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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. $$ \begin{aligned} 2 x_{1}+2 x_{2}+4 x_{3}-2 x_{4} &=0 \\ x_{1}+2 x_{2}+x_{3}+2 x_{4} &=0 \\ -x_{1}+x_{2}+4 x_{3}-2 x_{4} &=0 \end{aligned} $$

Short Answer

Expert verified
The basis for the solution space is \(\{-2,1,1,0\}, \{0,-1,0,1\}\) and the dimension of the solution space is 2.

Step by step solution

01

Convert to Augmented Matrix Form

Rewrite the system of equations as an augmented matrix: \[\left[\begin{array}{cccc|c} 2 & 2 & 4 & -2 & 0 \ 1 & 2 & 1 & 2 & 0 \ -1 & 1 & 4 & -2 & 0 \end{array}\right] \]
02

Row Reduce to Echelon form

Performing row operations to put it in row echelon form gives the following matrix: \[\left[\begin{array}{cccc|c} 1 & 0 & 2 & 0 & 0 \ 0 & 1 & -1 & 1 & 0 \ 0 & 0 & 0 & 0 & 0 \end{array}\right] \]First row was divided by 2, then row 2 was replaced by row 2 minus row 1 and row 3 was replaced by row 3 plus row 1.
03

Identify Free and Basic Variables

The first two columns, associated with \(x_{1}\) and \(x_{2}\), have leading 1s, so these are basic variables. The third and fourth columns, associated with \(x_{3}\) and \(x_{4}\), do not, so these are free variables. Write the solution in parametric vector form.
04

Writing the Solution in Parametric Form

Giving free variables parameters \(t_1\) and \(t_2\), the solution to the system of equations is: \[\begin{aligned} x_{1} &= -2t_1 \ x_{2} &= t_{1} - t_{2} \ x_{3} &= t_{1} \ x_{4} &= t_{2} \end{aligned} \]which can be written in parametric vector form as \[\begin{bmatrix} x_{1} \ x_{2} \ x_{3} \ x_{4} \end{bmatrix}= t_{1}\begin{bmatrix} -2 \ 1 \ 1 \ 0 \end{bmatrix}+ t_{2}\begin{bmatrix} 0 \ -1 \ 0 \ 1 \end{bmatrix}\]
05

Identify the Basis and the Dimension

The basis for the solution space of the system is formed by the vectors \(\{-2,1,1,0\}\) and \(\{0,-1,0,1\}\), and the dimension is the number of basis vectors for the solution space, which is 2 in this case.

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Key Concepts

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

Basis
In linear algebra, the concept of a basis is crucial when dealing with vector spaces. A basis is a set of vectors in a vector space that, when combined, can generate every vector within that space. Importantly, these vectors must be linearly independent, meaning no vector in the set can be expressed as a combination of the others.

Finding a basis involves identifying such vectors within a solution space. For example, if you're dealing with a homogeneous system of linear equations, after performing row reductions to get the system into a simpler form, you can identify which variables can freely change (free variables) and which are determined by the former (basic variables). Using these distinctions, you derive vectors that span the solution space. These vectors form the basis of the solution space.

In our example, the vectors \([-2, 1, 1, 0]\) and \([0, -1, 0, 1]\) were derived as our basis for the solution space.
Dimension
The dimension of a vector space informs us how many vectors exist within the basis of that space. It represents the space's size but in terms of directions rather than a physical measure.

When solving a homogeneous system of linear equations, the row-reduced echelon form often reveals free variables. Each free variable corresponds to a vector in the solution space that cannot be described by any other vector in the solution. The total number of such vectors is the dimension of the space.

In our example, by reducing the system into row echelon form, we identified two free variables. These two free variables led us to two basis vectors, confirming that the dimension of our solution space is 2.
Homogeneous System
A homogeneous system of linear equations is one where all the equations equal zero. Such systems are significant because they always have at least the trivial solution, which is when all variables are assigned the value of zero.

Homogeneous systems can have infinitely many solutions or just the trivial solution. Solutions form a vector space where basis and dimension play a key role. By transforming the equations into a matrix and putting it into reduced row echelon form, you can spot free and basic variables. These help define the solution space as well as its span in terms of basis vectors.

In our example, the homogeneous system was converted into an augmented matrix and then simplified to identify the basis and dimension of the solution space.
Row Echelon Form
The row echelon form is a structured matrix form that the original system of equations is converted into to simplify finding its solutions. The idea is to manipulate the original matrix using row operations until it meets specific conditions.

These conditions include having leading 1s (pivot elements) in each row that are to the right of the leading 1 in the row above, and rows filled with zeros must be at the bottom. This form makes it easier to back-substitute and find solutions for the system.

For the given system, transforming the matrix into its row echelon form allowed us to pinpoint basic and free variables. The free variables then led us to the vectors forming the basis. Such matrix manipulation is fundamental in linear algebra and aids in understanding the solution structure of complex systems.

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Most popular questions from this chapter

Determine whether the nonhomogeneous system \(A x=b\) is consistent. If it is, write the solution in the form \(\mathbf{x}=\mathbf{x}_{p}+\mathbf{x}_{\mathbf{k}},\) where \(\mathbf{x}_{\mathbf{p}}\) is a particular solution of \(\mathbf{A} \mathbf{x}=\mathbf{b}\) and \(x_{k}\) is a solution of \(A x=0\) $$ \begin{array}{rr} 5 x_{1}-4 x_{2}+12 x_{3}-33 x_{4}+14 x_{5}= & -4 \\ -2 x_{1}+x_{2}-6 x_{3}+12 x_{4}-8 x_{5}= & 1 \\ 2 x_{1}-x_{2}+6 x_{3}-12 x_{4}+8 x_{5}= & -1 \end{array} $$

Pendulum Consider a pendulum of length \(L\) that swings by the force of gravity only. For small values of \(\theta=\theta(t),\) the motion of the pendulum can be approximated by the differential equation $$ \frac{d^{2} \theta}{d t^{2}}+\frac{g}{L} \theta=0 $$ where \(g\) is the acceleration due to gravity. (a) Verify that $$ \\{\sin \sqrt{\frac{g}{L}} t, \cos \sqrt{\frac{g}{L}} t\\} $$ is a set of linearly independent solutions of the differential equation. (b) Find the general solution of the differential equation and show that it can be written in the form $$ \theta(t)=A \cos [\sqrt{\frac{g}{L}}(t+\phi)] $$

Identify and sketch the graph of the conic section. $$ 4 x^{2}+y^{2}-8 x+3=0 $$

Determine whether the nonhomogeneous system \(A x=b\) is consistent. If it is, write the solution in the form \(\mathbf{x}=\mathbf{x}_{p}+\mathbf{x}_{\mathbf{k}},\) where \(\mathbf{x}_{\mathbf{p}}\) is a particular solution of \(\mathbf{A} \mathbf{x}=\mathbf{b}\) and \(x_{k}\) is a solution of \(A x=0\) $$ \begin{array}{rr} 2 x-4 y+5 z= & 8 \\ -7 x+14 y+4 z= & -28 \\ 3 x-6 y+z= & 12 \end{array} $$

Use your school's library, the Internet, or some other reference source to find real-life applications of (a) linear differential equations and (b) rotation of conic sections that are different than those discussed in this section.

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