How do I solve 6b) and 6c) if my solution for 6a) is a consistent system of linear equations?
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
$$begincases
2x_1-x_2= dx_1 \
2x_1-x_2+x_3=dx_2 \
-2x_1+2x_2+x_3=dx_3
endcases
$$
a) Is it possible for the system to be inconsistent? Explain?
b) For what values of d will the system have infinitely many solutions?
c) Solve the system when it has infinitely many solutions?
For my solution in part a),
$$ left[
beginarrayc
1&0&0&0\
0&1&0&0\
0&0&1&0
endarray
right] $$
Hence, it is a unique set of solutions i.e. the system cannot be inconsistent. So how is it possible to get infinitely many solutions in 6b) and 6c)?
linear-algebra systems-of-equations
 |Â
show 1 more comment
up vote
1
down vote
favorite
$$begincases
2x_1-x_2= dx_1 \
2x_1-x_2+x_3=dx_2 \
-2x_1+2x_2+x_3=dx_3
endcases
$$
a) Is it possible for the system to be inconsistent? Explain?
b) For what values of d will the system have infinitely many solutions?
c) Solve the system when it has infinitely many solutions?
For my solution in part a),
$$ left[
beginarrayc
1&0&0&0\
0&1&0&0\
0&0&1&0
endarray
right] $$
Hence, it is a unique set of solutions i.e. the system cannot be inconsistent. So how is it possible to get infinitely many solutions in 6b) and 6c)?
linear-algebra systems-of-equations
Welcome to Math SE! Links to images are not accessible to screen readers. You can instead use MathJax to type mathematics on this site â in particular the 'Systems of equations' part.
â Toby Mak
Sep 1 at 4:22
For my solution in part a)
The rest of the question appears to center on this part, which you haven't posted. Please see How do I ask a good question? and add at least some minimal context.
â dxiv
Sep 1 at 4:28
Hi, I have edited my question accordingly, thank you for the suggestion!
â Cheryl
Sep 1 at 4:58
You should be able to tell that the system canâÂÂt be inconsistent by inspection: itâÂÂs homogeneous, so $x_1=x_2=x_3=0$ is always a solution.
â amd
Sep 1 at 8:59
How were you able to fully reduce the coefficient matrix without knowing what $d$ is? You might have divided by zero somewhere along the way, which would make the reduction invalid.
â amd
Sep 1 at 9:00
 |Â
show 1 more comment
up vote
1
down vote
favorite
up vote
1
down vote
favorite
$$begincases
2x_1-x_2= dx_1 \
2x_1-x_2+x_3=dx_2 \
-2x_1+2x_2+x_3=dx_3
endcases
$$
a) Is it possible for the system to be inconsistent? Explain?
b) For what values of d will the system have infinitely many solutions?
c) Solve the system when it has infinitely many solutions?
For my solution in part a),
$$ left[
beginarrayc
1&0&0&0\
0&1&0&0\
0&0&1&0
endarray
right] $$
Hence, it is a unique set of solutions i.e. the system cannot be inconsistent. So how is it possible to get infinitely many solutions in 6b) and 6c)?
linear-algebra systems-of-equations
$$begincases
2x_1-x_2= dx_1 \
2x_1-x_2+x_3=dx_2 \
-2x_1+2x_2+x_3=dx_3
endcases
$$
a) Is it possible for the system to be inconsistent? Explain?
b) For what values of d will the system have infinitely many solutions?
c) Solve the system when it has infinitely many solutions?
For my solution in part a),
$$ left[
beginarrayc
1&0&0&0\
0&1&0&0\
0&0&1&0
endarray
right] $$
Hence, it is a unique set of solutions i.e. the system cannot be inconsistent. So how is it possible to get infinitely many solutions in 6b) and 6c)?
linear-algebra systems-of-equations
linear-algebra systems-of-equations
edited Sep 1 at 4:57
asked Sep 1 at 4:19
Cheryl
243
243
Welcome to Math SE! Links to images are not accessible to screen readers. You can instead use MathJax to type mathematics on this site â in particular the 'Systems of equations' part.
â Toby Mak
Sep 1 at 4:22
For my solution in part a)
The rest of the question appears to center on this part, which you haven't posted. Please see How do I ask a good question? and add at least some minimal context.
â dxiv
Sep 1 at 4:28
Hi, I have edited my question accordingly, thank you for the suggestion!
â Cheryl
Sep 1 at 4:58
You should be able to tell that the system canâÂÂt be inconsistent by inspection: itâÂÂs homogeneous, so $x_1=x_2=x_3=0$ is always a solution.
â amd
Sep 1 at 8:59
How were you able to fully reduce the coefficient matrix without knowing what $d$ is? You might have divided by zero somewhere along the way, which would make the reduction invalid.
â amd
Sep 1 at 9:00
 |Â
show 1 more comment
Welcome to Math SE! Links to images are not accessible to screen readers. You can instead use MathJax to type mathematics on this site â in particular the 'Systems of equations' part.
â Toby Mak
Sep 1 at 4:22
For my solution in part a)
The rest of the question appears to center on this part, which you haven't posted. Please see How do I ask a good question? and add at least some minimal context.
â dxiv
Sep 1 at 4:28
Hi, I have edited my question accordingly, thank you for the suggestion!
â Cheryl
Sep 1 at 4:58
You should be able to tell that the system canâÂÂt be inconsistent by inspection: itâÂÂs homogeneous, so $x_1=x_2=x_3=0$ is always a solution.
â amd
Sep 1 at 8:59
How were you able to fully reduce the coefficient matrix without knowing what $d$ is? You might have divided by zero somewhere along the way, which would make the reduction invalid.
â amd
Sep 1 at 9:00
Welcome to Math SE! Links to images are not accessible to screen readers. You can instead use MathJax to type mathematics on this site â in particular the 'Systems of equations' part.
â Toby Mak
Sep 1 at 4:22
Welcome to Math SE! Links to images are not accessible to screen readers. You can instead use MathJax to type mathematics on this site â in particular the 'Systems of equations' part.
â Toby Mak
Sep 1 at 4:22
For my solution in part a)
The rest of the question appears to center on this part, which you haven't posted. Please see How do I ask a good question? and add at least some minimal context.â dxiv
Sep 1 at 4:28
For my solution in part a)
The rest of the question appears to center on this part, which you haven't posted. Please see How do I ask a good question? and add at least some minimal context.â dxiv
Sep 1 at 4:28
Hi, I have edited my question accordingly, thank you for the suggestion!
â Cheryl
Sep 1 at 4:58
Hi, I have edited my question accordingly, thank you for the suggestion!
â Cheryl
Sep 1 at 4:58
You should be able to tell that the system canâÂÂt be inconsistent by inspection: itâÂÂs homogeneous, so $x_1=x_2=x_3=0$ is always a solution.
â amd
Sep 1 at 8:59
You should be able to tell that the system canâÂÂt be inconsistent by inspection: itâÂÂs homogeneous, so $x_1=x_2=x_3=0$ is always a solution.
â amd
Sep 1 at 8:59
How were you able to fully reduce the coefficient matrix without knowing what $d$ is? You might have divided by zero somewhere along the way, which would make the reduction invalid.
â amd
Sep 1 at 9:00
How were you able to fully reduce the coefficient matrix without knowing what $d$ is? You might have divided by zero somewhere along the way, which would make the reduction invalid.
â amd
Sep 1 at 9:00
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
Using the Gaussian elimination method to solve a set of linear equations,
From the equations, you have given,
begincases
2x_1-x_2= dx_1 \
2x_1-x_2+x_3=dx_2 \
-2x_1+2x_2+x_3=dx_3
endcases
We can arrive at this augmented matrix,
beginbmatrix
beginarrayc
-d+2&-1&0&0\
2&-1-d&1&0\
-2&2&1-d&0
endarray
endbmatrix
Using row transformations,
beginbmatrix
beginarrayc
-2 & 2 & 1-d &0 \
2 & -1-d & 1 &0\
-d+2 & -1 & 0 &0\
endarray
endbmatrix
beginequation
downarrow
endequation
beginbmatrix
beginarrayc
-2 & 2 & 1-d &0 \
0 & 1-d & 2-d &0\
0 & 0 & (1-d)(2-d) -2(2-d) &0\
endarray
endbmatrix
For it to have infinite solutions,
beginequation
(1-d)(2-d) -2(2-d) = 0
endequation
beginequation
d = 2, -1
endequation
If $d$ takes the above value, then you will end up with a free variable ($x_3$)
beginequation
x_2 = frac(d-2)x_31-d
endequation
beginequation
x_1 = frac12(2x_2 + (1-d)x_3)
endequation
May I know how you derived your last row to that? I can't seem to eliminate my first 2 columns of the last row to = 0...
â Cheryl
Sep 1 at 9:39
Use $R_2 = R_1 + R_2$ ,$R_3 = (-d+2)R_1 + 2R_3$, then $R_3 = R_3 - 2R_2$ then it reduces to this form.
â Lokesh Kumar
Sep 1 at 9:49
Got it, thank you for your help!
â Cheryl
Sep 1 at 10:00
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Using the Gaussian elimination method to solve a set of linear equations,
From the equations, you have given,
begincases
2x_1-x_2= dx_1 \
2x_1-x_2+x_3=dx_2 \
-2x_1+2x_2+x_3=dx_3
endcases
We can arrive at this augmented matrix,
beginbmatrix
beginarrayc
-d+2&-1&0&0\
2&-1-d&1&0\
-2&2&1-d&0
endarray
endbmatrix
Using row transformations,
beginbmatrix
beginarrayc
-2 & 2 & 1-d &0 \
2 & -1-d & 1 &0\
-d+2 & -1 & 0 &0\
endarray
endbmatrix
beginequation
downarrow
endequation
beginbmatrix
beginarrayc
-2 & 2 & 1-d &0 \
0 & 1-d & 2-d &0\
0 & 0 & (1-d)(2-d) -2(2-d) &0\
endarray
endbmatrix
For it to have infinite solutions,
beginequation
(1-d)(2-d) -2(2-d) = 0
endequation
beginequation
d = 2, -1
endequation
If $d$ takes the above value, then you will end up with a free variable ($x_3$)
beginequation
x_2 = frac(d-2)x_31-d
endequation
beginequation
x_1 = frac12(2x_2 + (1-d)x_3)
endequation
May I know how you derived your last row to that? I can't seem to eliminate my first 2 columns of the last row to = 0...
â Cheryl
Sep 1 at 9:39
Use $R_2 = R_1 + R_2$ ,$R_3 = (-d+2)R_1 + 2R_3$, then $R_3 = R_3 - 2R_2$ then it reduces to this form.
â Lokesh Kumar
Sep 1 at 9:49
Got it, thank you for your help!
â Cheryl
Sep 1 at 10:00
add a comment |Â
up vote
0
down vote
accepted
Using the Gaussian elimination method to solve a set of linear equations,
From the equations, you have given,
begincases
2x_1-x_2= dx_1 \
2x_1-x_2+x_3=dx_2 \
-2x_1+2x_2+x_3=dx_3
endcases
We can arrive at this augmented matrix,
beginbmatrix
beginarrayc
-d+2&-1&0&0\
2&-1-d&1&0\
-2&2&1-d&0
endarray
endbmatrix
Using row transformations,
beginbmatrix
beginarrayc
-2 & 2 & 1-d &0 \
2 & -1-d & 1 &0\
-d+2 & -1 & 0 &0\
endarray
endbmatrix
beginequation
downarrow
endequation
beginbmatrix
beginarrayc
-2 & 2 & 1-d &0 \
0 & 1-d & 2-d &0\
0 & 0 & (1-d)(2-d) -2(2-d) &0\
endarray
endbmatrix
For it to have infinite solutions,
beginequation
(1-d)(2-d) -2(2-d) = 0
endequation
beginequation
d = 2, -1
endequation
If $d$ takes the above value, then you will end up with a free variable ($x_3$)
beginequation
x_2 = frac(d-2)x_31-d
endequation
beginequation
x_1 = frac12(2x_2 + (1-d)x_3)
endequation
May I know how you derived your last row to that? I can't seem to eliminate my first 2 columns of the last row to = 0...
â Cheryl
Sep 1 at 9:39
Use $R_2 = R_1 + R_2$ ,$R_3 = (-d+2)R_1 + 2R_3$, then $R_3 = R_3 - 2R_2$ then it reduces to this form.
â Lokesh Kumar
Sep 1 at 9:49
Got it, thank you for your help!
â Cheryl
Sep 1 at 10:00
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Using the Gaussian elimination method to solve a set of linear equations,
From the equations, you have given,
begincases
2x_1-x_2= dx_1 \
2x_1-x_2+x_3=dx_2 \
-2x_1+2x_2+x_3=dx_3
endcases
We can arrive at this augmented matrix,
beginbmatrix
beginarrayc
-d+2&-1&0&0\
2&-1-d&1&0\
-2&2&1-d&0
endarray
endbmatrix
Using row transformations,
beginbmatrix
beginarrayc
-2 & 2 & 1-d &0 \
2 & -1-d & 1 &0\
-d+2 & -1 & 0 &0\
endarray
endbmatrix
beginequation
downarrow
endequation
beginbmatrix
beginarrayc
-2 & 2 & 1-d &0 \
0 & 1-d & 2-d &0\
0 & 0 & (1-d)(2-d) -2(2-d) &0\
endarray
endbmatrix
For it to have infinite solutions,
beginequation
(1-d)(2-d) -2(2-d) = 0
endequation
beginequation
d = 2, -1
endequation
If $d$ takes the above value, then you will end up with a free variable ($x_3$)
beginequation
x_2 = frac(d-2)x_31-d
endequation
beginequation
x_1 = frac12(2x_2 + (1-d)x_3)
endequation
Using the Gaussian elimination method to solve a set of linear equations,
From the equations, you have given,
begincases
2x_1-x_2= dx_1 \
2x_1-x_2+x_3=dx_2 \
-2x_1+2x_2+x_3=dx_3
endcases
We can arrive at this augmented matrix,
beginbmatrix
beginarrayc
-d+2&-1&0&0\
2&-1-d&1&0\
-2&2&1-d&0
endarray
endbmatrix
Using row transformations,
beginbmatrix
beginarrayc
-2 & 2 & 1-d &0 \
2 & -1-d & 1 &0\
-d+2 & -1 & 0 &0\
endarray
endbmatrix
beginequation
downarrow
endequation
beginbmatrix
beginarrayc
-2 & 2 & 1-d &0 \
0 & 1-d & 2-d &0\
0 & 0 & (1-d)(2-d) -2(2-d) &0\
endarray
endbmatrix
For it to have infinite solutions,
beginequation
(1-d)(2-d) -2(2-d) = 0
endequation
beginequation
d = 2, -1
endequation
If $d$ takes the above value, then you will end up with a free variable ($x_3$)
beginequation
x_2 = frac(d-2)x_31-d
endequation
beginequation
x_1 = frac12(2x_2 + (1-d)x_3)
endequation
answered Sep 1 at 8:30
Lokesh Kumar
1363
1363
May I know how you derived your last row to that? I can't seem to eliminate my first 2 columns of the last row to = 0...
â Cheryl
Sep 1 at 9:39
Use $R_2 = R_1 + R_2$ ,$R_3 = (-d+2)R_1 + 2R_3$, then $R_3 = R_3 - 2R_2$ then it reduces to this form.
â Lokesh Kumar
Sep 1 at 9:49
Got it, thank you for your help!
â Cheryl
Sep 1 at 10:00
add a comment |Â
May I know how you derived your last row to that? I can't seem to eliminate my first 2 columns of the last row to = 0...
â Cheryl
Sep 1 at 9:39
Use $R_2 = R_1 + R_2$ ,$R_3 = (-d+2)R_1 + 2R_3$, then $R_3 = R_3 - 2R_2$ then it reduces to this form.
â Lokesh Kumar
Sep 1 at 9:49
Got it, thank you for your help!
â Cheryl
Sep 1 at 10:00
May I know how you derived your last row to that? I can't seem to eliminate my first 2 columns of the last row to = 0...
â Cheryl
Sep 1 at 9:39
May I know how you derived your last row to that? I can't seem to eliminate my first 2 columns of the last row to = 0...
â Cheryl
Sep 1 at 9:39
Use $R_2 = R_1 + R_2$ ,$R_3 = (-d+2)R_1 + 2R_3$, then $R_3 = R_3 - 2R_2$ then it reduces to this form.
â Lokesh Kumar
Sep 1 at 9:49
Use $R_2 = R_1 + R_2$ ,$R_3 = (-d+2)R_1 + 2R_3$, then $R_3 = R_3 - 2R_2$ then it reduces to this form.
â Lokesh Kumar
Sep 1 at 9:49
Got it, thank you for your help!
â Cheryl
Sep 1 at 10:00
Got it, thank you for your help!
â Cheryl
Sep 1 at 10:00
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2901362%2fhow-do-i-solve-6b-and-6c-if-my-solution-for-6a-is-a-consistent-system-of-line%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Welcome to Math SE! Links to images are not accessible to screen readers. You can instead use MathJax to type mathematics on this site â in particular the 'Systems of equations' part.
â Toby Mak
Sep 1 at 4:22
For my solution in part a)
The rest of the question appears to center on this part, which you haven't posted. Please see How do I ask a good question? and add at least some minimal context.â dxiv
Sep 1 at 4:28
Hi, I have edited my question accordingly, thank you for the suggestion!
â Cheryl
Sep 1 at 4:58
You should be able to tell that the system canâÂÂt be inconsistent by inspection: itâÂÂs homogeneous, so $x_1=x_2=x_3=0$ is always a solution.
â amd
Sep 1 at 8:59
How were you able to fully reduce the coefficient matrix without knowing what $d$ is? You might have divided by zero somewhere along the way, which would make the reduction invalid.
â amd
Sep 1 at 9:00