Can a 3x5 Matrix with 3 Free Variables Ever Have No Solution for Any Vector b?

Click For Summary
SUMMARY

A non-homogeneous system represented by a 3x5 matrix (A) with 3 free variables cannot have a solution for any vector b if the rank of the matrix is less than the number of equations. According to the rank theorem, for a matrix A with 5 columns, the rank must equal the number of pivot columns. In this case, the maximum rank is 3, leading to a scenario where the system is underdetermined, resulting in no solutions for certain vectors b. Therefore, if the rank of A is 2 or less, there are no solutions for any vector b.

PREREQUISITES
  • Understanding of linear algebra concepts, specifically matrix rank.
  • Familiarity with the rank theorem in the context of linear equations.
  • Knowledge of pivot columns and their significance in solving linear systems.
  • Basic comprehension of non-homogeneous systems of equations.
NEXT STEPS
  • Study the implications of the rank theorem in linear algebra.
  • Explore the concept of pivot columns in greater detail.
  • Investigate conditions under which a non-homogeneous system has no solutions.
  • Learn about the relationship between the rank of a matrix and the dimension of its null space.
USEFUL FOR

Students and educators in linear algebra, mathematicians analyzing systems of equations, and anyone seeking to understand the conditions for solvability in non-homogeneous systems.

sami23
Messages
69
Reaction score
1

Homework Statement


Suppose a non-homogeneous system, Ax = b, of 3 linear equations in 5 unknowns (3x5 matrix) and 3 free variables, prove there is no solution for any vector b.


Homework Equations


Using the rank theroem:
n = rank A + dim Nul(A) where n = # of columns; dim Nul(A) = # free variables


The Attempt at a Solution


rank A = n - dim Nul(A) = 5-3 = 2 (which represents the pivot columns)

How do I know there are no solutions for any vector b knowing there can be at most 3 pivot columns?
 
Physics news on Phys.org
This doesn't make sense. Most systems of 3 equations in 5 unknowns have many solutions.
 
sami23 said:

Homework Statement


Suppose a non-homogeneous system, Ax = b, of 3 linear equations in 5 unknowns (3x5 matrix) and 3 free variables, prove there is no solution for any vector b.


Homework Equations


Using the rank theroem:
n = rank A + dim Nul(A) where n = # of columns; dim Nul(A) = # free variables


The Attempt at a Solution


rank A = n - dim Nul(A) = 5-3 = 2 (which represents the pivot columns)

How do I know there are no solutions for any vector b knowing there can be at most 3 pivot columns?


If the matrix has rank 3 there are infinitely many solutions.

RGV
 

Similar threads

Replies
15
Views
2K
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
8K
Replies
15
Views
3K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 10 ·
Replies
10
Views
5K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K