Constraint on a linear system?

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Homework Help Overview

The discussion revolves around a linear system of equations and a nonlinear constraint. The original poster presents a homogeneous system defined by three equations and seeks to identify solutions that also meet the constraint given by a nonlinear equation.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the general solution of the linear system and discuss how it relates to the nonlinear constraint. There are attempts to verify specific solutions and to derive conditions under which the constraint holds.

Discussion Status

The conversation includes various interpretations of the solutions to the linear system and their compatibility with the nonlinear constraint. Some participants provide potential solutions while others question the validity of those solutions, indicating an ongoing exploration of the problem.

Contextual Notes

There is mention of a specific solution [0,0,0] and the general solution expressed in terms of a parameter. The discussion also highlights the need to satisfy both the linear and nonlinear equations, with some uncertainty about the implications of the quadratic equation derived from the constraint.

bonfire09
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The question goes like this.
Among all solutions that satisfy the homogeneous system
x + 2y + z = 0
2x + 4y + z = 0
x + 2y − z = 0
determine those that also satisfy the nonlinear constraint y − xy = 2z

I know that one of the solutions [0,0,0] but I'm not sure how to find the others. I row reduced the linear system and its general solution is x=y[-2,1,0].
 
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bonfire09 said:
The question goes like this.
Among all solutions that satisfy the homogeneous system
x + 2y + z = 0
2x + 4y + z = 0
x + 2y − z = 0
determine those that also satisfy the nonlinear constraint y − xy = 2z

I know that one of the solutions [0,0,0] but I'm not sure how to find the others. I row reduced the linear system and its general solution is x=y[-2,1,0].

So the general solution to your system of equations is t<-2, 1, 0>, where t is any real number. Another way to write this is <-2t, t, 0>. Does this satisfy your constraint for some value of t?
 
oh -1/2<-2,1,0>=<1,-1/2,0> is another solution to the constraint.
 
i think so.
 
You have already said that any solution to the linear equations must have x= -2t, y= t, z= 0. (I did not check that.)

Now you also want to require that y- xy= 2z which is just t- (2t)(t)= 0 or [itex]t- 2t^2= 0[/tex]. That's a quadratic equation and has <b>two</b> solutions.[/itex]
 

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