/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 18 The equation \(x_{1}+x_{2}+x_{3}... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 18

The equation \(x_{1}+x_{2}+x_{3}=1\) can be viewed as a linear system of one equation in three unknowns. Express its general solution as a particular solution plus the general solution of the corresponding homogeneous system. [Suggestion: Write the vectors in column form.]

Short Answer

Expert verified
General solution: \(\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} + x_2 \begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix} + x_3 \begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix}\).

Step by step solution

01

Interpret the Linear Equation

Consider the linear equation in three variables: \(x_1 + x_2 + x_3 = 1\). This equation represents a plane in three-dimensional space.
02

Rearrange the Variables

Express one variable in terms of the others. Let's write \(x_1 = 1 - x_2 - x_3\). Now, the equation becomes dependent on our choices for \(x_2\) and \(x_3\).
03

Determine a Particular Solution

Choose specific values for \(x_2\) and \(x_3\) to find a particular solution. A simple choice is \(x_2 = 0, x_3 = 0\), which gives us \(x_1 = 1\). Thus, one particular solution is \(\begin{pmatrix} 1 \ 0 \ 0 \end{pmatrix}\).
04

Write the General Solution of the Homogeneous System

Consider the homogeneous equation \(x_1 + x_2 + x_3 = 0\). This represents a plane through the origin. We can express the solution as a linear combination of vectors: \(\begin{pmatrix} -x_2 - x_3 \ x_2 \ x_3 \end{pmatrix}\), which is the general solution of the homogeneous equation.
05

Present the General Solution as a Sum

Combine the particular solution and a general solution to the homogeneous equation to express the general solution to the original equation. The general solution is given by:\[\begin{pmatrix} 1 \ 0 \ 0 \end{pmatrix} + x_2 \begin{pmatrix} -1 \ 1 \ 0 \end{pmatrix} + x_3 \begin{pmatrix} -1 \ 0 \ 1 \end{pmatrix}.\]This represents all possible solutions to \(x_1 + x_2 + x_3 = 1\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Linear Equations
Linear equations are fundamental in linear algebra. They are equations where each term is either a constant or the product of a constant and a single variable, summing up to a total. In the given exercise, we have the linear equation \(x_1 + x_2 + x_3 = 1\). This equation involves three unknowns: \(x_1\), \(x_2\), and \(x_3\). Linear equations can represent various geometric entities; in this case, a plane in a three-dimensional coordinate system.
Breaking it down further:
  • The coefficients (in this case, all 1s) determine the angle the plane forms with the axes.
  • The constant term (1 in our example) shifts the plane away from the origin.
  • When solving, rearranging the equation helps express one variable in terms of others, making it easier to identify particular and general solutions.
If this constant was zero, the plane would pass through the origin, forming a homogeneous system that we'll discuss soon.
Particular Solution
A particular solution is a specific set of values for the variables that satisfies the linear equation. For the equation \(x_1 + x_2 + x_3 = 1\), we need to pick values for two of the unknowns to solve for the third. Doing so gives us a particular instance, not the entire solution set. In our problem:
  • Choosing \(x_2 = 0\) and \(x_3 = 0\) allows us to solve for \(x_1\), yielding \(x_1 = 1\).
  • This results in the particular solution vector \(\begin{pmatrix} 1 \ 0 \ 0 \end{pmatrix}\).
This particular solution is a single point on the infinite plane defined by our equation. When adding it to solutions of the homogeneous system, it helps describe the entire set of solutions to the original equation.
Homogeneous System
A homogeneous system is a system of linear equations where all the constants are zero. In our case, the homogeneous counterpart is \(x_1 + x_2 + x_3 = 0\). Such systems always have at least one solution, the trivial solution, where all variables are zero, but they can have infinitely many solutions due to the presence of free variables.
Understanding the general solution for a homogeneous system:
  • The solution generates a vector space, often represented as a linear combination of vectors.
  • For the equation \(x_1 + x_2 + x_3 = 0\), let \(x_2\) and \(x_3\) be free variables. This gives a solution in the form \(\begin{pmatrix} -x_2 - x_3 \ x_2 \ x_3 \end{pmatrix}\).
  • These solutions create a plane through the origin and encompass all vectors that, when added to the particular solution, describe every solution to the original problem.
Combining the general solution of this homogeneous system with a particular solution solves the original equation, providing a complete overview of all possible solutions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Suppose that \(A\) is a \(3 \times 3\) matrix whose null space is a line through the origin in 3 -space. Can the row or column space of \(A\) also be a line through the origin? Explain.

In each part, determine whether \(T\) is a matrix transformation. (a) \(T(x, y, z)=(0,0)\) (b) \(T(x, y, z)=(1,1)\) (c) \(T(x, y, z)=(3 x-4 y, 2 x-5 z)\) (d) \(T(x, y, z)=\left\langle y^{2}, z\right\rangle\) (e) \(T(x, y, z)=(y-1, x)\)

(a) If \(P\) is a regular \(n \times n\) stochastic matrix with steady-state vector \(\mathbf{q},\) and if \(\mathbf{e}_{1}, \mathbf{e}_{2}, \ldots, \mathbf{e}_{n}\) are the standard unit vectors in column form, what can you say about the behavior of the sequence $$P \mathbf{e}_{i}, \quad P^{2} \mathbf{e}_{i}, \quad P^{3} \mathbf{e}_{i}, \ldots, \quad P^{k} \mathbf{e}_{i}, \ldots$$ as \(k \rightarrow \infty\) for each \(i=1,2, \ldots, n ?\) (b) What does this tell you about the behavior of the column vectors of \(P^{k}\) as \(k \rightarrow \infty ?\)

Find a \(3 \times 3\) matrix whose null space is (a) a point. (b) a line. (c) a plane.

Find a basis for the null space of \(A\) (a) \(A=\left[\begin{array}{rrr}1 & -1 & 3 \\ 5 & -4 & -4 \\ 7 & -6 & 2\end{array}\right]\) (b) \(A=\left[\begin{array}{rrr}2 & 0 & -1 \\ 4 & 0 & -2 \\ 0 & 0 & 0\end{array}\right]\) (c)\(A=\left[\begin{array}{rrrr}1 & 4 & 5 & 2 \\ 2 & 1 & 3 & 0 \\ -1 & 3 & 2 & 2\end{array}\right]\) (d) \(A=\left[\begin{array}{rrrrr}1 & 4 & 5 & 6 & 9 \\ 3 & -2 & 1 & 4 & -1 \\\ -1 & 0 & -1 & -2 & -1 \\ 2 & 3 & 5 & 7 & 8\end{array}\right]\) (e) \(A=\left[\begin{array}{rrrrr}1 & -3 & 2 & 2 & 1 \\ 0 & 3 & 6 & 0 & -3 \\\ 2 & -3 & -2 & 4 & 4 \\ 3 & -6 & 0 & 6 & 5 \\ -2 & 9 & 2 & -4 & -5\end{array}\right]\)
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Mechanics Maths

Read Explanation

Applied Mathematics

Read Explanation

Statistics

Read Explanation

Probability and Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.