/*! 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 14 Determine whether the homogeneou... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 14

Determine whether the homogeneous system has nontrivial solutions by inspection (without pencil and paper). $$\begin{aligned}x_{1}+3 x_{2}-x_{3} &=0 \\\x_{2}-8 x_{3} &=0 \\\4 x_{3} &=0\end{aligned}$$

Short Answer

Expert verified
The system has no nontrivial solutions; all variables must be zero.

Step by step solution

01

Recognize the Homogeneous System

A homogeneous system of linear equations is of the form \(A\mathbf{x} = \mathbf{0}\), where \(A\) is a matrix and \(\mathbf{x}\) is the vector of variables. The solution \(\mathbf{0}\) is always a solution, but we seek nontrivial solutions, where at least one variable is nonzero.
02

Analyze the Third Equation

Examine the third equation: \(4x_3 = 0\). By inspection, this equation implies \(x_3 = 0\). Since there is no other term in this equation, \(x_3\) must be zero in any solution.
03

Substitute \(x_3 = 0\) into Second Equation

Substitute \(x_3 = 0\) into the second equation: \(x_2 - 8x_3 = 0\). This simplifies to \(x_2 = 0\). Therefore, \(x_2\) must also be zero in any solution.
04

Substitute \(x_2 = 0\) and \(x_3 = 0\) into First Equation

Substitute both \(x_2 = 0\) and \(x_3 = 0\) into the first equation: \(x_1 + 3x_2 - x_3 = 0\). This simplifies to \(x_1 = 0\). Therefore, \(x_1\) must be zero as well.
05

Conclude the Existence of Solutions

Since \(x_1 = 0\), \(x_2 = 0\), and \(x_3 = 0\), the only solution to the system is the trivial solution \((0, 0, 0)\). Therefore, the system does not have nontrivial solutions.

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.

Matrix Equation
A matrix equation is a way of representing a system of linear equations in a compact form. Instead of writing each equation separately, you use a matrix to organize the coefficients and a vector to represent the variables. For example, a system like the one in the original exercise can be rewritten as:
  • The coefficients of the variables form a matrix \(A\).
  • The variables \(x_1, x_2, x_3\) are elements of a column vector \(\mathbf{x}\).
  • The equation is expressed as \(A\mathbf{x} = \mathbf{0}\).
This method simplifies the process of solving systems because the properties of matrices can be applied. It allows for more systematic approaches, especially when dealing with larger systems.
Trivial Solution
In the context of homogeneous systems, the trivial solution is a term that refers to one specific solution where all variables are zero. It is always a solution to any homogeneous equation because substituting zeros into any equation yields zero, which satisfies the equation:
  • For the system \(A\mathbf{x} = \mathbf{0}\), the trivial solution is \(\mathbf{x} = \mathbf{0}\).
  • It is called 'trivial' because it doesn't provide any new information鈥攊t merely states the obvious.
While trivial solutions are always present, our main interest usually lies in finding nontrivial solutions, which are less straightforward and more informative.
Nontrivial Solution
A nontrivial solution in a homogeneous system is a solution where at least one variable is not zero. Finding a nontrivial solution means identifying a set of values that still satisfy the matrix equation without resulting in all zero variables:
  • Nontrivial solutions exist only when the determinant of the coefficient matrix \(A\) is zero.
  • It implies that the system has an infinite number of solutions due to dependence among the equations.
In the given exercise, we found that all the equations simplify to requiring all variables to equal zero, indicating that no nontrivial solutions exist in this particular case.
Linear Equations Analysis
Analyzing a system of linear equations involves several steps to determine possible solutions. The analysis includes:
  • Recognizing the form of the system (e.g., homogeneous).
  • Inspecting individual equations to understand how they constrain variables. For example, visually checking simpler equations (like those ending in zero) helps quickly identify constraints.
  • Substituting known values from one equation to others progressively reduces unknowns.
  • Determining if the final system suggests only a trivial solution or if there's room for nontrivial ones.
The exercise showed how logical deductions from each equation led directly to concluding no nontrivial solutions exist without lengthy calculations.

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

In each part, find a \(6 \times 6\) matrix \(\left[a_{i j}\right]\) that satisfies the stated condition. Make your answers as general as possible by using letters rather than specific numbers for the nonzero entries. (a) \(a_{i j}=0\) if \(i \neq j\) (b) \(a_{i j}=0\) if \(i>j\) (c) \(a_{i j}=0\) if \(i1\) Answer: A. $$\left[\begin{array}{cccccc} a_{11} & 0 & 0 & 0 & 0 & 0 \\ 0 & a_{22} & 0 & 0 & 0 & 0 \\ 0 & 0 & a_{33} & 0 & 0 & 0 \\ 0 & 0 & 0 & a_{44} & 0 & 0 \\ 0 & 0 & 0 & 0 & a_{55} & 0 \\ 0 & 0 & 0 & 0 & 0 & a_{66} \end{array}\right]$$ B. $$\left[\begin{array}{cccccc} a_{11} & a_{12} & a_{13} & a_{14} & a_{15} & a_{16} \\ 0 & a_{22} & a_{23} & a_{24} & a_{25} & a_{26} \\ 0 & 0 & a_{33} & a_{34} & a_{35} & a_{36} \\ 0 & 0 & 0 & a_{44} & a_{45} & a_{46} \\ 0 & 0 & 0 & 0 & a_{55} & a_{56} \\ 0 & 0 & 0 & 0 & 0 & a_{66} \end{array}\right]$$ C. $$\left[\begin{array}{llllll} a_{11} & 0 & 0 & 0 & 0 & 0 \\ a_{21} & a_{22} & 0 & 0 & 0 & 0 \\ a_{31} & a_{32} & a_{33} & 0 & 0 & 0 \\ a_{41} & a_{42} & a_{43} & a_{44} & 0 & 0 \\ a_{51} & a_{52} & a_{53} & a_{54} & a_{55} & 0 \\ a_{61} & a_{62} & a_{63} & a_{64} & a_{65} & a_{66} \end{array}\right]$$ D. $$\left[\begin{array}{cccccc} a_{11} & a_{12} & 0 & 0 & 0 & 0 \\ a_{21} & a_{22} & a_{23} & 0 & 0 & 0 \\ 0 & a_{32} & a_{33} & a_{34} & 0 & 0 \\ 0 & 0 & a_{43} & a_{44} & a_{45} & 0 \\ 0 & 0 & 0 & a_{54} & a_{55} & a_{56} \\ 0 & 0 & 0 & 0 & a_{65} & a_{66} \end{array}\right]$$

A matrix \(B\) is said to be a square root of a matrix \(A\) if \(B B=A\) (a) Find two square roots of \(A=\left[\begin{array}{ll}2 & 2 \\ 2 & 2\end{array}\right]\) (b) How many different square roots can you find of \(A=\left[\begin{array}{ll}5 & 0 \\ 0 & 9\end{array}\right] ?\) (c) Do you think that every \(2 \times 2\) matrix has at least one square root? Explain your reasoning. Answer: A.$$\left[\begin{array}{ll} 1 & 1 \\ 1 & 1 \end{array}\right] \text { and }\left[\begin{array}{cc} -1 & -1 \\ -1 & -1 \end{array}\right]$$ B. $$\text { Four; }\left[\begin{array}{cc} \sqrt{5} & 0 \\ 0 & 3 \end{array}\right],\left[\begin{array}{cc} -\sqrt{5} & 0 \\ 0 & 3 \end{array}\right],\left[\begin{array}{cc} \sqrt{5} & 0 \\ 0 & -3 \end{array}\right],\left[\begin{array}{cc} -\sqrt{5} & 0 \\ 0 & -3 \end{array}\right]$$

Write the given matrix as a product of elementary matrices. $$\left[\begin{array}{lll} 1 & 1 & 0 \\ 1 & 1 & 1 \\ 0 & 1 & 1 \end{array}\right]$$

Solve the given homogeneous linear system by any method. $$\begin{aligned}&3 x_{1}+x_{2}+x_{3}+x_{4}=0\\\&5 x_{1}-x_{2}+x_{3}-x_{4}=0 \end{aligned}$$

Decide by inspection whether the given matrix is invertible. $$\left[\begin{array}{rrrr} 0 & 1 & -2 & 5 \\ 0 & 1 & 5 & 6 \\ 0 & 0 & -3 & 1 \\ 0 & 0 & 0 & 5 \end{array}\right]$$
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Statistics

Read Explanation

Decision Maths

Read Explanation

Calculus

Read Explanation

Logic and Functions

Read Explanation

Probability and Statistics

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.