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If a consistent system has more vari- 0 ables than equations, show that it has infinitely many solutions.

Short Answer

Expert verified
An underdetermined, consistent system with more variables than equations has infinitely many solutions.

Step by step solution

01

Understand the Problem

We are given a system of linear equations with fewer equations than variables, which implies it's an underdetermined system. We need to show that this system has infinitely many solutions.
02

Setup the Problem

Consider a system with \( m \) equations and \( n \) variables, where \( m < n \). Such a system is represented by the matrix equation \( A\mathbf{x} = \mathbf{b} \), where \( A \) is an \( m \times n \) matrix, \( \mathbf{x} \) is a vector of variables, and \( \mathbf{b} \) is a vector of constants.
03

Analyze the Consistency

The problem states the system is consistent, meaning there is at least one solution to \( A\mathbf{x} = \mathbf{b} \). This means there are values for \( \mathbf{x} \) that satisfy the equations.
04

Consider the Rank of the Matrix

The rank of a matrix \( A \) (denoted as \( ext{rank}(A) \)) provides the maximum number of linearly independent rows or columns. For a consistent system where \( m < n \), \( ext{rank}(A) \leq m \leq n-1 \). This means the nullity \( (n - ext{rank}(A)) \) is greater than zero.
05

Use the Rank-Nullity Theorem

The rank-nullity theorem states that for a matrix \( A \) of size \( m \times n \), \( ext{rank}(A) + ext{nullity}(A) = n \). Given \( ext{rank}(A) \leq m < n \), it follows that \( ext{nullity}(A) > 0 \), indicating there are free variables.
06

Interpret the Free Variables

The presence of free variables means there are infinitely many degrees of freedom in selecting the values for the variables in \( \mathbf{x} \). For each choice of these free variables, there exists a corresponding solution to the system.
07

Conclude with Infinite Solutions

Since each choice of values for the free variables results in a valid solution, and there are infinitely many ways to assign values to these free variables, the system has infinitely many solutions.

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Key Concepts

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

Linear Equations
Linear equations form the foundation of linear algebra. These equations involve variables with no exponents and have the general form: \(a_1x_1 + a_2x_2 + ext{...} + a_nx_n = b\). In our context, these equations make up a system that we wish to solve. A consistent system of linear equations means at least one set of variable values satisfies all equations.

For linear equations, solutions can be found using various methods such as
  • Substitution
  • Elimination
  • Matrix operations
When a system has more variables than equations, it typically results in the system being underdetermined, meaning there are infinitely many solutions. This is because there aren't enough constraints to pin down a unique solution.

Instead of determining a single point in the solution space, the system allows a whole range (or line, plane, or even higher-dimensional space) of solutions.
Matrix Theory
Matrix theory is indispensable in solving multiple linear equations simultaneously. A matrix is a rectangular arrangement of numbers into rows and columns, used to represent a system of linear equations. In our exercise, the system can be written in the form \(A\textbf{x} = \textbf{b}\), where

  • \(A\) is an \(m \times n\) matrix representing the coefficients of the linear equations,
  • \(\textbf{x}\) is a vector of unknowns, and
  • \(\textbf{b}\) represents the constant terms.
Matrix operations, such as row reduction, help transform these systems into simpler forms, making it easier to derive solutions. When the number of equations \(m\) is less than the number of unknowns \(n\), the system is underdetermined. Here, matrix theory provides tools such as the rank of \(A\) to determine properties like consistency and solutions.

Identifying free and pivot variables through Gaussian elimination can lead to sets of solutions, which exhibit dependencies based on the matrix structure.
Rank-Nullity Theorem
The rank-nullity theorem is a key principle in linear algebra. For any matrix \(A\) with dimensions \(m \times n\), the theorem states:

\[\text{rank}(A) + \text{nullity}(A) = n\] This means that the sum of the rank (the number of linearly independent rows or columns) and the nullity (the number of solutions in the solution set) equals the number of columns.

The rank tells us how many constraints the system imposes, while the nullity gives the number of free variables—variables that can take any value while still satisfying the system. For systems where the rank is less than the number of variables (\(n\)), the nullity is positive (
meaning free variables are present). These free variables enable us to find infinitely many solutions.

Thus, the theorem provides a direct method to understand the nature of the solutions to a system of equations. It's particularly important in illuminating the relationship between the number of equations, the rank, and the dimensionality of the solution space.

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Most popular questions from this chapter

Three Nissans, two Fords, and four Chevrolets can be rented for $$\$ 106$$ per day. At the same rates two Nissans, four Fords, and three Chevrolets cost $$\$ 107$$ per day, whereas four Nissans, three Fords, and two Chevrolets cost $$\$ 102$$ per day. Find the rental rates for all three kinds of cars.

In each of the following, find (if possible) conditions on \(a\) and \(b\) such that the system has no solution, one solution, and infinitely many solutions. a. \(\begin{aligned} & x-2 y=1 \\ & a x+b y=5 \end{aligned}\) \(\begin{aligned} \text { b. } & x+b y=-1 \\ & a x+2 y=5 \end{aligned}\) c. \(\quad x-b y=-1\) \(x+a y=3\) d. \(\quad a x+y=1\) \(2 x+y=b\)

Write a system of linear equations that has each of the following augmented matrices. $$ \text { a. }\left[\begin{array}{rrr|r} 1 & -1 & 6 & 0 \\ 0 & 1 & 0 & 3 \\ 2 & -1 & 0 & 1 \end{array}\right] \quad \text { b. }\left[\begin{array}{rrr|r} 2 & -1 & 0 & -1 \\ -3 & 2 & 1 & 0 \\ 0 & 1 & 1 & 3 \end{array}\right] $$

In each case either show that the statement is true, or give an example \(^{2}\) showing it is false. a. If a linear system has \(n\) variables and \(m\) equations, then the augmented matrix has \(n\) rows. b. A consistent linear system must have infinitely many solutions. c. If a row operation is done to a consistent linear system, the resulting system must be consistent. d. If a series of row operations on a linear system results in an inconsistent system, the original system is inconsistent.

Show that a system consisting of exactly one linear equation can have no solution, one solution, or infinitely many solutions. Give examples.

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