/*! 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 26 Suppose \(A \mathbf{x}=\mathbf{b... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Suppose \(A \mathbf{x}=\mathbf{b}\) has a solution. Explain why the solution is unique precisely when \(A \mathbf{x}=\mathbf{0}\) has only the trivial solution.

Short Answer

Expert verified
The solution is unique if \(A\mathbf{x} = \mathbf{0}\) only has the trivial solution \(\mathbf{x} = \mathbf{0}\), implying \(A\) is of full rank.

Step by step solution

01

Understanding Unique Solutions

A system of linear equations described by the matrix equation \(A\mathbf{x} = \mathbf{b}\) has a unique solution if and only if there is exactly one vector \(\mathbf{x}\) that satisfies this equation. This requires that the matrix \(A\) is invertible (non-singular), meaning its determinant is non-zero, and therefore its null space contains only the zero vector.
02

Analyzing Homogeneous System

The equation \(A\mathbf{x} = \mathbf{0}\) is a homogeneous system, which always has at least one solution: the trivial solution \(\mathbf{x} = \mathbf{0}\). For the solution to be unique for the non-homogeneous system, \(A\mathbf{x} = \mathbf{b}\), the homogeneous system must not have any other (non-trivial) solutions.
03

Relating Homogeneous and Non-Homogeneous Solutions

A unique solution occurs when the null space of matrix \(A\), i.e., the set of all \(\mathbf{x}\) satisfying \(A\mathbf{x} = \mathbf{0}\), contains only the trivial solution \(\mathbf{x} = \mathbf{0}\). If a non-trivial solution exists for \(A\mathbf{x} = \mathbf{0}\), then \(A\mathbf{x} = \mathbf{b}\) cannot have just one unique solution, because any scalar multiple of the non-trivial solution could also be added to \(\mathbf{x}\) serving as alternative solutions.
04

Conclusion

Therefore, the solution \(A\mathbf{x} = \mathbf{b}\) is unique precisely when the only solution to the system \(A\mathbf{x} = \mathbf{0}\) is the trivial solution \(\mathbf{x} = \mathbf{0}\). This means the matrix \(A\) must be full rank (its rank must equal the number of columns), indicating invertibility.

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 91Ó°ÊÓ!

Key Concepts

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

Homogeneous System
A homogeneous system of linear equations is one where the equation is set equal to zero, written as \( A\mathbf{x} = \mathbf{0} \).
Such systems always have at least the trivial solution, where \( \mathbf{x} = \mathbf{0} \).
This means every component of vector \( \mathbf{x} \) is zero, satisfying the equation naturally.

In the context of unique solutions, if a homogeneous system only admits this trivial solution,
  • there are no other vectors that satisfy \( A\mathbf{x} = \mathbf{0} \)
  • this absence of non-trivial solutions reflects back to the non-homogeneous system, \( A\mathbf{x} = \mathbf{b} \).
Therefore, if the homogeneous system lacks any non-trivial solution,
it ensures that \( A\mathbf{x} = \mathbf{b} \) has only one unique solution, as no other vectors satisfy the homogeneous system that could be added to potential solutions for the given \( \mathbf{b} \).
Matrix Invertibility
Matrix invertibility is a fundamental concept that determines whether a matrix has an inverse.
A matrix \( A \) is invertible—or non-singular—if it has a unique inverse;
an invertible matrix can "undo" the application of itself, akin to reversing an operation.
  • For \( A \) to be invertible, its determinant must be non-zero.
  • An invertible matrix means full rank, where the rank equals the number of its columns.

An invertible matrix implies that \( A\mathbf{x} = \mathbf{b} \) has precisely one solution, because the equation can be solved by applying the inverse: \( \mathbf{x} = A^{-1}\mathbf{b} \).
This characteristic directly relates to the unique solution condition for both non-homogeneous systems and homogeneous systems.

When the matrix \( A \) is invertible, the corresponding homogeneous system \( A\mathbf{x} = \mathbf{0} \) should only have the trivial solution.
Thus, matrix invertibility, unique solutions, and the nature of a homogeneous system's solutions are all interconnected.
Null Space
The null space of a matrix \( A \) consists of all the vector solutions \( \mathbf{x} \) to the equation \( A\mathbf{x} = \mathbf{0} \).
It provides critical insights into the behavior of a system.
For a system of equations, the dimension of the null space, also known as the nullity, indicates how many independent solutions exist.
  • If the null space contains only the zero vector, the nullity is zero, indicating a unique solution for the non-homogeneous equation \( A\mathbf{x} = \mathbf{b} \).
  • A non-zero nullity would suggest additional independent solutions, implying multiple possible solutions to \( A\mathbf{x} = \mathbf{b} \).
In simpler terms, the fewer solutions in the null space, the more likely the system has a unique solution.
An empty null space (except the trivial one) means \( A \) is invertible, ensuring a single outcome for any given \( \mathbf{b} \).
The null space serves as a bridge between matrix properties and the nature of solutions, ensuring a deeper understanding of linear system behaviors.

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

Each statement in Exercises 33–38 is either true (in all cases) or false (for at least one example). If false, construct a specific example to show that the statement is not always true. Such an example is called a counterexample to the statement. If a statement is true, give a justification. (One specific example cannot explain why a statement is always true. You will have to do more work here than in Exercises 21 and 22.) If \(\mathbf{v}_{1}\) and \(\mathbf{v}_{2}\) are in \(\mathbb{R}^{4}\) and \(\mathbf{v}_{2}\) is not a scalar multiple of \(\mathbf{v}_{1},\) then \(\left\\{\mathbf{v}_{1}, \mathbf{v}_{2}\right\\}\) is linearly independent.

[M] Budget® Rent A Car in Wichita, Kansas, has a fleet of about 500 cars, at three locations. A car rented at one location may be returned to any of the three locations. The various fractions of cars returned to the three locations are shown in the matrix below. Suppose that on Monday there are 295 cars at the airport (or rented from there), 55 cars at the east side office, and 150 cars at the west side office. What will be the approximate distribution of cars on Wednesday? \(\left[\begin{array}{ccc}{.97} & {.05} & {.10} \\ {.00} & {.90} & {.05} \\\ {.03} & {.05} & {.85}\end{array}\right]\)

In Exercises 29–32, find the elementary row operation that transforms the first matrix into the second, and then find the reverse row operation that transforms the second matrix into the first. $$ \left[\begin{array}{rrrr}{1} & {2} & {-5} & {0} \\ {0} & {1} & {-3} & {-2} \\\ {0} & {-3} & {9} & {5}\end{array}\right],\left[\begin{array}{rrrr}{1} & {2} & {-5} & {0} \\ {0} & {1} & {-3} & {-2} \\ {0} & {0} & {0} & {-1}\end{array}\right] $$

[M] Use as many columns of \(A\) as possible to construct a matrix \(B\) with the property that the equation \(B \mathbf{x}=\mathbf{0}\) has only the trivial solution. Solve \(B \mathbf{x}=\mathbf{0}\) to verify your work. \(A=\left[\begin{array}{rrrrrr}{12} & {10} & {-6} & {-3} & {7} & {10} \\ {-7} & {-6} & {4} & {7} & {-9} & {5} \\ {9} & {9} & {-9} & {-5} & {5} & {-1} \\\ {-4} & {-3} & {1} & {6} & {-8} & {9} \\ {8} & {7} & {-5} & {-9} & {11} & {-8}\end{array}\right]\)

Suppose an economy has only two sectors, Goods and Services. Each year, Goods sells 80\(\%\) of its output to Services and keeps the rest, while Services sells 70\(\%\) of its output to Goods and retains the rest. Find equilibrium prices for the annual outputs of the Goods and Services sectors that make each sector's income match its expenditures.

See all solutions

Recommended explanations on Math Textbooks

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.