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Finding a Basis and Dimension In Exercises \(43-48\), find (a) a basis for and (b) the dimension of the solution space of the homogeneous system of linear equations. $$ \begin{array}{r} x-2 y+3 z=0 \\ -3 x+6 y-9 z=0 \end{array} $$

Short Answer

Expert verified
The basis for the solution space of the given system of homogeneous equations is \( {(2,1,0), (-3,0,1)} \) and its dimension is 2.

Step by step solution

01

Simplification of Equations

First, note that the second equation is just the multiple of first equation. Therefore, this system of equations basically represents a single equation. The simplified system of equations is: \[\begin{array}{r} x-2y+3z=0 \end{array} \]
02

Describe the solution in parametric form

Next, write down the solution to the equation in parametric form. The free variables here are \( y \) and \( z \). So solve the equation for \( x \) to get \( x = 2y-3z \). Then the solution to the system can be written as: \( (x, y, z) = (2y-3z, y, z) = y(2,1,0) + z(-3,0,1) \).
03

Find the basis

The two vectors in front of \( y \) and \( z \) in the parametric solution are the basis vectors. Therefore, the basis for the solution space of the system of equations is \( {(2,1,0), (-3,0,1)} \).
04

Find the dimension

The dimension of the solution space of the system of equations is the number of basis vectors. So the dimension is 2.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Basis of Solution Space
In the context of systems of linear equations, the basis of the solution space represents the set of vectors that can be combined in various ways to produce all possible solutions to the system. Each vector provides a direction along which you can travel to reach different solutions. For a homogeneous system, which is a system of linear equations that equates to zero, we are looking for solutions that describe all possible combinations that result in the zero vector.
In our exercise, we've identified a basis by expressing the solutions in a parametric form. The solution (x, y, z) = (2y-3z, y, z) involves two parameters, y and z, meaning we can create an infinite number of solutions by choosing different values for y and z. These solutions are linear combinations of the vectors (2, 1, 0) and (-3, 0, 1), which represent the basis. Each vector stands for a distinct direction in the solution space.

Why is a basis important?

The concept of a basis is crucial because it offers the most efficient set of building blocks to describe all solutions. To ensure the entire solution space is spanned without redundancy, each vector in the basis must be linearly independent, meaning no vector can be written as a combination of the others. In our example, this is satisfied by the vectors (2, 1, 0) and (-3, 0, 1). The simplicity of a basis is what makes it a powerful tool in solving and understanding homogeneous systems.
Dimension of Solution Space
The dimension of the solution space is essentially the number of vectors that make up the basis of that space. In linear algebra terms, it quantifies the freedom we have in choosing solutions to the system. For a homogeneous system, the dimension also reflects the number of free variables present.
In the given exercise, since we have two vectors in our basis set, the dimension of our solution space is 2. This means that there are two degrees of freedom when forming solutions, corresponding to the free variables y and z. The dimension tells us that the solution space is a two-dimensional plane within the three-dimensional space.

Connection between Dimension and Equations

  • One must note that the dimension is directly linked to the number of linearly independent equations in the system.
  • In this case, we only have one unique equation as the second equation is a multiple of the first, which is why we do not gain any new information from it.
As such, the system does not restrict us to a single line (which would be a one-dimensional space) but allows for a plane of solutions. Understanding the dimension of the solution space is vital as it helps in visualizing and dealing with abstract concepts in higher-dimensional spaces.
Parametric Form of Solutions
The parametric form of solutions is a way to express all potential solutions to a system of linear equations using parameters, which represent free variables. In the solution to our exercise, the free variables are y and z, and we've written the general solution as (x, y, z) = (2y-3z, y, z). This means that for each value of y and z, we can calculate a corresponding value of x, leading to an infinite amount of solutions.
This form is powerful for several reasons. Firstly, it makes the structure of the solution space explicit—showing exactly how the free variables y and z contribute to forming different solutions. Secondly, the parametric form can simplify the process of understanding and working with solutions, by focusing on the roles of these parameters rather than specific numeric answers.

Practical Usage

  • Parametric form is commonly used in both theoretical and applied mathematics.
  • It can be particularly useful in engineering and physics where systems of linear equations often arise, and solutions represent physical phenomena such as forces, movements, or fields.
By mastering the parametric form of solutions, students can enhance their problem-solving skills and better handle complex systems of linear equations in practical scenarios.

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Most popular questions from this chapter

Show that the set of solutions of a second-order linear homogeneous differential equation is linearly independent. $$ \left\\{e^{a x}, x e^{a x}\right\\} $$

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In Exercises \(43-48\) find (a) a basis for and (b) the dimension of the solution space of the homogeneous system of linear equations.. $$ \begin{aligned} 3 x_{1}+3 x_{2}+15 x_{3}+11 x_{4} &=0 \\ x_{1}-3 x_{2}+x_{3}+x_{4} &=0 \\ 2 x_{1}+3 x_{2}+11 x_{3}+8 x_{4} &=0 \end{aligned} $$

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