Understanding Solutions to Systems of Linear Equations

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SUMMARY

The discussion centers on solving a system of linear equations represented by the matrix equation A * X = 0, where A = \begin{bmatrix} 1 & 0 & 0 \\ -1 & 0 & 0 \\ 3 & -5 & 0 \end{bmatrix}. The general solution involves identifying free variables, with z being the free variable when x = 0 and y = 0. The solutions derived include X = \begin{bmatrix} 5/3 \\ 1 \\ 0 \end{bmatrix} y + \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} z and X = \begin{bmatrix} 1 \\ 3/5 \\ 0 \end{bmatrix} x + \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} z. The key takeaway is understanding how to determine which variable to treat as free based on the equations.

PREREQUISITES
  • Understanding of linear algebra concepts, specifically systems of linear equations.
  • Familiarity with matrix representation of equations.
  • Knowledge of free and basic variables in linear systems.
  • Ability to perform Gaussian elimination or row reduction techniques.
NEXT STEPS
  • Study Gaussian elimination techniques for solving linear systems.
  • Learn about the rank of a matrix and its implications on the solution set.
  • Explore the concept of vector spaces and linear combinations in linear algebra.
  • Practice solving systems of equations with varying numbers of variables and equations.
USEFUL FOR

Students studying linear algebra, educators teaching systems of equations, and anyone seeking to improve their problem-solving skills in mathematics.

roam
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Homework Statement



This is from a solved problem:

Here is a system:

[tex]\begin{bmatrix} {1 & 0&0 \\ -1&0&0\\3&-5&0 \end{bmatrix} \begin{bmatrix} {x \\ y\\z \end{bmatrix}= \begin{bmatrix} {0\\ 0\\0\end{bmatrix}[/tex]

A general solution of this system is

[tex]x= \begin{bmatrix} {0 \\ 0\\t \end{bmatrix}= \begin{bmatrix} {0 \\ 0\\1 \end{bmatrix} t[/tex]

So, how did they get this solution? I tried solving this and I just don't get it!

The Attempt at a Solution



The matrix corresponds to the set of equations:

x=0
-x=0
3x-5y=0

If I take "y" to be the free variable, I have x=5/3y therefore the solution to the system will be

[tex]\begin{bmatrix} {x \\ y\\z \end{bmatrix} = \begin{bmatrix} {5/3 \\ 1\\0 \end{bmatrix} y + \begin{bmatrix} {0 \\ 0\\1 \end{bmatrix} z[/tex]

But if I take "x" to be the free variable I get y=3/5x, so I get the following solution:

[tex]\begin{bmatrix} {x \\ y\\z \end{bmatrix} = \begin{bmatrix} {1 \\ 3/5\\0 \end{bmatrix} x + \begin{bmatrix} {0 \\ 0\\1 \end{bmatrix} z[/tex]

How do I know which one (x or y) should be taken as the free variable?

Could anyone please explain to me how to get the right answer? :frown: I just can't see how to deal with this type of problems.
 
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roam said:

Homework Statement



This is from a solved problem:



So, how did they get this solution? I tried solving this and I just don't get it!

The Attempt at a Solution



The matrix corresponds to the set of equations:

x=0
-x=0
3x-5y=0
From these equations, x is obviously 0. With x = 0, the 3rd equation implies that y = 0. This means that z is the free variable.
roam said:
If I take "y" to be the free variable, I have x=5/3y therefore the solution to the system will be

[tex]\begin{bmatrix} {x \\ y\\z \end{bmatrix} = \begin{bmatrix} {5/3 \\ 1\\0 \end{bmatrix} y + \begin{bmatrix} {0 \\ 0\\1 \end{bmatrix} z[/tex]

But if I take "x" to be the free variable I get y=3/5x, so I get the following solution:

[tex]\begin{bmatrix} {x \\ y\\z \end{bmatrix} = \begin{bmatrix} {1 \\ 3/5\\0 \end{bmatrix} x + \begin{bmatrix} {0 \\ 0\\1 \end{bmatrix} z[/tex]

How do I know which one (x or y) should be taken as the free variable?

Could anyone please explain to me how to get the right answer? :frown: I just can't see how to deal with this type of problems.
 

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