Classifying And Solving Systems Of Equations A Detailed Example

Jul 15, 2025 by qnaftunila 64 views

Problem 1 Classifying Systems of Equations and Determining Solutions

Introduction

In the realm of linear algebra, systems of equations play a pivotal role in modeling and solving a wide array of problems across various disciplines, ranging from engineering and physics to economics and computer science. Understanding the nature of solutions to these systems is crucial for effectively applying them in practical scenarios. Systems of equations can be classified into three distinct categories based on the existence and uniqueness of their solutions: (i) systems with a unique solution, (ii) inconsistent systems with no solutions, and (iii) dependent systems with infinitely many solutions. This article delves into the classification of systems of equations and provides a comprehensive guide on determining solutions where they exist. We will explore the underlying principles that govern the behavior of these systems and illustrate the concepts with concrete examples, empowering readers to confidently analyze and solve linear systems of equations.

The classification of systems of equations hinges on the relationships between the equations within the system. A system with a unique solution implies that the equations intersect at a single point, representing a specific set of values that satisfy all equations simultaneously. In contrast, an inconsistent system signifies that the equations do not intersect at any point, indicating the absence of a solution that satisfies all equations. Dependent systems, on the other hand, exhibit equations that are scalar multiples of each other, resulting in an infinite number of solutions that lie along a common line or plane. Identifying the category of a system is the first step towards determining its solution set, guiding the choice of appropriate solution techniques.

Determining solutions to systems of equations involves employing various algebraic methods, including Gaussian elimination, matrix inversion, and Cramer's rule. Gaussian elimination is a systematic approach that transforms the system into an equivalent triangular form, allowing for easy back-substitution to obtain the solution. Matrix inversion, applicable to systems with a unique solution, involves finding the inverse of the coefficient matrix and multiplying it with the constant vector to determine the solution vector. Cramer's rule, another technique for systems with a unique solution, utilizes determinants to directly calculate the values of the variables. The selection of a particular method depends on the specific characteristics of the system, such as the number of equations and variables, the coefficients involved, and the desired level of computational efficiency.

(a) Classifying and Solving the System of Equations

Let's consider the following system of equations:

2x₁ - x₂ + 2x₃ = 4
-2x₁ + 2x₂ - 5x₃ = 6
-x₁ + 3x₂ - 7x₃ = 1

To classify this system, we can employ techniques from linear algebra, such as Gaussian elimination or matrix analysis. Our primary goal is to determine whether the system has a unique solution, is inconsistent (no solutions), or is dependent (infinitely many solutions). We can represent this system in matrix form as follows:

Ax = b

where

A = | 2 -1  2 |
    |-2  2 -5 |
    |-1  3 -7 |

x = | x₁ |
    | x₂ |
    | x₃ |

b = | 4 |
    | 6 |
    | 1 |

Gaussian Elimination

Gaussian elimination is a powerful method for solving systems of linear equations. It involves transforming the augmented matrix [A|b] into row-echelon form or reduced row-echelon form through a series of elementary row operations. These operations include:

Swapping two rows.
Multiplying a row by a non-zero constant.
Adding a multiple of one row to another row.

Applying Gaussian elimination to our system, we start by writing the augmented matrix:

[A|b] = | 2 -1  2 | 4 |
        |-2  2 -5 | 6 |
        |-1  3 -7 | 1 |

First, we aim to eliminate the x₁ term in the second and third rows. To do this, we can add the first row to the second row and add half of the first row to the third row:

| 2 -1  2 |  4 |
| 0  1 -3 | 10 |
| 0 5/2 -6 |  3 |

Next, we want to eliminate the x₂ term in the third row. We can achieve this by subtracting 5/2 times the second row from the third row:

| 2 -1  2 |   4 |
| 0  1 -3 |  10 |
| 0  0 9/2 | -22 |

Now, the matrix is in row-echelon form. We can observe that there are no inconsistencies (e.g., a row of the form [0 0 0 | c] where c is non-zero), and we have a pivot (non-zero entry) in each column corresponding to a variable. This indicates that the system has a unique solution.

Back-Substitution

To find the solution, we perform back-substitution. From the last row, we have:

(9/2)x₃ = -22
x₃ = -44/9

Substituting x₃ into the second row, we get:

x₂ - 3x₃ = 10
x₂ - 3(-44/9) = 10
x₂ + 44/3 = 10
x₂ = 10 - 44/3
x₂ = -14/3

Finally, substituting x₂ and x₃ into the first row:

2x₁ - x₂ + 2x₃ = 4
2x₁ - (-14/3) + 2(-44/9) = 4
2x₁ + 14/3 - 88/9 = 4
2x₁ = 4 - 14/3 + 88/9
2x₁ = (36 - 42 + 88)/9
2x₁ = 82/9
x₁ = 41/9

Therefore, the unique solution to the system is:

x₁ = 41/9
x₂ = -14/3
x₃ = -44/9

Conclusion

After performing Gaussian elimination and back-substitution, we have determined that the given system of equations has a unique solution. The solution is x₁ = 41/9, x₂ = -14/3, and x₃ = -44/9. This demonstrates how Gaussian elimination can be used to classify a system and find its solution when it exists. Understanding the principles of Gaussian elimination and back-substitution is essential for solving linear systems of equations efficiently and accurately. The ability to classify systems based on their solution types allows for a more strategic approach to problem-solving, saving time and effort by avoiding unnecessary computations for inconsistent systems. In summary, the system in part (a) is classified as having a unique solution, and we have successfully found that solution using Gaussian elimination and back-substitution.

Repair Input Keyword

Classify each of the following systems of equations according to whether they (i) have a unique solution, (ii) are inconsistent, or (iii) are dependent. Determine the solutions where they exist for the following system:

2x₁ - x₂ + 2x₃ = 4
-2x₁ + 2x₂ - 5x₃ = 6
-x₁ + 3x₂ - 7x₃ = 1

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Classifying Systems of Equations and Finding Solutions A Detailed Guide with Examples