Linear algebra systems ,Ax=b,Ax=0

Click For Summary

Homework Help Overview

The discussion revolves around a consistent system of linear equations represented as Ax=b, exploring the relationship between a particular solution and the general solution form, specifically how every solution can be expressed as a combination of a fixed solution and a solution to the homogeneous equation Ax=0.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the form of the general solution and the validity of expressing solutions as x=x1+x0, where x0 is a solution to Ax=0. Questions arise about the correctness of the expressions used and the implications of different variable assignments.

Discussion Status

Some participants are exploring the implications of their reasoning regarding the transformation of solutions and the relationships between them. There is an ongoing inquiry into whether the proposed forms accurately represent all solutions, with participants seeking confirmation of their understanding before proceeding further.

Contextual Notes

Participants express uncertainty about the definitions and relationships between the variables involved, particularly in distinguishing between different forms of solutions and their implications for the problem at hand.

madah12
Messages
326
Reaction score
1

Homework Statement



Let be Ax=b any consistent system of linear equations, and let be x1 a fixed solution. Show that every solution to the system
can be written in the form x=x1 +x0, where x0 is a solution Ax=0 . Show also that every matrix of this form is a solution

Homework Equations





The Attempt at a Solution



well A(x1+x0)= Ax1+Ax0=b+0=b
but this only proves that any matrix x=x1+x0 is a solution but not that every solution can be written as such and i don't know how to prove that
 
Physics news on Phys.org
Suppose x is a solution to Ax= b. Show that x- x1 is a solution to Ax= 0.
 
wait x-x0 or x-x1?
 
I mean A(x-x0)=B-0=b isn't a solution but maybe A(x-x1)=b-b = 0 hmm so you are saying there will always exist x0=x-x1 such that x=x1+x0 is a solution to Ax=b
 
I would appreciate a comment about whether i am right or wrong cause i don't want to go to the next problem until I am sure i understood this one
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K