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Conditions so that a system of equation is incompatible
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I am trying to find a real parameter $m$ so that the following system is incompatible $$begincases
x+y+mz = 1 \
x-2y+z=m \
mx+y+z=0
endcases$$ I did the determinant of the system $$Delta=2+m+m+2m^2-1-1=2m^2+2m-4$$ By putting $Delta =0$ $$m^2+m-2=0rightarrow(m-1)(m+2)=0$$ So I have either $m=1$ or $m=-2$, but how I proceed now? And what are the conditions for a system to be incompatible?
linear-algebra systems-of-equations
edited Jul 23 at 11:49
Harry Peter
5,47311438
asked Jul 16 at 15:53
Sonkun
4979
add a comment |Â
up vote
1
down vote
favorite
I am trying to find a real parameter $m$ so that the following system is incompatible $$begincases
x+y+mz = 1 \
x-2y+z=m \
mx+y+z=0
endcases$$ I did the determinant of the system $$Delta=2+m+m+2m^2-1-1=2m^2+2m-4$$ By putting $Delta =0$ $$m^2+m-2=0rightarrow(m-1)(m+2)=0$$ So I have either $m=1$ or $m=-2$, but how I proceed now? And what are the conditions for a system to be incompatible?
linear-algebra systems-of-equations
edited Jul 23 at 11:49
Harry Peter
5,47311438
asked Jul 16 at 15:53
Sonkun
4979
Insert the values of $m$ ($1$ and $-2$) at the system and use Gauss elimination to see what kind of system you get!
– Arnaldo
Jul 16 at 15:58
Can I use Crammer Rule? Because I don't know Gauss elimination..
– Sonkun
Jul 16 at 15:59
Cramer just works for a system with an unique solution
– Arnaldo
Jul 16 at 16:00
The point is that if the determinant is nonzero then there is a possibility that the system is incompatible, because row reduction will create a row of zeros in the echelon form. But it is possible that the right hand side is chosen "just right" so that in fact the system is still compatible anyway (in this case reducing the augmented matrix results in a full row of zeros). So you have to attempt to solve the system with determinant zero by hand to see if it has a solution or not.
– Ian
Jul 16 at 16:08
@Sonkun: If you undelete your question math.stackexchange.com/questions/2882832/…, I'll post an answer.
– quasi
Aug 14 at 23:08
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am trying to find a real parameter $m$ so that the following system is incompatible $$begincases
x+y+mz = 1 \
x-2y+z=m \
mx+y+z=0
endcases$$ I did the determinant of the system $$Delta=2+m+m+2m^2-1-1=2m^2+2m-4$$ By putting $Delta =0$ $$m^2+m-2=0rightarrow(m-1)(m+2)=0$$ So I have either $m=1$ or $m=-2$, but how I proceed now? And what are the conditions for a system to be incompatible?
linear-algebra systems-of-equations
edited Jul 23 at 11:49
Harry Peter
5,47311438
asked Jul 16 at 15:53
Sonkun
4979
I am trying to find a real parameter $m$ so that the following system is incompatible $$begincases
x+y+mz = 1 \
x-2y+z=m \
mx+y+z=0
endcases$$ I did the determinant of the system $$Delta=2+m+m+2m^2-1-1=2m^2+2m-4$$ By putting $Delta =0$ $$m^2+m-2=0rightarrow(m-1)(m+2)=0$$ So I have either $m=1$ or $m=-2$, but how I proceed now? And what are the conditions for a system to be incompatible?
linear-algebra systems-of-equations
edited Jul 23 at 11:49
Harry Peter
5,47311438
asked Jul 16 at 15:53
Sonkun
4979
edited Jul 23 at 11:49
Harry Peter
5,47311438
edited Jul 23 at 11:49
Harry Peter
5,47311438
edited Jul 23 at 11:49
Harry Peter
5,47311438
5,47311438
asked Jul 16 at 15:53
Sonkun
4979
asked Jul 16 at 15:53
Sonkun
4979
asked Jul 16 at 15:53
Sonkun
4979
4979
Insert the values of $m$ ($1$ and $-2$) at the system and use Gauss elimination to see what kind of system you get!
– Arnaldo
Jul 16 at 15:58
Can I use Crammer Rule? Because I don't know Gauss elimination..
– Sonkun
Jul 16 at 15:59
Cramer just works for a system with an unique solution
– Arnaldo
Jul 16 at 16:00
The point is that if the determinant is nonzero then there is a possibility that the system is incompatible, because row reduction will create a row of zeros in the echelon form. But it is possible that the right hand side is chosen "just right" so that in fact the system is still compatible anyway (in this case reducing the augmented matrix results in a full row of zeros). So you have to attempt to solve the system with determinant zero by hand to see if it has a solution or not.
– Ian
Jul 16 at 16:08
@Sonkun: If you undelete your question math.stackexchange.com/questions/2882832/…, I'll post an answer.
– quasi
Aug 14 at 23:08
add a comment |Â
Insert the values of $m$ ($1$ and $-2$) at the system and use Gauss elimination to see what kind of system you get!
– Arnaldo
Jul 16 at 15:58
Can I use Crammer Rule? Because I don't know Gauss elimination..
– Sonkun
Jul 16 at 15:59
Cramer just works for a system with an unique solution
– Arnaldo
Jul 16 at 16:00
The point is that if the determinant is nonzero then there is a possibility that the system is incompatible, because row reduction will create a row of zeros in the echelon form. But it is possible that the right hand side is chosen "just right" so that in fact the system is still compatible anyway (in this case reducing the augmented matrix results in a full row of zeros). So you have to attempt to solve the system with determinant zero by hand to see if it has a solution or not.
– Ian
Jul 16 at 16:08
@Sonkun: If you undelete your question math.stackexchange.com/questions/2882832/…, I'll post an answer.
– quasi
Aug 14 at 23:08
Insert the values of $m$ ($1$ and $-2$) at the system and use Gauss elimination to see what kind of system you get!
– Arnaldo
Jul 16 at 15:58
Insert the values of $m$ ($1$ and $-2$) at the system and use Gauss elimination to see what kind of system you get!
– Arnaldo
Jul 16 at 15:58
Can I use Crammer Rule? Because I don't know Gauss elimination..
– Sonkun
Jul 16 at 15:59
Can I use Crammer Rule? Because I don't know Gauss elimination..
– Sonkun
Jul 16 at 15:59
Cramer just works for a system with an unique solution
– Arnaldo
Jul 16 at 16:00
Cramer just works for a system with an unique solution
– Arnaldo
Jul 16 at 16:00
The point is that if the determinant is nonzero then there is a possibility that the system is incompatible, because row reduction will create a row of zeros in the echelon form. But it is possible that the right hand side is chosen "just right" so that in fact the system is still compatible anyway (in this case reducing the augmented matrix results in a full row of zeros). So you have to attempt to solve the system with determinant zero by hand to see if it has a solution or not.
– Ian
Jul 16 at 16:08
The point is that if the determinant is nonzero then there is a possibility that the system is incompatible, because row reduction will create a row of zeros in the echelon form. But it is possible that the right hand side is chosen "just right" so that in fact the system is still compatible anyway (in this case reducing the augmented matrix results in a full row of zeros). So you have to attempt to solve the system with determinant zero by hand to see if it has a solution or not.
– Ian
Jul 16 at 16:08
@Sonkun: If you undelete your question math.stackexchange.com/questions/2882832/…, I'll post an answer.
– quasi
Aug 14 at 23:08
@Sonkun: If you undelete your question math.stackexchange.com/questions/2882832/…, I'll post an answer.
– quasi
Aug 14 at 23:08
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
You have to replace the values of $m$ at the system and use Gauss elimination. For example, for $m=1$ you get
begincases
x+y+z = 1 \
x-2y+z=1 \
x+y+z=0
endcases
See that the first and the third equations are incompatible.
Do the same for $m=-2$.
Can you finish?
answered Jul 16 at 16:02
Arnaldo
18k42146
$$begincases x+y-2z = 1 \ x-2y+z=-2 \ -2x+y+z=0 endcases$$ I think I see something. If I add all three of them I get $0=-1$ Is that it?
– Sonkun
Jul 16 at 16:05
Thank you! But Is there a general conditions? Maybe in future I will find harder systems.
– Sonkun
Jul 16 at 16:08
@Sonkun: the general condition is use Gauss Elimination: en.wikipedia.org/wiki/Gaussian_elimination
– Arnaldo
Jul 16 at 17:51
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
You have to replace the values of $m$ at the system and use Gauss elimination. For example, for $m=1$ you get
begincases
x+y+z = 1 \
x-2y+z=1 \
x+y+z=0
endcases
See that the first and the third equations are incompatible.
Do the same for $m=-2$.
Can you finish?
answered Jul 16 at 16:02
Arnaldo
18k42146
$$begincases x+y-2z = 1 \ x-2y+z=-2 \ -2x+y+z=0 endcases$$ I think I see something. If I add all three of them I get $0=-1$ Is that it?
– Sonkun
Jul 16 at 16:05
Thank you! But Is there a general conditions? Maybe in future I will find harder systems.
– Sonkun
Jul 16 at 16:08
@Sonkun: the general condition is use Gauss Elimination: en.wikipedia.org/wiki/Gaussian_elimination
– Arnaldo
Jul 16 at 17:51
add a comment |Â
up vote
1
down vote
You have to replace the values of $m$ at the system and use Gauss elimination. For example, for $m=1$ you get
begincases
x+y+z = 1 \
x-2y+z=1 \
x+y+z=0
endcases
See that the first and the third equations are incompatible.
Do the same for $m=-2$.
Can you finish?
answered Jul 16 at 16:02
Arnaldo
18k42146
$$begincases x+y-2z = 1 \ x-2y+z=-2 \ -2x+y+z=0 endcases$$ I think I see something. If I add all three of them I get $0=-1$ Is that it?
– Sonkun
Jul 16 at 16:05
Thank you! But Is there a general conditions? Maybe in future I will find harder systems.
– Sonkun
Jul 16 at 16:08
@Sonkun: the general condition is use Gauss Elimination: en.wikipedia.org/wiki/Gaussian_elimination
– Arnaldo
Jul 16 at 17:51
add a comment |Â
up vote
1
down vote
up vote
1
down vote
You have to replace the values of $m$ at the system and use Gauss elimination. For example, for $m=1$ you get
begincases
x+y+z = 1 \
x-2y+z=1 \
x+y+z=0
endcases
See that the first and the third equations are incompatible.
Do the same for $m=-2$.
Can you finish?
answered Jul 16 at 16:02
Arnaldo
18k42146
You have to replace the values of $m$ at the system and use Gauss elimination. For example, for $m=1$ you get
begincases
x+y+z = 1 \
x-2y+z=1 \
x+y+z=0
endcases
See that the first and the third equations are incompatible.
Do the same for $m=-2$.
Can you finish?
answered Jul 16 at 16:02
Arnaldo
18k42146
answered Jul 16 at 16:02
Arnaldo
18k42146
answered Jul 16 at 16:02
Arnaldo
18k42146
answered Jul 16 at 16:02
Arnaldo
18k42146
18k42146
$$begincases x+y-2z = 1 \ x-2y+z=-2 \ -2x+y+z=0 endcases$$ I think I see something. If I add all three of them I get $0=-1$ Is that it?
– Sonkun
Jul 16 at 16:05
Thank you! But Is there a general conditions? Maybe in future I will find harder systems.
– Sonkun
Jul 16 at 16:08
@Sonkun: the general condition is use Gauss Elimination: en.wikipedia.org/wiki/Gaussian_elimination
– Arnaldo
Jul 16 at 17:51
add a comment |Â
$$begincases x+y-2z = 1 \ x-2y+z=-2 \ -2x+y+z=0 endcases$$ I think I see something. If I add all three of them I get $0=-1$ Is that it?
– Sonkun
Jul 16 at 16:05
Thank you! But Is there a general conditions? Maybe in future I will find harder systems.
– Sonkun
Jul 16 at 16:08
@Sonkun: the general condition is use Gauss Elimination: en.wikipedia.org/wiki/Gaussian_elimination
– Arnaldo
Jul 16 at 17:51
$$begincases x+y-2z = 1 \ x-2y+z=-2 \ -2x+y+z=0 endcases$$ I think I see something. If I add all three of them I get $0=-1$ Is that it?
– Sonkun
Jul 16 at 16:05
$$begincases x+y-2z = 1 \ x-2y+z=-2 \ -2x+y+z=0 endcases$$ I think I see something. If I add all three of them I get $0=-1$ Is that it?
– Sonkun
Jul 16 at 16:05
Thank you! But Is there a general conditions? Maybe in future I will find harder systems.
– Sonkun
Jul 16 at 16:08
Thank you! But Is there a general conditions? Maybe in future I will find harder systems.
– Sonkun
Jul 16 at 16:08
@Sonkun: the general condition is use Gauss Elimination: en.wikipedia.org/wiki/Gaussian_elimination
– Arnaldo
Jul 16 at 17:51
@Sonkun: the general condition is use Gauss Elimination: en.wikipedia.org/wiki/Gaussian_elimination
– Arnaldo
Jul 16 at 17:51
add a comment |Â
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Insert the values of $m$ ($1$ and $-2$) at the system and use Gauss elimination to see what kind of system you get!
– Arnaldo
Jul 16 at 15:58
Can I use Crammer Rule? Because I don't know Gauss elimination..
– Sonkun
Jul 16 at 15:59
Cramer just works for a system with an unique solution
– Arnaldo
Jul 16 at 16:00
The point is that if the determinant is nonzero then there is a possibility that the system is incompatible, because row reduction will create a row of zeros in the echelon form. But it is possible that the right hand side is chosen "just right" so that in fact the system is still compatible anyway (in this case reducing the augmented matrix results in a full row of zeros). So you have to attempt to solve the system with determinant zero by hand to see if it has a solution or not.
– Ian
Jul 16 at 16:08
@Sonkun: If you undelete your question math.stackexchange.com/questions/2882832/…, I'll post an answer.
– quasi
Aug 14 at 23:08