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Why does adding a multiple of an equation to another equation in linear algebra result in row equivalence?
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Why is number 3 true? In theory, why does this work? Why are these systems equivalent?
I also don't understand what the above is trying to get at. How do you figure out 1?
linear-algebra
asked Jul 15 at 23:04
Jwan622
1,61211224
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up vote
0
down vote
favorite
Why is number 3 true? In theory, why does this work? Why are these systems equivalent?
I also don't understand what the above is trying to get at. How do you figure out 1?
linear-algebra
asked Jul 15 at 23:04
Jwan622
1,61211224
Definitions are your friends. To reason mathematically (or explain) why two systems of equations are equivalent (as you ask), one needs to begin with a definition of when two systems of equations are equivalent. The large images of textbook material are a poor substitute for using your own words to explain what you want help with.
– hardmath
Jul 16 at 13:49
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Why is number 3 true? In theory, why does this work? Why are these systems equivalent?
I also don't understand what the above is trying to get at. How do you figure out 1?
linear-algebra
asked Jul 15 at 23:04
Jwan622
1,61211224
Why is number 3 true? In theory, why does this work? Why are these systems equivalent?
I also don't understand what the above is trying to get at. How do you figure out 1?
linear-algebra
asked Jul 15 at 23:04
Jwan622
1,61211224
asked Jul 15 at 23:04
Jwan622
1,61211224
asked Jul 15 at 23:04
Jwan622
1,61211224
asked Jul 15 at 23:04
Jwan622
1,61211224
1,61211224
Definitions are your friends. To reason mathematically (or explain) why two systems of equations are equivalent (as you ask), one needs to begin with a definition of when two systems of equations are equivalent. The large images of textbook material are a poor substitute for using your own words to explain what you want help with.
– hardmath
Jul 16 at 13:49
add a comment |Â
Definitions are your friends. To reason mathematically (or explain) why two systems of equations are equivalent (as you ask), one needs to begin with a definition of when two systems of equations are equivalent. The large images of textbook material are a poor substitute for using your own words to explain what you want help with.
– hardmath
Jul 16 at 13:49
Definitions are your friends. To reason mathematically (or explain) why two systems of equations are equivalent (as you ask), one needs to begin with a definition of when two systems of equations are equivalent. The large images of textbook material are a poor substitute for using your own words to explain what you want help with.
– hardmath
Jul 16 at 13:49
Definitions are your friends. To reason mathematically (or explain) why two systems of equations are equivalent (as you ask), one needs to begin with a definition of when two systems of equations are equivalent. The large images of textbook material are a poor substitute for using your own words to explain what you want help with.
– hardmath
Jul 16 at 13:49
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
3
down vote
accepted
When you have an equation you essentially have the statement $a = b$. When two equations are involved, you also have $c = d$.
Now let's assume both these statements are true. Then we can start with one:
$$a = b$$
Add $rc$ to both sides (this is obviously true), with arbitrary real $r$:
$$a + rc = b + rc$$
And finally use $c = d$ to replace the $c$ on the right hand side:
$$a + rc = b + rd$$
That is, in a system of equations, you can add an arbitrary multiple of an equation to another equation and still have an equivalent system. What is meant by that is iff one system is true then the other must be as well.
answered Jul 15 at 23:14
orlp
6,6051228
This is also an excellent way to teach this material.
– Randall
Jul 16 at 13:36
@Randall Thanks :)
– orlp
Jul 16 at 18:21
add a comment |Â
up vote
1
down vote
For example if we have the equations $3y+2x=6$ and $5y-2x=10$ we can easily find the values of $x$ and $y$ by adding those two equations to get $y$ and then find $x$.
But if we have the equations like $3x+y=9$ and $5x+4y=22$ we cannot just add or subtract these two equations. we need to multiply with the multiple of the other equations in order to make the $x$ or $y$ values equal.
Note that every homogeneous system of linear equations is consistent. If the system has fewer equations than variables, then it is said to have an infinite number of solutions.
answered Jul 15 at 23:17
Key Flex
4,416525
add a comment |Â
up vote
0
down vote
Since you said "in theory"... :)
For a more abstract viewpoint (which may be helpful once you get further in your linear algebra studies), note that a system of linear equations
$$
a_11 x_1 + a_12 x_2 + cdots + a_1m x_m = b_1 \
a_21 x_1 + a_22 x_2 + cdots + a_2m x_m = b_2 \
vdots \
a_n1 x_1 + a_n2 x_2 + cdots + a_nm x_m = b_n
$$
can be rewritten as a matrix equation
$$
A vecx = vecb,
$$
and that each elementary row operation is equivalent to multiplication on both sides of this equation by an elementary matrix, i.e. changing $A vecx = vecb$ to the modified system
$$
E A vecx = E vecb.
$$
From this point it is easy to see that if $vecx$ is a solution set to the original system, then it is also a solution to the modified system.
Now you may be asking why we care about solutions to the modified system --- we were trying to solve $A vecx = vecb$, not $E A vecx = E vecb$. Well, it is a basic exercise in linear algebra to show that each type of elementary matrix is invertible (in fact they generate all invertible matrices!), so that if $vecx$ solves the modified system, we can "undo" the elementary matrices and see that $vecx$ solves the original system as well:
$$
E A vecx = Evecb iff E^-1 E A vecx = E^-1 E vecb iff A vecx = vecb.
$$
In short, multiplication by elementary (or invertible) matrices preserves the solution set of the matrix equation. This is the more abstract / theoretical viewpoint of why Gaussian elimination works.
answered Jul 16 at 0:04
Adam Williams
35115
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
When you have an equation you essentially have the statement $a = b$. When two equations are involved, you also have $c = d$.
Now let's assume both these statements are true. Then we can start with one:
$$a = b$$
Add $rc$ to both sides (this is obviously true), with arbitrary real $r$:
$$a + rc = b + rc$$
And finally use $c = d$ to replace the $c$ on the right hand side:
$$a + rc = b + rd$$
That is, in a system of equations, you can add an arbitrary multiple of an equation to another equation and still have an equivalent system. What is meant by that is iff one system is true then the other must be as well.
answered Jul 15 at 23:14
orlp
6,6051228
This is also an excellent way to teach this material.
– Randall
Jul 16 at 13:36
@Randall Thanks :)
– orlp
Jul 16 at 18:21
add a comment |Â
up vote
3
down vote
accepted
When you have an equation you essentially have the statement $a = b$. When two equations are involved, you also have $c = d$.
Now let's assume both these statements are true. Then we can start with one:
$$a = b$$
Add $rc$ to both sides (this is obviously true), with arbitrary real $r$:
$$a + rc = b + rc$$
And finally use $c = d$ to replace the $c$ on the right hand side:
$$a + rc = b + rd$$
That is, in a system of equations, you can add an arbitrary multiple of an equation to another equation and still have an equivalent system. What is meant by that is iff one system is true then the other must be as well.
answered Jul 15 at 23:14
orlp
6,6051228
This is also an excellent way to teach this material.
– Randall
Jul 16 at 13:36
@Randall Thanks :)
– orlp
Jul 16 at 18:21
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
When you have an equation you essentially have the statement $a = b$. When two equations are involved, you also have $c = d$.
Now let's assume both these statements are true. Then we can start with one:
$$a = b$$
Add $rc$ to both sides (this is obviously true), with arbitrary real $r$:
$$a + rc = b + rc$$
And finally use $c = d$ to replace the $c$ on the right hand side:
$$a + rc = b + rd$$
That is, in a system of equations, you can add an arbitrary multiple of an equation to another equation and still have an equivalent system. What is meant by that is iff one system is true then the other must be as well.
answered Jul 15 at 23:14
orlp
6,6051228
When you have an equation you essentially have the statement $a = b$. When two equations are involved, you also have $c = d$.
Now let's assume both these statements are true. Then we can start with one:
$$a = b$$
Add $rc$ to both sides (this is obviously true), with arbitrary real $r$:
$$a + rc = b + rc$$
And finally use $c = d$ to replace the $c$ on the right hand side:
$$a + rc = b + rd$$
That is, in a system of equations, you can add an arbitrary multiple of an equation to another equation and still have an equivalent system. What is meant by that is iff one system is true then the other must be as well.
answered Jul 15 at 23:14
orlp
6,6051228
edited Jul 15 at 23:23
answered Jul 15 at 23:14
orlp
6,6051228
answered Jul 15 at 23:14
orlp
6,6051228
answered Jul 15 at 23:14
orlp
6,6051228
6,6051228
This is also an excellent way to teach this material.
– Randall
Jul 16 at 13:36
@Randall Thanks :)
– orlp
Jul 16 at 18:21
add a comment |Â
This is also an excellent way to teach this material.
– Randall
Jul 16 at 13:36
@Randall Thanks :)
– orlp
Jul 16 at 18:21
This is also an excellent way to teach this material.
– Randall
Jul 16 at 13:36
This is also an excellent way to teach this material.
– Randall
Jul 16 at 13:36
@Randall Thanks :)
– orlp
Jul 16 at 18:21
@Randall Thanks :)
– orlp
Jul 16 at 18:21
add a comment |Â
up vote
1
down vote
For example if we have the equations $3y+2x=6$ and $5y-2x=10$ we can easily find the values of $x$ and $y$ by adding those two equations to get $y$ and then find $x$.
But if we have the equations like $3x+y=9$ and $5x+4y=22$ we cannot just add or subtract these two equations. we need to multiply with the multiple of the other equations in order to make the $x$ or $y$ values equal.
Note that every homogeneous system of linear equations is consistent. If the system has fewer equations than variables, then it is said to have an infinite number of solutions.
answered Jul 15 at 23:17
Key Flex
4,416525
add a comment |Â
up vote
1
down vote
For example if we have the equations $3y+2x=6$ and $5y-2x=10$ we can easily find the values of $x$ and $y$ by adding those two equations to get $y$ and then find $x$.
But if we have the equations like $3x+y=9$ and $5x+4y=22$ we cannot just add or subtract these two equations. we need to multiply with the multiple of the other equations in order to make the $x$ or $y$ values equal.
Note that every homogeneous system of linear equations is consistent. If the system has fewer equations than variables, then it is said to have an infinite number of solutions.
answered Jul 15 at 23:17
Key Flex
4,416525
add a comment |Â
up vote
1
down vote
up vote
1
down vote
For example if we have the equations $3y+2x=6$ and $5y-2x=10$ we can easily find the values of $x$ and $y$ by adding those two equations to get $y$ and then find $x$.
But if we have the equations like $3x+y=9$ and $5x+4y=22$ we cannot just add or subtract these two equations. we need to multiply with the multiple of the other equations in order to make the $x$ or $y$ values equal.
Note that every homogeneous system of linear equations is consistent. If the system has fewer equations than variables, then it is said to have an infinite number of solutions.
answered Jul 15 at 23:17
Key Flex
4,416525
For example if we have the equations $3y+2x=6$ and $5y-2x=10$ we can easily find the values of $x$ and $y$ by adding those two equations to get $y$ and then find $x$.
But if we have the equations like $3x+y=9$ and $5x+4y=22$ we cannot just add or subtract these two equations. we need to multiply with the multiple of the other equations in order to make the $x$ or $y$ values equal.
Note that every homogeneous system of linear equations is consistent. If the system has fewer equations than variables, then it is said to have an infinite number of solutions.
answered Jul 15 at 23:17
Key Flex
4,416525
answered Jul 15 at 23:17
Key Flex
4,416525
answered Jul 15 at 23:17
Key Flex
4,416525
answered Jul 15 at 23:17
Key Flex
4,416525
4,416525
add a comment |Â
add a comment |Â
up vote
0
down vote
Since you said "in theory"... :)
For a more abstract viewpoint (which may be helpful once you get further in your linear algebra studies), note that a system of linear equations
$$
a_11 x_1 + a_12 x_2 + cdots + a_1m x_m = b_1 \
a_21 x_1 + a_22 x_2 + cdots + a_2m x_m = b_2 \
vdots \
a_n1 x_1 + a_n2 x_2 + cdots + a_nm x_m = b_n
$$
can be rewritten as a matrix equation
$$
A vecx = vecb,
$$
and that each elementary row operation is equivalent to multiplication on both sides of this equation by an elementary matrix, i.e. changing $A vecx = vecb$ to the modified system
$$
E A vecx = E vecb.
$$
From this point it is easy to see that if $vecx$ is a solution set to the original system, then it is also a solution to the modified system.
Now you may be asking why we care about solutions to the modified system --- we were trying to solve $A vecx = vecb$, not $E A vecx = E vecb$. Well, it is a basic exercise in linear algebra to show that each type of elementary matrix is invertible (in fact they generate all invertible matrices!), so that if $vecx$ solves the modified system, we can "undo" the elementary matrices and see that $vecx$ solves the original system as well:
$$
E A vecx = Evecb iff E^-1 E A vecx = E^-1 E vecb iff A vecx = vecb.
$$
In short, multiplication by elementary (or invertible) matrices preserves the solution set of the matrix equation. This is the more abstract / theoretical viewpoint of why Gaussian elimination works.
answered Jul 16 at 0:04
Adam Williams
35115
add a comment |Â
up vote
0
down vote
Since you said "in theory"... :)
For a more abstract viewpoint (which may be helpful once you get further in your linear algebra studies), note that a system of linear equations
$$
a_11 x_1 + a_12 x_2 + cdots + a_1m x_m = b_1 \
a_21 x_1 + a_22 x_2 + cdots + a_2m x_m = b_2 \
vdots \
a_n1 x_1 + a_n2 x_2 + cdots + a_nm x_m = b_n
$$
can be rewritten as a matrix equation
$$
A vecx = vecb,
$$
and that each elementary row operation is equivalent to multiplication on both sides of this equation by an elementary matrix, i.e. changing $A vecx = vecb$ to the modified system
$$
E A vecx = E vecb.
$$
From this point it is easy to see that if $vecx$ is a solution set to the original system, then it is also a solution to the modified system.
Now you may be asking why we care about solutions to the modified system --- we were trying to solve $A vecx = vecb$, not $E A vecx = E vecb$. Well, it is a basic exercise in linear algebra to show that each type of elementary matrix is invertible (in fact they generate all invertible matrices!), so that if $vecx$ solves the modified system, we can "undo" the elementary matrices and see that $vecx$ solves the original system as well:
$$
E A vecx = Evecb iff E^-1 E A vecx = E^-1 E vecb iff A vecx = vecb.
$$
In short, multiplication by elementary (or invertible) matrices preserves the solution set of the matrix equation. This is the more abstract / theoretical viewpoint of why Gaussian elimination works.
answered Jul 16 at 0:04
Adam Williams
35115
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Since you said "in theory"... :)
For a more abstract viewpoint (which may be helpful once you get further in your linear algebra studies), note that a system of linear equations
$$
a_11 x_1 + a_12 x_2 + cdots + a_1m x_m = b_1 \
a_21 x_1 + a_22 x_2 + cdots + a_2m x_m = b_2 \
vdots \
a_n1 x_1 + a_n2 x_2 + cdots + a_nm x_m = b_n
$$
can be rewritten as a matrix equation
$$
A vecx = vecb,
$$
and that each elementary row operation is equivalent to multiplication on both sides of this equation by an elementary matrix, i.e. changing $A vecx = vecb$ to the modified system
$$
E A vecx = E vecb.
$$
From this point it is easy to see that if $vecx$ is a solution set to the original system, then it is also a solution to the modified system.
Now you may be asking why we care about solutions to the modified system --- we were trying to solve $A vecx = vecb$, not $E A vecx = E vecb$. Well, it is a basic exercise in linear algebra to show that each type of elementary matrix is invertible (in fact they generate all invertible matrices!), so that if $vecx$ solves the modified system, we can "undo" the elementary matrices and see that $vecx$ solves the original system as well:
$$
E A vecx = Evecb iff E^-1 E A vecx = E^-1 E vecb iff A vecx = vecb.
$$
In short, multiplication by elementary (or invertible) matrices preserves the solution set of the matrix equation. This is the more abstract / theoretical viewpoint of why Gaussian elimination works.
answered Jul 16 at 0:04
Adam Williams
35115
Since you said "in theory"... :)
For a more abstract viewpoint (which may be helpful once you get further in your linear algebra studies), note that a system of linear equations
$$
a_11 x_1 + a_12 x_2 + cdots + a_1m x_m = b_1 \
a_21 x_1 + a_22 x_2 + cdots + a_2m x_m = b_2 \
vdots \
a_n1 x_1 + a_n2 x_2 + cdots + a_nm x_m = b_n
$$
can be rewritten as a matrix equation
$$
A vecx = vecb,
$$
and that each elementary row operation is equivalent to multiplication on both sides of this equation by an elementary matrix, i.e. changing $A vecx = vecb$ to the modified system
$$
E A vecx = E vecb.
$$
From this point it is easy to see that if $vecx$ is a solution set to the original system, then it is also a solution to the modified system.
Now you may be asking why we care about solutions to the modified system --- we were trying to solve $A vecx = vecb$, not $E A vecx = E vecb$. Well, it is a basic exercise in linear algebra to show that each type of elementary matrix is invertible (in fact they generate all invertible matrices!), so that if $vecx$ solves the modified system, we can "undo" the elementary matrices and see that $vecx$ solves the original system as well:
$$
E A vecx = Evecb iff E^-1 E A vecx = E^-1 E vecb iff A vecx = vecb.
$$
In short, multiplication by elementary (or invertible) matrices preserves the solution set of the matrix equation. This is the more abstract / theoretical viewpoint of why Gaussian elimination works.
answered Jul 16 at 0:04
Adam Williams
35115
edited Jul 16 at 0:12
answered Jul 16 at 0:04
Adam Williams
35115
answered Jul 16 at 0:04
Adam Williams
35115
answered Jul 16 at 0:04
Adam Williams
35115
35115
add a comment |Â
add a comment |Â
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Definitions are your friends. To reason mathematically (or explain) why two systems of equations are equivalent (as you ask), one needs to begin with a definition of when two systems of equations are equivalent. The large images of textbook material are a poor substitute for using your own words to explain what you want help with.
– hardmath
Jul 16 at 13:49