin{array}{cc|c}
    2 & -1 & 5 \
    1 & 1 & 1
    end{array}right]
    &to
    left[begin{array}{cc|c}
    1 & -1/2 & 5/2 \
    1 & 1 & 1
    end{array}right]
    && R_1getstfrac{1}{2}R_1
    \[6px] &to
    left[begin{array}{cc|c}
    1 & -1/2 & 5/2 \
    0 & 3/2 & -3/2
    end{array}right]
    && R_2gets R_2-R_1
    \[6px] &to
    left[begin{array}{cc|c}
    1 & -1/2 & 5/2 \
    0 & 1 & -1
    end{array}right]
    && R_2gets tfrac{2}{3}R_2
    \[6px] &to
    left[begin{array}{cc|c}
    1 & 0 & 2 \
    0 & 1 & -1
    end{array}right]
    && R_1gets R_1+tfrac{1}{2}R_2
    end{align}

    Therefore $x=2$ and $y=-1$.






    share|cite|improve this answer




























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      3
      down vote



      accepted










      There is something wrong with the given solution, indeed we have that




      • $2x - y = 5$

      • $x + y = 1 iff 2x+2y=2$


      then subtracting the first one from the second we obtain the equivalent system




      • $2x - y = 5$

      • $0+3y=-3 implies y=-1$


      and then




      • $x=2$

      • $y=-1$






      share|cite|improve this answer

























        up vote
        3
        down vote



        accepted










        There is something wrong with the given solution, indeed we have that




        • $2x - y = 5$

        • $x + y = 1 iff 2x+2y=2$


        then subtracting the first one from the second we obtain the equivalent system




        • $2x - y = 5$

        • $0+3y=-3 implies y=-1$


        and then




        • $x=2$

        • $y=-1$






        share|cite|improve this answer























          up vote
          3
          down vote



          accepted







          up vote
          3
          down vote



          accepted






          There is something wrong with the given solution, indeed we have that




          • $2x - y = 5$

          • $x + y = 1 iff 2x+2y=2$


          then subtracting the first one from the second we obtain the equivalent system




          • $2x - y = 5$

          • $0+3y=-3 implies y=-1$


          and then




          • $x=2$

          • $y=-1$






          share|cite|improve this answer












          There is something wrong with the given solution, indeed we have that




          • $2x - y = 5$

          • $x + y = 1 iff 2x+2y=2$


          then subtracting the first one from the second we obtain the equivalent system




          • $2x - y = 5$

          • $0+3y=-3 implies y=-1$


          and then




          • $x=2$

          • $y=-1$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 2 at 10:41









          gimusi

          90k74495




          90k74495






















              up vote
              3
              down vote













              It depends on what kind of Gaussian elimination you use.



              If you use Gauss-Crout with also backward elimination, the steps are
              begin{align}
              left[begin{array}{cc|c}
              2 & -1 & 5 \
              1 & 1 & 1
              end{array}right]
              &to
              left[begin{array}{cc|c}
              1 & -1/2 & 5/2 \
              1 & 1 & 1
              end{array}right]
              && R_1getstfrac{1}{2}R_1
              \[6px] &to
              left[begin{array}{cc|c}
              1 & -1/2 & 5/2 \
              0 & 3/2 & -3/2
              end{array}right]
              && R_2gets R_2-R_1
              \[6px] &to
              left[begin{array}{cc|c}
              1 & -1/2 & 5/2 \
              0 & 1 & -1
              end{array}right]
              && R_2gets tfrac{2}{3}R_2
              \[6px] &to
              left[begin{array}{cc|c}
              1 & 0 & 2 \
              0 & 1 & -1
              end{array}right]
              && R_1gets R_1+tfrac{1}{2}R_2
              end{align}

              Therefore $x=2$ and $y=-1$.






              share|cite|improve this answer

























                up vote
                3
                down vote













                It depends on what kind of Gaussian elimination you use.



                If you use Gauss-Crout with also backward elimination, the steps are
                begin{align}
                left[begin{array}{cc|c}
                2 & -1 & 5 \
                1 & 1 & 1
                end{array}right]
                &to
                left[begin{array}{cc|c}
                1 & -1/2 & 5/2 \
                1 & 1 & 1
                end{array}right]
                && R_1getstfrac{1}{2}R_1
                \[6px] &to
                left[begin{array}{cc|c}
                1 & -1/2 & 5/2 \
                0 & 3/2 & -3/2
                end{array}right]
                && R_2gets R_2-R_1
                \[6px] &to
                left[begin{array}{cc|c}
                1 & -1/2 & 5/2 \
                0 & 1 & -1
                end{array}right]
                && R_2gets tfrac{2}{3}R_2
                \[6px] &to
                left[begin{array}{cc|c}
                1 & 0 & 2 \
                0 & 1 & -1
                end{array}right]
                && R_1gets R_1+tfrac{1}{2}R_2
                end{align}

                Therefore $x=2$ and $y=-1$.






                share|cite|improve this answer























                  up vote
                  3
                  down vote










                  up vote
                  3
                  down vote









                  It depends on what kind of Gaussian elimination you use.



                  If you use Gauss-Crout with also backward elimination, the steps are
                  begin{align}
                  left[begin{array}{cc|c}
                  2 & -1 & 5 \
                  1 & 1 & 1
                  end{array}right]
                  &to
                  left[begin{array}{cc|c}
                  1 & -1/2 & 5/2 \
                  1 & 1 & 1
                  end{array}right]
                  && R_1getstfrac{1}{2}R_1
                  \[6px] &to
                  left[begin{array}{cc|c}
                  1 & -1/2 & 5/2 \
                  0 & 3/2 & -3/2
                  end{array}right]
                  && R_2gets R_2-R_1
                  \[6px] &to
                  left[begin{array}{cc|c}
                  1 & -1/2 & 5/2 \
                  0 & 1 & -1
                  end{array}right]
                  && R_2gets tfrac{2}{3}R_2
                  \[6px] &to
                  left[begin{array}{cc|c}
                  1 & 0 & 2 \
                  0 & 1 & -1
                  end{array}right]
                  && R_1gets R_1+tfrac{1}{2}R_2
                  end{align}

                  Therefore $x=2$ and $y=-1$.






                  share|cite|improve this answer












                  It depends on what kind of Gaussian elimination you use.



                  If you use Gauss-Crout with also backward elimination, the steps are
                  begin{align}
                  left[begin{array}{cc|c}
                  2 & -1 & 5 \
                  1 & 1 & 1
                  end{array}right]
                  &to
                  left[begin{array}{cc|c}
                  1 & -1/2 & 5/2 \
                  1 & 1 & 1
                  end{array}right]
                  && R_1getstfrac{1}{2}R_1
                  \[6px] &to
                  left[begin{array}{cc|c}
                  1 & -1/2 & 5/2 \
                  0 & 3/2 & -3/2
                  end{array}right]
                  && R_2gets R_2-R_1
                  \[6px] &to
                  left[begin{array}{cc|c}
                  1 & -1/2 & 5/2 \
                  0 & 1 & -1
                  end{array}right]
                  && R_2gets tfrac{2}{3}R_2
                  \[6px] &to
                  left[begin{array}{cc|c}
                  1 & 0 & 2 \
                  0 & 1 & -1
                  end{array}right]
                  && R_1gets R_1+tfrac{1}{2}R_2
                  end{align}

                  Therefore $x=2$ and $y=-1$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 2 at 11:17









                  egreg

                  175k1383198




                  175k1383198















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