Find a basis for the solution space of the given homogeneous system.

Click For Summary
SUMMARY

The discussion centers on finding a basis for the solution space of a given homogeneous system represented by a matrix. The correct reduced row echelon form of the matrix is identified as:

1 0 0 3 | 0
0 1 0 0 | 0
0 0 1 0 | 0

This indicates that the solution can be expressed as a linear combination of the vector (-3, 0, 0, 1), confirming that the basis for the solution space is indeed a single vector derived from the free variable x4.

PREREQUISITES
  • Understanding of linear algebra concepts, specifically homogeneous systems.
  • Familiarity with matrix operations, including row reduction to reduced row echelon form.
  • Knowledge of vector spaces and basis vectors.
  • Experience with solving systems of equations using matrices.
NEXT STEPS
  • Study the process of row reduction in detail, focusing on achieving reduced row echelon form.
  • Learn about the implications of free variables in homogeneous systems.
  • Explore the concept of vector spaces and how to determine bases for various dimensions.
  • Practice solving additional homogeneous systems to reinforce understanding of solution spaces.
USEFUL FOR

Students and educators in linear algebra, mathematicians, and anyone involved in solving systems of linear equations or studying vector spaces.

memo_juentes
Messages
8
Reaction score
0

Homework Statement


Find a basis for the solution space of the given homogeneous system.

x1 x2 x3 x4
1 2 -1 3 | 0
2 2 -1 6 | 0
1 0 0 3 | 0



The Attempt at a Solution


When I reduced to reduced row echelon form i get the following matrix:

1 0 0 3 | 0
0 1 0 0 | 0
0 0 1 0 | 0

Which I thought it meant that the basis for the solution space would be:

1
0
0
-3

But apparently it isn't...what am I doing wrong?
 
Physics news on Phys.org
You didn't reduce the matrix correctly. Fix that first.
 
I'm sorry I actually typed the matrix wrong when making this thread. The correct matrix is:

1 2 -1 3 |0
2 2 -1 6 |0
1 0 3 3 |0
 
memo_juentes said:
When I reduced to reduced row echelon form i get the following matrix:

1 0 0 3 | 0
0 1 0 0 | 0
0 0 1 0 | 0

Which I thought it meant that the basis for the solution space would be:

1
0
0
-3

But apparently it isn't...what am I doing wrong?
Your reduced matrix says this:
x1 = -3x4
x2 = 0
x3 = 0
x4 = x4

This means that any vector x in the solution space is a constant multiple of what vector?
 
Ohh got it, seems pretty obvious now that i see it

Thanks a lot btw
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 24 ·
Replies
24
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
4
Views
2K