Why is addition with a real not an elementary row operation?
$begingroup$
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
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add a comment |
$begingroup$
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
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1
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What is "addition by a real"? $(a,b,c) mapsto (a + r, b+r, c+r)$?
$endgroup$
– Billy Rubina
1 hour ago
add a comment |
$begingroup$
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
$endgroup$
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
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
add a comment |
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
add a comment |
2 Answers
2
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$begingroup$
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!
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
active
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$begingroup$
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!
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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!
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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!
$endgroup$
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!
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
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.
edited 33 mins ago
answered 40 mins ago
timtfjtimtfj
1,638318
1,638318
add a comment |
add a comment |
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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