How do I solve 6b) and 6c) if my solution for 6a) is a consistent system of linear equations?

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$$begincases
2x_1-x_2= dx_1 \
2x_1-x_2+x_3=dx_2 \
-2x_1+2x_2+x_3=dx_3
endcases
$$



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)?










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  • Welcome to Math SE! Links to images are not accessible to screen readers. You can instead use MathJax to type mathematics on this site – in particular the 'Systems of equations' part.
    – 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.
    – dxiv
    Sep 1 at 4:28











  • Hi, I have edited my question accordingly, thank you for the suggestion!
    – Cheryl
    Sep 1 at 4:58










  • You should be able to tell that the system can’t be inconsistent by inspection: it’s homogeneous, so $x_1=x_2=x_3=0$ is always a solution.
    – 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.
    – amd
    Sep 1 at 9:00














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)?










share|cite|improve this question























  • Welcome to Math SE! Links to images are not accessible to screen readers. You can instead use MathJax to type mathematics on this site – in particular the 'Systems of equations' part.
    – 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.
    – dxiv
    Sep 1 at 4:28











  • Hi, I have edited my question accordingly, thank you for the suggestion!
    – Cheryl
    Sep 1 at 4:58










  • You should be able to tell that the system can’t be inconsistent by inspection: it’s homogeneous, so $x_1=x_2=x_3=0$ is always a solution.
    – 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.
    – amd
    Sep 1 at 9:00












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)?










share|cite|improve this question















$$begincases
2x_1-x_2= dx_1 \
2x_1-x_2+x_3=dx_2 \
-2x_1+2x_2+x_3=dx_3
endcases
$$



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






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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 – in particular the 'Systems of equations' part.
    – 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.
    – dxiv
    Sep 1 at 4:28











  • Hi, I have edited my question accordingly, thank you for the suggestion!
    – Cheryl
    Sep 1 at 4:58










  • You should be able to tell that the system can’t be inconsistent by inspection: it’s homogeneous, so $x_1=x_2=x_3=0$ is always a solution.
    – 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.
    – 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 – in particular the 'Systems of equations' part.
    – 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.
    – dxiv
    Sep 1 at 4:28











  • Hi, I have edited my question accordingly, thank you for the suggestion!
    – Cheryl
    Sep 1 at 4:58










  • You should be able to tell that the system can’t be inconsistent by inspection: it’s homogeneous, so $x_1=x_2=x_3=0$ is always a solution.
    – 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.
    – 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 – in particular the 'Systems of equations' part.
– 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 – in particular the 'Systems of equations' part.
– 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.
– 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.
– dxiv
Sep 1 at 4:28













Hi, I have edited my question accordingly, thank you for the suggestion!
– Cheryl
Sep 1 at 4:58




Hi, I have edited my question accordingly, thank you for the suggestion!
– Cheryl
Sep 1 at 4:58












You should be able to tell that the system can’t be inconsistent by inspection: it’s homogeneous, so $x_1=x_2=x_3=0$ is always a solution.
– amd
Sep 1 at 8:59




You should be able to tell that the system can’t be inconsistent by inspection: it’s homogeneous, so $x_1=x_2=x_3=0$ is always a solution.
– 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.
– 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.
– amd
Sep 1 at 9:00










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






share|cite|improve this answer




















  • 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...
    – 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.
    – Lokesh Kumar
    Sep 1 at 9:49










  • Got it, thank you for your help!
    – Cheryl
    Sep 1 at 10:00










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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






share|cite|improve this answer




















  • 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...
    – 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.
    – Lokesh Kumar
    Sep 1 at 9:49










  • Got it, thank you for your help!
    – Cheryl
    Sep 1 at 10:00














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






share|cite|improve this answer




















  • 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...
    – 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.
    – Lokesh Kumar
    Sep 1 at 9:49










  • Got it, thank you for your help!
    – Cheryl
    Sep 1 at 10:00












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






share|cite|improve this answer












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







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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...
    – 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.
    – Lokesh Kumar
    Sep 1 at 9:49










  • Got it, thank you for your help!
    – 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...
    – 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.
    – Lokesh Kumar
    Sep 1 at 9:49










  • Got it, thank you for your help!
    – 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...
– 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...
– 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.
– 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.
– Lokesh Kumar
Sep 1 at 9:49












Got it, thank you for your help!
– Cheryl
Sep 1 at 10:00




Got it, thank you for your help!
– Cheryl
Sep 1 at 10:00

















 

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