Solving Linear Equations A Step By Step Guide

Introduction to Linear Systems

In the realm of mathematics, particularly in linear algebra, systems of linear equations play a pivotal role. These systems arise in diverse applications, ranging from physics and engineering to computer science and economics. Understanding how to solve these systems is therefore crucial for anyone working in these fields. This article delves into the intricacies of solving a specific system of linear equations, providing a step-by-step guide and offering insights into the underlying principles. This involves matrix representation, Gaussian elimination, and the interpretation of solutions. Linear equations are at the heart of many mathematical models, representing relationships between variables in a straightforward manner. A system of linear equations, then, is a collection of such equations, all considered simultaneously. The goal is often to find values for the variables that satisfy all equations in the system. Such systems can have one solution, no solutions, or infinitely many solutions, depending on the relationships between the equations. The matrix representation of a linear system is a compact and efficient way to represent the equations. It allows us to use matrix operations to solve the system, making the process more streamlined. Gaussian elimination is a fundamental algorithm for solving linear systems represented in matrix form. It involves transforming the matrix into an upper triangular form, which can then be easily solved using back-substitution. The nature of the solutions to a linear system—whether it's unique, non-existent, or infinite—depends on the properties of the coefficient matrix and the constant vector. These properties can be analyzed to gain insights into the system's behavior. In practical applications, systems of linear equations can model a wide range of phenomena, from electrical circuits to economic models. Therefore, understanding how to solve these systems is not just a theoretical exercise but a practical skill with broad applicability. Throughout this article, we will explore these concepts in detail, focusing on a specific example to illustrate the methods and principles involved. By the end of this discussion, you will have a solid understanding of how to approach and solve linear systems, and you'll appreciate the power and versatility of linear algebra in solving real-world problems.

Problem Statement: Defining the System

The problem at hand involves a system of linear equations expressed in matrix form. We are given the matrix equation:

$$\left[\begin{array}{ccc}-1 & -1 & -1 \\ -1 & -2 & 0 \\ -1 & 0 & -2\end{array}\right]\left[\begin{array}{l}v_1 \\ v_2 \\ v_3\end{array}\right]=\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]$$

This equation represents a homogeneous system of linear equations, where the right-hand side is the zero vector. The coefficient matrix is a 3x3 matrix, and the vector extbf{[v1, v2, v3]} represents the unknowns we aim to find. The zero vector on the right-hand side indicates that we are looking for non-trivial solutions, meaning solutions other than the trivial solution where v1 = v2 = v3 = 0. To solve this system, we will employ methods from linear algebra, specifically Gaussian elimination or row reduction, to transform the coefficient matrix into a simpler form. This will allow us to determine the relationships between the variables and identify the solution space. The coefficient matrix, in this case, is a square matrix, which means we can also consider its determinant and eigenvalues to gain further insights into the system's properties. The fact that the system is homogeneous simplifies the solution process somewhat, as we don't need to worry about inconsistencies that can arise in non-homogeneous systems. However, it also means that we need to be careful to identify the correct number of free variables and express the solution in terms of these free variables. The problem statement is clear and well-defined, providing us with a specific system of equations to solve. The matrix representation is standard and allows us to apply a variety of techniques to find the solution. In the following sections, we will discuss the steps involved in solving this system, starting with Gaussian elimination and row reduction.

Solving the System: Gaussian Elimination

To solve the given system of linear equations, we will utilize Gaussian elimination, a fundamental technique in linear algebra. Gaussian elimination involves transforming the coefficient matrix into an upper triangular form through a series of elementary row operations. These operations include swapping rows, multiplying a row by a non-zero scalar, and adding a multiple of one row to another. The goal is to systematically eliminate variables from the equations until the system is in a form that can be easily solved using back-substitution. Let's start with the given matrix:

$$\left[\begin{array}{ccc}-1 & -1 & -1 \\ -1 & -2 & 0 \\ -1 & 0 & -2\end{array}\right]$$

We can begin by multiplying the first row by -1 to make the leading entry positive:

$$\left[\begin{array}{ccc}1 & 1 & 1 \\ -1 & -2 & 0 \\ -1 & 0 & -2\end{array}\right]$$

Next, we add the first row to the second row and the first row to the third row to eliminate the -1 entries in the first column:

$$\left[\begin{array}{ccc}1 & 1 & 1 \\ 0 & -1 & 1 \\ 0 & 1 & -1\end{array}\right]$$

Now, we multiply the second row by -1 to make the leading entry positive:

$$\left[\begin{array}{ccc}1 & 1 & 1 \\ 0 & 1 & -1 \\ 0 & 1 & -1\end{array}\right]$$

Finally, we subtract the second row from the third row:

$$\left[\begin{array}{ccc}1 & 1 & 1 \\ 0 & 1 & -1 \\ 0 & 0 & 0\end{array}\right]$$

This is the row-echelon form of the matrix. We can see that the third row is all zeros, which indicates that the system has infinitely many solutions or a non-trivial solution. The matrix now corresponds to the following system of equations:

$$v_1 + v_2 + v_3 = 0 \\ v_2 - v_3 = 0$$

This simplified system allows us to express the solutions in terms of a free variable, which we will discuss in the next section.

Solution Space: Free Variables and General Solution

After performing Gaussian elimination, we arrived at the following system of equations:

$$v_1 + v_2 + v_3 = 0 \\ v_2 - v_3 = 0$$

From the second equation, we can express v2 in terms of v3:

$$v_2 = v_3$$

Substituting this into the first equation, we get:

$$v_1 + v_3 + v_3 = 0 \\ v_1 = -2v_3$$

Here, v3 is a free variable, meaning it can take any value, and v1 and v2 are expressed in terms of v3. Let's denote v3 as t, where t is any real number. Then, the solution can be written as:

$$v_1 = -2t \\ v_2 = t \\ v_3 = t$$

In vector form, the general solution is:

$$\left[\begin{array}{l}v_1 \\ v_2 \\ v_3\end{array}\right] = t \left[\begin{array}{c}-2 \\ 1 \\ 1\end{array}\right]$$

This means that the solution space is a line in three-dimensional space, passing through the origin and spanned by the vector [-2, 1, 1]. The existence of a free variable indicates that the system has infinitely many solutions. This is because we have fewer independent equations than unknowns. The general solution provides a complete description of all possible solutions to the system. By varying the value of the parameter t, we can generate any solution in the solution space. The concept of free variables is crucial in understanding the solution structure of linear systems. When a system has free variables, it means that there are degrees of freedom in the solution, leading to an infinite number of possibilities. The vector [-2, 1, 1] is a basis for the solution space, meaning that any solution can be expressed as a scalar multiple of this vector. This geometric interpretation of the solution space provides valuable insights into the nature of the system. In summary, the solution to the given system of linear equations is a one-dimensional subspace of R3, spanned by the vector [-2, 1, 1]. This solution was obtained by systematically applying Gaussian elimination and identifying the free variable in the system.

Conclusion: Insights and Implications

In this article, we tackled the problem of solving a system of linear equations represented in matrix form. We employed Gaussian elimination to reduce the coefficient matrix to row-echelon form, which allowed us to identify the relationships between the variables. We found that the system had infinitely many solutions, characterized by a free variable. The general solution was expressed in vector form, revealing the solution space as a line in three-dimensional space. This exercise demonstrates the power of linear algebra in solving systems of equations and provides valuable insights into the structure of solution spaces. The key takeaway is that the number of solutions to a linear system is determined by the rank of the coefficient matrix and the number of unknowns. In this case, the rank of the matrix was less than the number of variables, leading to infinitely many solutions. The existence of a non-trivial solution to a homogeneous system implies that the determinant of the coefficient matrix is zero. This is a fundamental property of linear systems and is often used to determine the existence of non-trivial solutions. The methods used here, such as Gaussian elimination and the identification of free variables, are widely applicable to a variety of linear systems. They form the foundation for solving more complex problems in linear algebra and related fields. Furthermore, the geometric interpretation of the solution space as a line in R3 provides a visual understanding of the solutions. This geometric perspective is essential for gaining a deeper appreciation of linear systems and their properties. Understanding the concepts discussed in this article is crucial for anyone working with mathematical models that involve linear equations. From engineering to economics, systems of equations arise in numerous applications, making the ability to solve them a valuable skill. By mastering these techniques, one can gain a more profound understanding of the underlying mathematical principles and apply them to real-world problems effectively. In conclusion, solving systems of linear equations is a cornerstone of mathematics, with wide-ranging implications and applications. This article has provided a comprehensive guide to solving a specific system, highlighting the key concepts and techniques involved.