Solve linear equation
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
asked 43 mins ago
user1238097
487
add a comment |
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
asked 43 mins ago
user1238097
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
add a comment |
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
asked 43 mins ago
user1238097
487
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
asked 43 mins ago
user1238097
487
asked 43 mins ago
user1238097
487
edited 18 mins ago
asked 43 mins ago
user1238097
487
asked 43 mins ago
user1238097
487
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
add a comment |
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
add a comment |
3 Answers
3
active
oldest
votes
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.
answered 29 mins ago
marty cohen
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
add a comment |
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?
answered 36 mins ago
ImNotTheGuy
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.
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
add a comment |
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.
answered 16 mins ago
Bernard
118k638111
Thanks for the hint I need to go back to look at homogenous systems section in the book.
– user1238097
8 mins ago
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
answered 29 mins ago
marty cohen
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
add a comment |
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.
answered 29 mins ago
marty cohen
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
add a comment |
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.
answered 29 mins ago
marty cohen
72.3k549127
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.
answered 29 mins ago
marty cohen
72.3k549127
answered 29 mins ago
marty cohen
72.3k549127
answered 29 mins ago
marty cohen
72.3k549127
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
add a comment |
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
add a comment |
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?
answered 36 mins ago
ImNotTheGuy
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.
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
add a comment |
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?
answered 36 mins ago
ImNotTheGuy
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.
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
add a comment |
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?
answered 36 mins ago
ImNotTheGuy
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.
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?
answered 36 mins ago
ImNotTheGuy
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.
answered 36 mins ago
ImNotTheGuy
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.
answered 36 mins ago
ImNotTheGuy
712
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
add a comment |
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
add a comment |
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.
answered 16 mins ago
Bernard
118k638111
Thanks for the hint I need to go back to look at homogenous systems section in the book.
– user1238097
8 mins ago
add a comment |
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.
answered 16 mins ago
Bernard
118k638111
Thanks for the hint I need to go back to look at homogenous systems section in the book.
– user1238097
8 mins ago
add a comment |
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.
answered 16 mins ago
Bernard
118k638111
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.
answered 16 mins ago
Bernard
118k638111
answered 16 mins ago
Bernard
118k638111
answered 16 mins ago
Bernard
118k638111
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
add a comment |
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
add a comment |
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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