Linear Algebra solution to a system of equations

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SUMMARY

The discussion focuses on solving a system of linear equations using matrix methods, specifically through the process of reducing to Row Reduced Echelon Form (RREF). The equations provided are: x + y + z = 0, 3x + 2y - 2z = 0, 4x + 3y - z = 0, and 6x + 5y + z = 0. The solution reveals that there are two independent equations, leading to one free variable, represented as t, where x = 4t, y = -5t, and z = t. The rank of the coefficient matrix is determined to be 2, confirming the dimension of the nullspace is 1, indicating the presence of a single free variable.

PREREQUISITES
  • Understanding of linear equations and systems
  • Familiarity with matrix operations and RREF
  • Knowledge of concepts like rank and nullspace in linear algebra
  • Ability to interpret free variables in the context of linear systems
NEXT STEPS
  • Study the process of performing elementary row operations on matrices
  • Learn about the rank-nullity theorem in linear algebra
  • Explore the concept of basis vectors and their significance in vector spaces
  • Practice solving systems of equations with varying numbers of equations and unknowns
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Students and educators in mathematics, particularly those studying linear algebra, as well as anyone interested in solving systems of equations using matrix methods.

csgirl504
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Homework Statement



x + y+ z = 0
3x + 2y -2z = 0
4x + 3y -z = 0
6x + 5y + z = 0

Homework Equations





The Attempt at a Solution



I put the equations into a matrix and reduced to RREF. This is what I end up with:

x - 4z = 0
y + 5z = 0

The other two rows in the matrix are all zeroes.

I've never solved a system that had more equations than unknowns, so I'm confused on how many free variables I will need. Right now I have this as my solution:

x = 4t
y = -5t
z = t
 
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csgirl504 said:

Homework Statement



x + y+ z = 0
3x + 2y -2z = 0
4x + 3y -z = 0
6x + 5y + z = 0

Homework Equations


The Attempt at a Solution



I put the equations into a matrix and reduced to RREF. This is what I end up with:

x - 4z = 0
y + 5z = 0

The other two rows in the matrix are all zeroes.

I've never solved a system that had more equations than unknowns, so I'm confused on how many free variables I will need. Right now I have this as my solution:

x = 4t
y = -5t
z = t

Homework Statement


Homework Equations


The Attempt at a Solution


And that is just fine as a solution. There are really only two independent equations in there. The third one is the sum of the first two equations. The fourth is three times the first equation added to the second. They are redundant, as your RREF showed you.
 
Last edited:
If you write the system of the equations in matrix form and you perform elementary row operations and put the matrix into row reduced echelon form then what is the rank of the matrix ? (Hint...which columns are independent and which are dependent ?) If the rank is r and the number of columns is n then n-r = the dimension of the nullspace for the coeficient matrix = the number of free variables. In this case the dimension for the nullspace is 1 since n - r = 3 - 2 = 1 and a basis for the nullspace is the vector c(4, -5, 1). So if c = 1 then x = 4, y = -5, z=1 is a solution. In other words you are correct.
 

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