Solution to systems of linear equations formed by linear combinations of another system of linear equations.

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I have recently began reading a text on linear algebra in preparation the upcoming fall semester. This is the first time I've looked at linear algebra proper in a few years. Early on I was reading about some basic concepts regarding systems of linear equations. In particular you can take a given system of linear equations call it system A and if we form linear combinations out of these equations we can construct another system of linear equations call it system B. Clearly any solution to system A is a solution to system B.

My question is that the text stated that there may be cases where a solution to system B may not be a solution to system A, but as I have been writing out simple systems to try to generate a case where this happens I have been unable to do so. If there is a trivial case to this then it escapes me. If someone could help me understand why this is so obvious and if there are trivial cases give me an example it would greatly help.

Also please note that I understand that in the case that all the solutions of B are solutions of A then we call the systems equivalent.

linear-algebra systems-of-equations

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asked Jul 22 at 3:09

Christian Harris

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I have recently began reading a text on linear algebra in preparation the upcoming fall semester. This is the first time I've looked at linear algebra proper in a few years. Early on I was reading about some basic concepts regarding systems of linear equations. In particular you can take a given system of linear equations call it system A and if we form linear combinations out of these equations we can construct another system of linear equations call it system B. Clearly any solution to system A is a solution to system B.

My question is that the text stated that there may be cases where a solution to system B may not be a solution to system A, but as I have been writing out simple systems to try to generate a case where this happens I have been unable to do so. If there is a trivial case to this then it escapes me. If someone could help me understand why this is so obvious and if there are trivial cases give me an example it would greatly help.

Also please note that I understand that in the case that all the solutions of B are solutions of A then we call the systems equivalent.

linear-algebra systems-of-equations

share|cite|improve this question

asked Jul 22 at 3:09

Christian Harris

191111

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

I have recently began reading a text on linear algebra in preparation the upcoming fall semester. This is the first time I've looked at linear algebra proper in a few years. Early on I was reading about some basic concepts regarding systems of linear equations. In particular you can take a given system of linear equations call it system A and if we form linear combinations out of these equations we can construct another system of linear equations call it system B. Clearly any solution to system A is a solution to system B.

My question is that the text stated that there may be cases where a solution to system B may not be a solution to system A, but as I have been writing out simple systems to try to generate a case where this happens I have been unable to do so. If there is a trivial case to this then it escapes me. If someone could help me understand why this is so obvious and if there are trivial cases give me an example it would greatly help.

Also please note that I understand that in the case that all the solutions of B are solutions of A then we call the systems equivalent.

linear-algebra systems-of-equations

share|cite|improve this question

asked Jul 22 at 3:09

Christian Harris

191111

I have recently began reading a text on linear algebra in preparation the upcoming fall semester. This is the first time I've looked at linear algebra proper in a few years. Early on I was reading about some basic concepts regarding systems of linear equations. In particular you can take a given system of linear equations call it system A and if we form linear combinations out of these equations we can construct another system of linear equations call it system B. Clearly any solution to system A is a solution to system B.

My question is that the text stated that there may be cases where a solution to system B may not be a solution to system A, but as I have been writing out simple systems to try to generate a case where this happens I have been unable to do so. If there is a trivial case to this then it escapes me. If someone could help me understand why this is so obvious and if there are trivial cases give me an example it would greatly help.

Also please note that I understand that in the case that all the solutions of B are solutions of A then we call the systems equivalent.

linear-algebra systems-of-equations

share|cite|improve this question

asked Jul 22 at 3:09

Christian Harris

191111

share|cite|improve this question

share|cite|improve this question

asked Jul 22 at 3:09

Christian Harris

191111

asked Jul 22 at 3:09

Christian Harris

191111

asked Jul 22 at 3:09

Christian Harris

191111

191111

add a comment | 

add a comment | 

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Consider the equations
$$x+y=1tag1$$
$$x-y=1tag2$$
which has as unique solution $x=1,y=0$. If we take the linear equation $2(1)+(2)$ (twice of first equation plus second) and $4(1)+2(2)$ we get
$$3x+y=3$$
$$6x+2y=6$$
which has an infinite number of solutions beside $x=1,y=0$, one of which is $x=0,y=3$.

Such additional solutions, at least in two variables, arise when some of the new equations are multiples of each other, so that one degree of specification is lost. Later in the course you're taking I suppose that matrices will be introduced, and it can be shown that such generation of additional solutions arises when the determinant of the matrix of linear combinations – $beginbmatrix2&1\4&2endbmatrix$ here – is zero.

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answered Jul 22 at 3:26

Parcly Taxel

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1 Answer
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active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

Consider the equations
$$x+y=1tag1$$
$$x-y=1tag2$$
which has as unique solution $x=1,y=0$. If we take the linear equation $2(1)+(2)$ (twice of first equation plus second) and $4(1)+2(2)$ we get
$$3x+y=3$$
$$6x+2y=6$$
which has an infinite number of solutions beside $x=1,y=0$, one of which is $x=0,y=3$.

Such additional solutions, at least in two variables, arise when some of the new equations are multiples of each other, so that one degree of specification is lost. Later in the course you're taking I suppose that matrices will be introduced, and it can be shown that such generation of additional solutions arises when the determinant of the matrix of linear combinations – $beginbmatrix2&1\4&2endbmatrix$ here – is zero.

share|cite|improve this answer

answered Jul 22 at 3:26

Parcly Taxel

33.6k136588

add a comment | 

up vote
1
down vote



accepted

Consider the equations
$$x+y=1tag1$$
$$x-y=1tag2$$
which has as unique solution $x=1,y=0$. If we take the linear equation $2(1)+(2)$ (twice of first equation plus second) and $4(1)+2(2)$ we get
$$3x+y=3$$
$$6x+2y=6$$
which has an infinite number of solutions beside $x=1,y=0$, one of which is $x=0,y=3$.

Such additional solutions, at least in two variables, arise when some of the new equations are multiples of each other, so that one degree of specification is lost. Later in the course you're taking I suppose that matrices will be introduced, and it can be shown that such generation of additional solutions arises when the determinant of the matrix of linear combinations – $beginbmatrix2&1\4&2endbmatrix$ here – is zero.

share|cite|improve this answer

answered Jul 22 at 3:26

Parcly Taxel

33.6k136588

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

Consider the equations
$$x+y=1tag1$$
$$x-y=1tag2$$
which has as unique solution $x=1,y=0$. If we take the linear equation $2(1)+(2)$ (twice of first equation plus second) and $4(1)+2(2)$ we get
$$3x+y=3$$
$$6x+2y=6$$
which has an infinite number of solutions beside $x=1,y=0$, one of which is $x=0,y=3$.

Such additional solutions, at least in two variables, arise when some of the new equations are multiples of each other, so that one degree of specification is lost. Later in the course you're taking I suppose that matrices will be introduced, and it can be shown that such generation of additional solutions arises when the determinant of the matrix of linear combinations – $beginbmatrix2&1\4&2endbmatrix$ here – is zero.

share|cite|improve this answer

answered Jul 22 at 3:26

Parcly Taxel

33.6k136588

Consider the equations
$$x+y=1tag1$$
$$x-y=1tag2$$
which has as unique solution $x=1,y=0$. If we take the linear equation $2(1)+(2)$ (twice of first equation plus second) and $4(1)+2(2)$ we get
$$3x+y=3$$
$$6x+2y=6$$
which has an infinite number of solutions beside $x=1,y=0$, one of which is $x=0,y=3$.

Such additional solutions, at least in two variables, arise when some of the new equations are multiples of each other, so that one degree of specification is lost. Later in the course you're taking I suppose that matrices will be introduced, and it can be shown that such generation of additional solutions arises when the determinant of the matrix of linear combinations – $beginbmatrix2&1\4&2endbmatrix$ here – is zero.

share|cite|improve this answer

answered Jul 22 at 3:26

Parcly Taxel

33.6k136588

share|cite|improve this answer

share|cite|improve this answer

answered Jul 22 at 3:26

Parcly Taxel

33.6k136588

answered Jul 22 at 3:26

Parcly Taxel

33.6k136588

answered Jul 22 at 3:26

Parcly Taxel

33.6k136588

33.6k136588

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