Why is addition with a real not an elementary row operation?












2












$begingroup$


Many linear algebra courses build upon the concept of elementary row operations, and further develop on their properties and relevance.



My question is, why isn't addition by a real considered an elementary row operation and how can this be argued at the level of the beginning of the course? I know that this operation doesn't share some properties of the elementary row operations. However, linear algebra courses present elementary row operations as operations that don't change the solution set. Well, addition with a real doesn't change the solution set.






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  • 1




    $begingroup$
    What is "addition by a real"? $(a,b,c) mapsto (a + r, b+r, c+r)$?
    $endgroup$
    – Billy Rubina
    1 hour ago


















2












$begingroup$


Many linear algebra courses build upon the concept of elementary row operations, and further develop on their properties and relevance.



My question is, why isn't addition by a real considered an elementary row operation and how can this be argued at the level of the beginning of the course? I know that this operation doesn't share some properties of the elementary row operations. However, linear algebra courses present elementary row operations as operations that don't change the solution set. Well, addition with a real doesn't change the solution set.






linear-algebra







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    What is "addition by a real"? $(a,b,c) mapsto (a + r, b+r, c+r)$?
    $endgroup$
    – Billy Rubina
    1 hour ago
















2












2








2





$begingroup$


Many linear algebra courses build upon the concept of elementary row operations, and further develop on their properties and relevance.



My question is, why isn't addition by a real considered an elementary row operation and how can this be argued at the level of the beginning of the course? I know that this operation doesn't share some properties of the elementary row operations. However, linear algebra courses present elementary row operations as operations that don't change the solution set. Well, addition with a real doesn't change the solution set.






linear-algebra







share|cite|improve this question









$endgroup$




Many linear algebra courses build upon the concept of elementary row operations, and further develop on their properties and relevance.



My question is, why isn't addition by a real considered an elementary row operation and how can this be argued at the level of the beginning of the course? I know that this operation doesn't share some properties of the elementary row operations. However, linear algebra courses present elementary row operations as operations that don't change the solution set. Well, addition with a real doesn't change the solution set.






linear-algebra




linear-algebra






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asked 1 hour ago









Paul92Paul92

31014




31014








  • 1




    $begingroup$
    What is "addition by a real"? $(a,b,c) mapsto (a + r, b+r, c+r)$?
    $endgroup$
    – Billy Rubina
    1 hour ago
















  • 1




    $begingroup$
    What is "addition by a real"? $(a,b,c) mapsto (a + r, b+r, c+r)$?
    $endgroup$
    – Billy Rubina
    1 hour ago










1




1




$begingroup$
What is "addition by a real"? $(a,b,c) mapsto (a + r, b+r, c+r)$?
$endgroup$
– Billy Rubina
1 hour ago






$begingroup$
What is "addition by a real"? $(a,b,c) mapsto (a + r, b+r, c+r)$?
$endgroup$
– Billy Rubina
1 hour ago












2 Answers
2






active

oldest

votes


















6












$begingroup$

Take for example the augmented matrix
$$(begin{array}{cc|c} 1 & -1 & 0 end{array}).$$
This has a solution of $(x, y) = t(1, 1)$.



Add the real number $1$ to the row:
$$(begin{array}{cc|c} 2 & 0 & 1 end{array}).$$
This has a solution of $(x, y) = (1/2, 0) + t(0, 1)$, which is a completely different line of solutions.



So, adding a real does indeed change the solution set.



More generally, there are other operations that do preserve the solution set, but only three are called "elementary". The reason why we restrict ourselves to just three operations is because these three operations are sufficient to transform between any two matrices of the same dimensions with the same solution set. Every row operation that preserves solution sets is a combination of elementary row operations!






share|cite|improve this answer











$endgroup$













  • $begingroup$
    So the point is: adding $1$ to the row is equivalent to adding $x+y$ to one side of the equation (by adding $1$ to the coefficients of $x$ and $y$), while adding $1$ to the other side.
    $endgroup$
    – timtfj
    1 hour ago





















0












$begingroup$

Expanding on the first part of Theo Bendit's answer: if we represent the equation $$ax+by=c$$ by the matrix row $$(begin{array}{cc|c} a & b & c end{array})$$
and add a constant $k$ to the row, the result
$$(begin{array}{cc|c} a+k & b+k & c+k end{array})$$
is equivalent to the equation



$$(a+k)x+(b+k)y=c+k.$$



Comparing with the original equation, we see we've added $k(x+y)$ to one side and $k$ to the other—producing a completely different equation.



So adding a constant isn't a valid row operation.






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









    6












    $begingroup$

    Take for example the augmented matrix
    $$(begin{array}{cc|c} 1 & -1 & 0 end{array}).$$
    This has a solution of $(x, y) = t(1, 1)$.



    Add the real number $1$ to the row:
    $$(begin{array}{cc|c} 2 & 0 & 1 end{array}).$$
    This has a solution of $(x, y) = (1/2, 0) + t(0, 1)$, which is a completely different line of solutions.



    So, adding a real does indeed change the solution set.



    More generally, there are other operations that do preserve the solution set, but only three are called "elementary". The reason why we restrict ourselves to just three operations is because these three operations are sufficient to transform between any two matrices of the same dimensions with the same solution set. Every row operation that preserves solution sets is a combination of elementary row operations!






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      So the point is: adding $1$ to the row is equivalent to adding $x+y$ to one side of the equation (by adding $1$ to the coefficients of $x$ and $y$), while adding $1$ to the other side.
      $endgroup$
      – timtfj
      1 hour ago


















    6












    $begingroup$

    Take for example the augmented matrix
    $$(begin{array}{cc|c} 1 & -1 & 0 end{array}).$$
    This has a solution of $(x, y) = t(1, 1)$.



    Add the real number $1$ to the row:
    $$(begin{array}{cc|c} 2 & 0 & 1 end{array}).$$
    This has a solution of $(x, y) = (1/2, 0) + t(0, 1)$, which is a completely different line of solutions.



    So, adding a real does indeed change the solution set.



    More generally, there are other operations that do preserve the solution set, but only three are called "elementary". The reason why we restrict ourselves to just three operations is because these three operations are sufficient to transform between any two matrices of the same dimensions with the same solution set. Every row operation that preserves solution sets is a combination of elementary row operations!






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      So the point is: adding $1$ to the row is equivalent to adding $x+y$ to one side of the equation (by adding $1$ to the coefficients of $x$ and $y$), while adding $1$ to the other side.
      $endgroup$
      – timtfj
      1 hour ago
















    6












    6








    6





    $begingroup$

    Take for example the augmented matrix
    $$(begin{array}{cc|c} 1 & -1 & 0 end{array}).$$
    This has a solution of $(x, y) = t(1, 1)$.



    Add the real number $1$ to the row:
    $$(begin{array}{cc|c} 2 & 0 & 1 end{array}).$$
    This has a solution of $(x, y) = (1/2, 0) + t(0, 1)$, which is a completely different line of solutions.



    So, adding a real does indeed change the solution set.



    More generally, there are other operations that do preserve the solution set, but only three are called "elementary". The reason why we restrict ourselves to just three operations is because these three operations are sufficient to transform between any two matrices of the same dimensions with the same solution set. Every row operation that preserves solution sets is a combination of elementary row operations!






    share|cite|improve this answer











    $endgroup$



    Take for example the augmented matrix
    $$(begin{array}{cc|c} 1 & -1 & 0 end{array}).$$
    This has a solution of $(x, y) = t(1, 1)$.



    Add the real number $1$ to the row:
    $$(begin{array}{cc|c} 2 & 0 & 1 end{array}).$$
    This has a solution of $(x, y) = (1/2, 0) + t(0, 1)$, which is a completely different line of solutions.



    So, adding a real does indeed change the solution set.



    More generally, there are other operations that do preserve the solution set, but only three are called "elementary". The reason why we restrict ourselves to just three operations is because these three operations are sufficient to transform between any two matrices of the same dimensions with the same solution set. Every row operation that preserves solution sets is a combination of elementary row operations!







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited 1 hour ago

























    answered 1 hour ago









    Theo BenditTheo Bendit

    17.4k12149




    17.4k12149












    • $begingroup$
      So the point is: adding $1$ to the row is equivalent to adding $x+y$ to one side of the equation (by adding $1$ to the coefficients of $x$ and $y$), while adding $1$ to the other side.
      $endgroup$
      – timtfj
      1 hour ago




















    • $begingroup$
      So the point is: adding $1$ to the row is equivalent to adding $x+y$ to one side of the equation (by adding $1$ to the coefficients of $x$ and $y$), while adding $1$ to the other side.
      $endgroup$
      – timtfj
      1 hour ago


















    $begingroup$
    So the point is: adding $1$ to the row is equivalent to adding $x+y$ to one side of the equation (by adding $1$ to the coefficients of $x$ and $y$), while adding $1$ to the other side.
    $endgroup$
    – timtfj
    1 hour ago






    $begingroup$
    So the point is: adding $1$ to the row is equivalent to adding $x+y$ to one side of the equation (by adding $1$ to the coefficients of $x$ and $y$), while adding $1$ to the other side.
    $endgroup$
    – timtfj
    1 hour ago













    0












    $begingroup$

    Expanding on the first part of Theo Bendit's answer: if we represent the equation $$ax+by=c$$ by the matrix row $$(begin{array}{cc|c} a & b & c end{array})$$
    and add a constant $k$ to the row, the result
    $$(begin{array}{cc|c} a+k & b+k & c+k end{array})$$
    is equivalent to the equation



    $$(a+k)x+(b+k)y=c+k.$$



    Comparing with the original equation, we see we've added $k(x+y)$ to one side and $k$ to the other—producing a completely different equation.



    So adding a constant isn't a valid row operation.






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      Expanding on the first part of Theo Bendit's answer: if we represent the equation $$ax+by=c$$ by the matrix row $$(begin{array}{cc|c} a & b & c end{array})$$
      and add a constant $k$ to the row, the result
      $$(begin{array}{cc|c} a+k & b+k & c+k end{array})$$
      is equivalent to the equation



      $$(a+k)x+(b+k)y=c+k.$$



      Comparing with the original equation, we see we've added $k(x+y)$ to one side and $k$ to the other—producing a completely different equation.



      So adding a constant isn't a valid row operation.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        Expanding on the first part of Theo Bendit's answer: if we represent the equation $$ax+by=c$$ by the matrix row $$(begin{array}{cc|c} a & b & c end{array})$$
        and add a constant $k$ to the row, the result
        $$(begin{array}{cc|c} a+k & b+k & c+k end{array})$$
        is equivalent to the equation



        $$(a+k)x+(b+k)y=c+k.$$



        Comparing with the original equation, we see we've added $k(x+y)$ to one side and $k$ to the other—producing a completely different equation.



        So adding a constant isn't a valid row operation.






        share|cite|improve this answer











        $endgroup$



        Expanding on the first part of Theo Bendit's answer: if we represent the equation $$ax+by=c$$ by the matrix row $$(begin{array}{cc|c} a & b & c end{array})$$
        and add a constant $k$ to the row, the result
        $$(begin{array}{cc|c} a+k & b+k & c+k end{array})$$
        is equivalent to the equation



        $$(a+k)x+(b+k)y=c+k.$$



        Comparing with the original equation, we see we've added $k(x+y)$ to one side and $k$ to the other—producing a completely different equation.



        So adding a constant isn't a valid row operation.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 32 mins ago

























        answered 39 mins ago









        timtfjtimtfj

        1,638318




        1,638318






























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