Solve linear equation












2














I am working on some practice linear algebra problems but I am not understanding how the answers where gotten for the following problem.



Problem:
find all values of $a$ for which the resulting linear system has (a) no solution, (b) a unique solution, and (c) infinitely many solutions.



begin{cases} x + y - z =2\ x + 2y + z = 3 \ x + y + (a^2 -5)z = a end{cases}



Answer:



(a) $a= -2,;$ (b) $;aneq 2,;$ (c) $a=2$



I can't for the life figure how they got the answer in the book. I only got as far as the row echelon matrix.



$begin{bmatrix}1 & 1 & -1 & 2\0 & 1 & 2 & 1\ 0 & 0 & a^2 - 4 & a-2end{bmatrix}$



Can anyone recommend an approach for solving the problem.






linear-algebra







share|cite|improve this question
























  • What are $b$ and $c$?
    – Bernard
    31 mins ago










  • @Bernard thats what I am wondering. It is from the text Introductory Linear Algebra with applications 7th edition.
    – user1238097
    29 mins ago
















2














I am working on some practice linear algebra problems but I am not understanding how the answers where gotten for the following problem.



Problem:
find all values of $a$ for which the resulting linear system has (a) no solution, (b) a unique solution, and (c) infinitely many solutions.



begin{cases} x + y - z =2\ x + 2y + z = 3 \ x + y + (a^2 -5)z = a end{cases}



Answer:



(a) $a= -2,;$ (b) $;aneq 2,;$ (c) $a=2$



I can't for the life figure how they got the answer in the book. I only got as far as the row echelon matrix.



$begin{bmatrix}1 & 1 & -1 & 2\0 & 1 & 2 & 1\ 0 & 0 & a^2 - 4 & a-2end{bmatrix}$



Can anyone recommend an approach for solving the problem.






linear-algebra







share|cite|improve this question
























  • What are $b$ and $c$?
    – Bernard
    31 mins ago










  • @Bernard thats what I am wondering. It is from the text Introductory Linear Algebra with applications 7th edition.
    – user1238097
    29 mins ago














2












2








2







I am working on some practice linear algebra problems but I am not understanding how the answers where gotten for the following problem.



Problem:
find all values of $a$ for which the resulting linear system has (a) no solution, (b) a unique solution, and (c) infinitely many solutions.



begin{cases} x + y - z =2\ x + 2y + z = 3 \ x + y + (a^2 -5)z = a end{cases}



Answer:



(a) $a= -2,;$ (b) $;aneq 2,;$ (c) $a=2$



I can't for the life figure how they got the answer in the book. I only got as far as the row echelon matrix.



$begin{bmatrix}1 & 1 & -1 & 2\0 & 1 & 2 & 1\ 0 & 0 & a^2 - 4 & a-2end{bmatrix}$



Can anyone recommend an approach for solving the problem.






linear-algebra







share|cite|improve this question















I am working on some practice linear algebra problems but I am not understanding how the answers where gotten for the following problem.



Problem:
find all values of $a$ for which the resulting linear system has (a) no solution, (b) a unique solution, and (c) infinitely many solutions.



begin{cases} x + y - z =2\ x + 2y + z = 3 \ x + y + (a^2 -5)z = a end{cases}



Answer:



(a) $a= -2,;$ (b) $;aneq 2,;$ (c) $a=2$



I can't for the life figure how they got the answer in the book. I only got as far as the row echelon matrix.



$begin{bmatrix}1 & 1 & -1 & 2\0 & 1 & 2 & 1\ 0 & 0 & a^2 - 4 & a-2end{bmatrix}$



Can anyone recommend an approach for solving the problem.






linear-algebra




linear-algebra






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 18 mins ago

























asked 43 mins ago









user1238097

487




487












  • What are $b$ and $c$?
    – Bernard
    31 mins ago










  • @Bernard thats what I am wondering. It is from the text Introductory Linear Algebra with applications 7th edition.
    – user1238097
    29 mins ago


















  • What are $b$ and $c$?
    – Bernard
    31 mins ago










  • @Bernard thats what I am wondering. It is from the text Introductory Linear Algebra with applications 7th edition.
    – user1238097
    29 mins ago
















What are $b$ and $c$?
– Bernard
31 mins ago




What are $b$ and $c$?
– Bernard
31 mins ago












@Bernard thats what I am wondering. It is from the text Introductory Linear Algebra with applications 7th edition.
– user1238097
29 mins ago




@Bernard thats what I am wondering. It is from the text Introductory Linear Algebra with applications 7th edition.
– user1238097
29 mins ago










3 Answers
3






active

oldest

votes


















3














The standard start:



The matrix is



$begin{bmatrix}
1 & 1 & -1 & 2\
1 & 2 & 1 & 3\
1 & 1 & a^2-5 & a
end{bmatrix}
$



Its determinant is
$2(a^2-5)+1-1-(1+(a^2-5)-2)
=a^2-4
$
so,
if $a^2 ne 4$,
there is a unique solution.






share|cite|improve this answer





















  • So I don't have to the linear equations to solve this wow.
    – user1238097
    28 mins ago










  • I take issue with this answer, as it is a bit sloppy; you cannot take the determinant of a $3times 4$ matrix. The approach is nice, but you should clean up your explanation.
    – ImNotTheGuy
    24 mins ago










  • @ImNotTheGuy it is definitely a bit more confusing.
    – user1238097
    23 mins ago



















2














Having the matrix in something approaching row echelon form is definitely the right way to go, and you are nearly done once you have it. Unfortunately I think you did your final row operation incorrectly; my row echelon form matrix looks like this:
$$
begin{bmatrix}1 & 1 & -1 & 2\0 & 1 & 2 & 1\ 0 & 0 & a^2-4 & a-2end{bmatrix}
$$
Is this enough to show you the way?






share|cite|improve this answer








New contributor




ImNotTheGuy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.


















  • I think I get where the a^2 - 4 gives the + or - 2. The not equal is because its a quadratic equation?
    – user1238097
    20 mins ago






  • 1




    @user1238097 The distinction has to come from how we interpret each row of this augmented matrix; when $a=-2$, the final row reads as $[0,0,0,-4]$, which in the language of equations reads $0=-4$. Of course, it cannot be that $0=-4$, and that tells us the system of equations we started with was inconsistent (that it had no solutions). Instead, if $a=2$, the final row is $[0,0,0,0]$, which reads $0=0$, which is true but uninformative. This means there is a free variable, and we will have infinitely many solutions.
    – ImNotTheGuy
    18 mins ago










  • @user1238097 When $aneq pm 2$, what happens? We could go from row echelon form to reduced row echelon form (since we know we can divide by $a^2-4$ safely), and we will end up with an identity matrix in the left hand portion of the augmented matrix, which means there is a unique solution.
    – ImNotTheGuy
    16 mins ago





















1














Hint:



A non-homogeneous linear system has solutions if and only if the matrix of the homogeneous side and the augmented matrix have the same rank. Furthermore, this rank is the codimension of the affine space of solutions.



Thus here, there is a unique solution if and only if the matrix $;smash{begin{bmatrix}1&1&-1\ 1&2&1\1&1&a^2-5end{bmatrix}}$ has rank $3$. You can obtain its rank with row reduction.






share|cite|improve this answer





















  • Thanks for the hint I need to go back to look at homogenous systems section in the book.
    – user1238097
    8 mins ago











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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









3














The standard start:



The matrix is



$begin{bmatrix}
1 & 1 & -1 & 2\
1 & 2 & 1 & 3\
1 & 1 & a^2-5 & a
end{bmatrix}
$



Its determinant is
$2(a^2-5)+1-1-(1+(a^2-5)-2)
=a^2-4
$
so,
if $a^2 ne 4$,
there is a unique solution.






share|cite|improve this answer





















  • So I don't have to the linear equations to solve this wow.
    – user1238097
    28 mins ago










  • I take issue with this answer, as it is a bit sloppy; you cannot take the determinant of a $3times 4$ matrix. The approach is nice, but you should clean up your explanation.
    – ImNotTheGuy
    24 mins ago










  • @ImNotTheGuy it is definitely a bit more confusing.
    – user1238097
    23 mins ago
















3














The standard start:



The matrix is



$begin{bmatrix}
1 & 1 & -1 & 2\
1 & 2 & 1 & 3\
1 & 1 & a^2-5 & a
end{bmatrix}
$



Its determinant is
$2(a^2-5)+1-1-(1+(a^2-5)-2)
=a^2-4
$
so,
if $a^2 ne 4$,
there is a unique solution.






share|cite|improve this answer





















  • So I don't have to the linear equations to solve this wow.
    – user1238097
    28 mins ago










  • I take issue with this answer, as it is a bit sloppy; you cannot take the determinant of a $3times 4$ matrix. The approach is nice, but you should clean up your explanation.
    – ImNotTheGuy
    24 mins ago










  • @ImNotTheGuy it is definitely a bit more confusing.
    – user1238097
    23 mins ago














3












3








3






The standard start:



The matrix is



$begin{bmatrix}
1 & 1 & -1 & 2\
1 & 2 & 1 & 3\
1 & 1 & a^2-5 & a
end{bmatrix}
$



Its determinant is
$2(a^2-5)+1-1-(1+(a^2-5)-2)
=a^2-4
$
so,
if $a^2 ne 4$,
there is a unique solution.






share|cite|improve this answer












The standard start:



The matrix is



$begin{bmatrix}
1 & 1 & -1 & 2\
1 & 2 & 1 & 3\
1 & 1 & a^2-5 & a
end{bmatrix}
$



Its determinant is
$2(a^2-5)+1-1-(1+(a^2-5)-2)
=a^2-4
$
so,
if $a^2 ne 4$,
there is a unique solution.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 29 mins ago









marty cohen

72.3k549127




72.3k549127












  • So I don't have to the linear equations to solve this wow.
    – user1238097
    28 mins ago










  • I take issue with this answer, as it is a bit sloppy; you cannot take the determinant of a $3times 4$ matrix. The approach is nice, but you should clean up your explanation.
    – ImNotTheGuy
    24 mins ago










  • @ImNotTheGuy it is definitely a bit more confusing.
    – user1238097
    23 mins ago


















  • So I don't have to the linear equations to solve this wow.
    – user1238097
    28 mins ago










  • I take issue with this answer, as it is a bit sloppy; you cannot take the determinant of a $3times 4$ matrix. The approach is nice, but you should clean up your explanation.
    – ImNotTheGuy
    24 mins ago










  • @ImNotTheGuy it is definitely a bit more confusing.
    – user1238097
    23 mins ago
















So I don't have to the linear equations to solve this wow.
– user1238097
28 mins ago




So I don't have to the linear equations to solve this wow.
– user1238097
28 mins ago












I take issue with this answer, as it is a bit sloppy; you cannot take the determinant of a $3times 4$ matrix. The approach is nice, but you should clean up your explanation.
– ImNotTheGuy
24 mins ago




I take issue with this answer, as it is a bit sloppy; you cannot take the determinant of a $3times 4$ matrix. The approach is nice, but you should clean up your explanation.
– ImNotTheGuy
24 mins ago












@ImNotTheGuy it is definitely a bit more confusing.
– user1238097
23 mins ago




@ImNotTheGuy it is definitely a bit more confusing.
– user1238097
23 mins ago











2














Having the matrix in something approaching row echelon form is definitely the right way to go, and you are nearly done once you have it. Unfortunately I think you did your final row operation incorrectly; my row echelon form matrix looks like this:
$$
begin{bmatrix}1 & 1 & -1 & 2\0 & 1 & 2 & 1\ 0 & 0 & a^2-4 & a-2end{bmatrix}
$$
Is this enough to show you the way?






share|cite|improve this answer








New contributor




ImNotTheGuy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.


















  • I think I get where the a^2 - 4 gives the + or - 2. The not equal is because its a quadratic equation?
    – user1238097
    20 mins ago






  • 1




    @user1238097 The distinction has to come from how we interpret each row of this augmented matrix; when $a=-2$, the final row reads as $[0,0,0,-4]$, which in the language of equations reads $0=-4$. Of course, it cannot be that $0=-4$, and that tells us the system of equations we started with was inconsistent (that it had no solutions). Instead, if $a=2$, the final row is $[0,0,0,0]$, which reads $0=0$, which is true but uninformative. This means there is a free variable, and we will have infinitely many solutions.
    – ImNotTheGuy
    18 mins ago










  • @user1238097 When $aneq pm 2$, what happens? We could go from row echelon form to reduced row echelon form (since we know we can divide by $a^2-4$ safely), and we will end up with an identity matrix in the left hand portion of the augmented matrix, which means there is a unique solution.
    – ImNotTheGuy
    16 mins ago


















2














Having the matrix in something approaching row echelon form is definitely the right way to go, and you are nearly done once you have it. Unfortunately I think you did your final row operation incorrectly; my row echelon form matrix looks like this:
$$
begin{bmatrix}1 & 1 & -1 & 2\0 & 1 & 2 & 1\ 0 & 0 & a^2-4 & a-2end{bmatrix}
$$
Is this enough to show you the way?






share|cite|improve this answer








New contributor




ImNotTheGuy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.


















  • I think I get where the a^2 - 4 gives the + or - 2. The not equal is because its a quadratic equation?
    – user1238097
    20 mins ago






  • 1




    @user1238097 The distinction has to come from how we interpret each row of this augmented matrix; when $a=-2$, the final row reads as $[0,0,0,-4]$, which in the language of equations reads $0=-4$. Of course, it cannot be that $0=-4$, and that tells us the system of equations we started with was inconsistent (that it had no solutions). Instead, if $a=2$, the final row is $[0,0,0,0]$, which reads $0=0$, which is true but uninformative. This means there is a free variable, and we will have infinitely many solutions.
    – ImNotTheGuy
    18 mins ago










  • @user1238097 When $aneq pm 2$, what happens? We could go from row echelon form to reduced row echelon form (since we know we can divide by $a^2-4$ safely), and we will end up with an identity matrix in the left hand portion of the augmented matrix, which means there is a unique solution.
    – ImNotTheGuy
    16 mins ago
















2












2








2






Having the matrix in something approaching row echelon form is definitely the right way to go, and you are nearly done once you have it. Unfortunately I think you did your final row operation incorrectly; my row echelon form matrix looks like this:
$$
begin{bmatrix}1 & 1 & -1 & 2\0 & 1 & 2 & 1\ 0 & 0 & a^2-4 & a-2end{bmatrix}
$$
Is this enough to show you the way?






share|cite|improve this answer








New contributor




ImNotTheGuy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









Having the matrix in something approaching row echelon form is definitely the right way to go, and you are nearly done once you have it. Unfortunately I think you did your final row operation incorrectly; my row echelon form matrix looks like this:
$$
begin{bmatrix}1 & 1 & -1 & 2\0 & 1 & 2 & 1\ 0 & 0 & a^2-4 & a-2end{bmatrix}
$$
Is this enough to show you the way?







share|cite|improve this answer








New contributor




ImNotTheGuy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this answer



share|cite|improve this answer






New contributor




ImNotTheGuy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









answered 36 mins ago









ImNotTheGuy

712




712




New contributor




ImNotTheGuy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





ImNotTheGuy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






ImNotTheGuy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • I think I get where the a^2 - 4 gives the + or - 2. The not equal is because its a quadratic equation?
    – user1238097
    20 mins ago






  • 1




    @user1238097 The distinction has to come from how we interpret each row of this augmented matrix; when $a=-2$, the final row reads as $[0,0,0,-4]$, which in the language of equations reads $0=-4$. Of course, it cannot be that $0=-4$, and that tells us the system of equations we started with was inconsistent (that it had no solutions). Instead, if $a=2$, the final row is $[0,0,0,0]$, which reads $0=0$, which is true but uninformative. This means there is a free variable, and we will have infinitely many solutions.
    – ImNotTheGuy
    18 mins ago










  • @user1238097 When $aneq pm 2$, what happens? We could go from row echelon form to reduced row echelon form (since we know we can divide by $a^2-4$ safely), and we will end up with an identity matrix in the left hand portion of the augmented matrix, which means there is a unique solution.
    – ImNotTheGuy
    16 mins ago




















  • I think I get where the a^2 - 4 gives the + or - 2. The not equal is because its a quadratic equation?
    – user1238097
    20 mins ago






  • 1




    @user1238097 The distinction has to come from how we interpret each row of this augmented matrix; when $a=-2$, the final row reads as $[0,0,0,-4]$, which in the language of equations reads $0=-4$. Of course, it cannot be that $0=-4$, and that tells us the system of equations we started with was inconsistent (that it had no solutions). Instead, if $a=2$, the final row is $[0,0,0,0]$, which reads $0=0$, which is true but uninformative. This means there is a free variable, and we will have infinitely many solutions.
    – ImNotTheGuy
    18 mins ago










  • @user1238097 When $aneq pm 2$, what happens? We could go from row echelon form to reduced row echelon form (since we know we can divide by $a^2-4$ safely), and we will end up with an identity matrix in the left hand portion of the augmented matrix, which means there is a unique solution.
    – ImNotTheGuy
    16 mins ago


















I think I get where the a^2 - 4 gives the + or - 2. The not equal is because its a quadratic equation?
– user1238097
20 mins ago




I think I get where the a^2 - 4 gives the + or - 2. The not equal is because its a quadratic equation?
– user1238097
20 mins ago




1




1




@user1238097 The distinction has to come from how we interpret each row of this augmented matrix; when $a=-2$, the final row reads as $[0,0,0,-4]$, which in the language of equations reads $0=-4$. Of course, it cannot be that $0=-4$, and that tells us the system of equations we started with was inconsistent (that it had no solutions). Instead, if $a=2$, the final row is $[0,0,0,0]$, which reads $0=0$, which is true but uninformative. This means there is a free variable, and we will have infinitely many solutions.
– ImNotTheGuy
18 mins ago




@user1238097 The distinction has to come from how we interpret each row of this augmented matrix; when $a=-2$, the final row reads as $[0,0,0,-4]$, which in the language of equations reads $0=-4$. Of course, it cannot be that $0=-4$, and that tells us the system of equations we started with was inconsistent (that it had no solutions). Instead, if $a=2$, the final row is $[0,0,0,0]$, which reads $0=0$, which is true but uninformative. This means there is a free variable, and we will have infinitely many solutions.
– ImNotTheGuy
18 mins ago












@user1238097 When $aneq pm 2$, what happens? We could go from row echelon form to reduced row echelon form (since we know we can divide by $a^2-4$ safely), and we will end up with an identity matrix in the left hand portion of the augmented matrix, which means there is a unique solution.
– ImNotTheGuy
16 mins ago






@user1238097 When $aneq pm 2$, what happens? We could go from row echelon form to reduced row echelon form (since we know we can divide by $a^2-4$ safely), and we will end up with an identity matrix in the left hand portion of the augmented matrix, which means there is a unique solution.
– ImNotTheGuy
16 mins ago













1














Hint:



A non-homogeneous linear system has solutions if and only if the matrix of the homogeneous side and the augmented matrix have the same rank. Furthermore, this rank is the codimension of the affine space of solutions.



Thus here, there is a unique solution if and only if the matrix $;smash{begin{bmatrix}1&1&-1\ 1&2&1\1&1&a^2-5end{bmatrix}}$ has rank $3$. You can obtain its rank with row reduction.






share|cite|improve this answer





















  • Thanks for the hint I need to go back to look at homogenous systems section in the book.
    – user1238097
    8 mins ago
















1














Hint:



A non-homogeneous linear system has solutions if and only if the matrix of the homogeneous side and the augmented matrix have the same rank. Furthermore, this rank is the codimension of the affine space of solutions.



Thus here, there is a unique solution if and only if the matrix $;smash{begin{bmatrix}1&1&-1\ 1&2&1\1&1&a^2-5end{bmatrix}}$ has rank $3$. You can obtain its rank with row reduction.






share|cite|improve this answer





















  • Thanks for the hint I need to go back to look at homogenous systems section in the book.
    – user1238097
    8 mins ago














1












1








1






Hint:



A non-homogeneous linear system has solutions if and only if the matrix of the homogeneous side and the augmented matrix have the same rank. Furthermore, this rank is the codimension of the affine space of solutions.



Thus here, there is a unique solution if and only if the matrix $;smash{begin{bmatrix}1&1&-1\ 1&2&1\1&1&a^2-5end{bmatrix}}$ has rank $3$. You can obtain its rank with row reduction.






share|cite|improve this answer












Hint:



A non-homogeneous linear system has solutions if and only if the matrix of the homogeneous side and the augmented matrix have the same rank. Furthermore, this rank is the codimension of the affine space of solutions.



Thus here, there is a unique solution if and only if the matrix $;smash{begin{bmatrix}1&1&-1\ 1&2&1\1&1&a^2-5end{bmatrix}}$ has rank $3$. You can obtain its rank with row reduction.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 16 mins ago









Bernard

118k638111




118k638111












  • Thanks for the hint I need to go back to look at homogenous systems section in the book.
    – user1238097
    8 mins ago


















  • Thanks for the hint I need to go back to look at homogenous systems section in the book.
    – user1238097
    8 mins ago
















Thanks for the hint I need to go back to look at homogenous systems section in the book.
– user1238097
8 mins ago




Thanks for the hint I need to go back to look at homogenous systems section in the book.
– user1238097
8 mins ago


















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