Linear System with Parameter: Solutions and Homogeneity

In summary: Yes, using the Rouché–Capelli theorem to determine the values of ##\lambda## for which the system has solutions was a correct approach.
  • #1
Felafel
171
0

Homework Statement



I have done this exercise, but I don't have a file with the solutions. COuld you please check it?
Thank you in advance :)

Given the following system:
##\lambda \in \mathbb{R}##

##x − z = \lambda##
##x + y + 2z + t = 0 ##
##y + 3z = ##
##x + z + t = 0##

1-find ##rk(A_{\lambda}) and rk(A_{\lambda}, B_{\lambda})## according to the different values of ##\lambda##

2- for which values of ##\lambda## does tha system have solutions?

3- what is its solutions set?

4- for which values is the system homogeneous ?

The Attempt at a Solution



This is the matrix ##(A_{\lambda}, B_{\lambda})##
( 1 0 1 -1 λ)
( λ 1 2 1 0)
( 0 1 3 0 λ)
( 1 0 1 λ 0)

doing some row reduction i get:

( 1 0 -1 0 λ )
( 0 0 2 λ -λ )
(λ-1 0 0 1 -2λ)
( 0 1 3 0 λ)

and i see rk(A)=rk(A,B) for any ##\lambda##, so according to Rouchè-Capelli's theorem the system has solutions.

So, answers to 1 and 2 are: 1- rk(A)=rk(A,B) for any ##\lambda \in \mathbb{R}## 2-##\forany \lambda \in \mathbb{R}##

Now, if ##\lambda##=1 i get:
(1 0 -1 0 1)
(0 0 2 1 -1)
(0 0 0 1 -2)
(0 1 3 0 1) thus: x=3/2, y= -1/2, z=1/2, t=-2 is the only solution

Doing the same, if ##\lambda##=0 i get (x, y, z, t)=(0, 0, 0, 0)
(IS IT ACCEPTABLE AS A SOLUTION?)

If ##\lambda## is different from 0 and 1,

x-z=##\lambda##
2z+λt=-λ
(λ-1)x+t=-2λ
3z+y=λ

with z= ##\alpha##
##\Sigma##: (x,y,z,t)= (λ+##\alpha##, λ-3 ##\alpha, \alpha, \frac{-λ-2\alpha}{λ}##)

so answer to nu,ber 3: the solutions sets are those above
4- the system is homogeneous for λ=0. for its solutions set, see above.
 
Physics news on Phys.org
  • #2
I know it is a very long exercise, so I'll just ask again for the most urging question: could anyone please tell me if ##\lambda=0## is right?
thank you :)
 
  • #3
Felafel said:

Homework Statement



I have done this exercise, but I don't have a file with the solutions. COuld you please check it?
Thank you in advance :)

Given the following system:
##\lambda \in \mathbb{R}##

##x − z = \lambda##
##x + y + 2z + t = 0 ##
##y + 3z = ##
##x + z + t = 0##
What's on the right side of the third equation? λ?
Felafel said:
1-find ##rk(A_{\lambda}) and rk(A_{\lambda}, B_{\lambda})## according to the different values of ##\lambda##

2- for which values of ##\lambda## does tha system have solutions?

3- what is its solutions set?

4- for which values is the system homogeneous ?

The Attempt at a Solution



This is the matrix ##(A_{\lambda}, B_{\lambda})##
( 1 0 1 -1 λ)
( λ 1 2 1 0)
( 0 1 3 0 λ)
( 1 0 1 λ 0)
Your matrix doesn't look right to me. I'm assuming this is the augmented matrix that represents your system. If so, the first row in the matrix should be:
1 0 -1 0 λ

You have
1 0 1 -1 λ

Felafel said:
doing some row reduction i get:

( 1 0 -1 0 λ )
( 0 0 2 λ -λ )
(λ-1 0 0 1 -2λ)
( 0 1 3 0 λ)

and i see rk(A)=rk(A,B) for any ##\lambda##, so according to Rouchè-Capelli's theorem the system has solutions.

So, answers to 1 and 2 are: 1- rk(A)=rk(A,B) for any ##\lambda \in \mathbb{R}## 2-##\forany \lambda \in \mathbb{R}##

Now, if ##\lambda##=1 i get:
(1 0 -1 0 1)
(0 0 2 1 -1)
(0 0 0 1 -2)
(0 1 3 0 1) thus: x=3/2, y= -1/2, z=1/2, t=-2 is the only solution

Doing the same, if ##\lambda##=0 i get (x, y, z, t)=(0, 0, 0, 0)
(IS IT ACCEPTABLE AS A SOLUTION?)

If ##\lambda## is different from 0 and 1,

x-z=##\lambda##
2z+λt=-λ
(λ-1)x+t=-2λ
3z+y=λ

with z= ##\alpha##
##\Sigma##: (x,y,z,t)= (λ+##\alpha##, λ-3 ##\alpha, \alpha, \frac{-λ-2\alpha}{λ}##)

so answer to nu,ber 3: the solutions sets are those above
4- the system is homogeneous for λ=0. for its solutions set, see above.
 
  • #4
oops, yes, it should be as you said, but I think it's just a typo, because I've solved it on a piece of paper, so the rest of the calculation should be right and done on the correct matrix
 
  • #5
and on the right side of the third equation there is a λ, yes
 
  • #6
Felafel said:
I know it is a very long exercise, so I'll just ask again for the most urging question: could anyone please tell me if ##\lambda=0## is right?
thank you :)
If you mean "is [itex]\lambda= 0[/itex] the correct answer to 'for what values is the system homogeneous,'" then, by the definition of "homogenous system", yes.
 
  • #7
thank you!
and was it correct to use rouchè-capelli's thoerem?
 

Related to Linear System with Parameter: Solutions and Homogeneity

1. What is a linear system with parameter?

A linear system with parameter is a type of mathematical model that describes the relationship between a set of variables using linear equations. The parameter in the model represents a variable that can vary and affect the behavior of the system.

2. What is the difference between a linear system and a nonlinear system?

A linear system is one in which the output is directly proportional to the input, while a nonlinear system is one in which the output is not directly proportional to the input. In other words, in a linear system, the relationship between the variables can be described by a straight line, while in a nonlinear system, the relationship is more complex and cannot be represented by a straight line.

3. What are some real-world applications of linear systems with parameters?

Linear systems with parameters are commonly used in fields such as physics, engineering, and economics to model and predict the behavior of various systems. Some examples include predicting the trajectory of a projectile, analyzing the stability of an electrical circuit, and determining the optimal production levels for a manufacturing process.

4. How are parameters determined in a linear system?

Parameters in a linear system are typically determined through experimentation or data analysis. By collecting data and performing calculations, scientists can determine the values of the parameters that best fit the observed behavior of the system.

5. Can linear systems with parameters be solved analytically?

Yes, linear systems with parameters can be solved analytically by using techniques such as substitution, elimination, or matrix operations. However, in some cases, numerical methods may be necessary to find a solution due to the complexity of the system.

Similar threads

  • Calculus and Beyond Homework Help
Replies
7
Views
891
  • Calculus and Beyond Homework Help
Replies
10
Views
619
  • Calculus and Beyond Homework Help
Replies
5
Views
390
  • Calculus and Beyond Homework Help
Replies
2
Views
608
  • Calculus and Beyond Homework Help
Replies
4
Views
915
  • Calculus and Beyond Homework Help
Replies
2
Views
876
  • Calculus and Beyond Homework Help
Replies
7
Views
640
  • Calculus and Beyond Homework Help
Replies
2
Views
571
  • Calculus and Beyond Homework Help
Replies
1
Views
757
  • Calculus and Beyond Homework Help
Replies
8
Views
728
Back
Top