Solving egin{bmatrix}1 & 1 & 1 \ 1 & 2 & 0 \ 1 & 0 & 2 \end{bmatrix}egin{bmatrix}v_1 \ V_2 \ V_3\end{bmatrix}=egin{bmatrix}0 \ 0 \ 0\end{bmatrix} A Comprehensive Guide

In the realm of linear algebra, solving systems of linear equations is a fundamental task. These systems appear in various applications, ranging from circuit analysis in electrical engineering to economic modeling and computer graphics. A system of linear equations can have a unique solution, infinitely many solutions, or no solution at all. The matrix representation of a system of equations provides a concise and powerful method for analyzing and solving these systems. This article delves into the process of solving a specific system of linear equations represented in matrix form, offering a step-by-step guide and exploring the underlying mathematical concepts.

This article aims to provide a detailed explanation of how to solve the given system of linear equations, which is represented in matrix form as:

$$\begin{bmatrix}1 & 1 & 1 \\ 1 & 2 & 0 \\ 1 & 0 & 2\end{bmatrix}\begin{bmatrix}v_1 \\ v_2 \\ v_3\end{bmatrix}=\begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix}$$

We will explore the methods for finding the solution vector egin{bmatrix}v_1 \ v_2 \ v_3\end{bmatrix} that satisfies this equation. The matrix equation represents a homogeneous system of linear equations, where the right-hand side is a zero vector. Understanding how to solve such systems is crucial in various fields of mathematics, physics, and engineering. The process involves techniques such as Gaussian elimination, finding the determinant, and analyzing the rank of the matrix. Through a detailed explanation and step-by-step calculations, this article will provide a comprehensive guide for solving this specific system and similar systems of linear equations.

Before diving into the solution, let's understand the components of the matrix equation. The equation is given by $Ax = 0$, where $A$ is the coefficient matrix, $x$ is the vector of unknowns, and $0$ is the zero vector. Specifically, the coefficient matrix $A$ is:

$$A = \begin{bmatrix}1 & 1 & 1 \\ 1 & 2 & 0 \\ 1 & 0 & 2\end{bmatrix}$$

The vector of unknowns $x$ is:

$$x = \begin{bmatrix}v_1 \\ v_2 \\ v_3\end{bmatrix}$$

And the zero vector is:

$$0 = \begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix}$$

Thus, the matrix equation represents the following system of linear equations:

$$\begin{cases} v_1 + v_2 + v_3 = 0 \\ v_1 + 2v_2 = 0 \\ v_1 + 2v_3 = 0 \end{cases}$$

The goal is to find the values of $v_1$, $v_2$, and $v_3$ that satisfy all three equations simultaneously. This system of equations is homogeneous because the right-hand side of each equation is zero. Homogeneous systems always have at least one solution, the trivial solution, where all variables are zero. However, we are interested in determining whether there are non-trivial solutions as well.

To determine if non-trivial solutions exist, we can analyze the determinant of the coefficient matrix $A$. If the determinant is non-zero, the system has only the trivial solution. If the determinant is zero, the system has infinitely many solutions. Additionally, we can use methods like Gaussian elimination to solve the system and find the specific solutions.

Understanding the structure of the matrix equation and the underlying system of linear equations is crucial for selecting the appropriate method to solve it. The nature of the system, whether it's homogeneous or non-homogeneous, and the properties of the coefficient matrix, such as its determinant and rank, provide valuable insights into the solution space. In the following sections, we will explore these concepts and apply them to solve the given system of equations.

There are several methods to solve a system of linear equations represented in matrix form. These methods include Gaussian elimination, finding the determinant, and using the concept of the rank of the matrix. Each method provides a different perspective on the system and its solutions. Let's explore each of these methods in detail to solve the given system.

1. Gaussian Elimination

Gaussian elimination is a method for solving systems of linear equations by transforming the augmented matrix into row-echelon form or reduced row-echelon form. The augmented matrix is formed by appending the column vector of constants (in this case, the zero vector) to the coefficient matrix. The goal is to perform elementary row operations to simplify the matrix and solve for the unknowns. The elementary row operations include:

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

For the given system, the augmented matrix is:

$$\begin{bmatrix}1 & 1 & 1 & | & 0 \\ 1 & 2 & 0 & | & 0 \\ 1 & 0 & 2 & | & 0\end{bmatrix}$$

First, we subtract the first row from the second and third rows to eliminate the first variable in the second and third equations:

$$R_2 \rightarrow R_2 - R_1, R_3 \rightarrow R_3 - R_1$$

$$\begin{bmatrix}1 & 1 & 1 & | & 0 \\ 0 & 1 & -1 & | & 0 \\ 0 & -1 & 1 & | & 0\end{bmatrix}$$

Next, we add the second row to the third row to eliminate the second variable in the third equation:

$$R_3 \rightarrow R_3 + R_2$$

$$\begin{bmatrix}1 & 1 & 1 & | & 0 \\ 0 & 1 & -1 & | & 0 \\ 0 & 0 & 0 & | & 0\end{bmatrix}$$

The matrix is now in row-echelon form. We can see that the third row is all zeros, which indicates that the system has infinitely many solutions. To find the solutions, we can express the variables in terms of a free variable. Let $v_3 = t$, where $t$ is a parameter. From the second row, we have:

$$v_2 - v_3 = 0 \Rightarrow v_2 = v_3 = t$$

From the first row, we have:

$$v_1 + v_2 + v_3 = 0 \Rightarrow v_1 = -v_2 - v_3 = -t - t = -2t$$

Thus, the solutions are given by:

$$\begin{bmatrix}v_1 \\ v_2 \\ v_3\end{bmatrix} = \begin{bmatrix}-2t \\ t \\ t\end{bmatrix} = t\begin{bmatrix}-2 \\ 1 \\ 1\end{bmatrix}$$

This shows that the system has infinitely many solutions, which are multiples of the vector $\begin{bmatrix}-2 \ 1 \ 1\end{bmatrix}$.

2. Determinant of the Matrix

The determinant of the coefficient matrix provides information about the uniqueness of the solution. If the determinant is non-zero, the system has a unique solution. If the determinant is zero, the system has either infinitely many solutions or no solution. For a homogeneous system, a zero determinant implies infinitely many solutions.

The determinant of matrix $A$ is calculated as:

$$det(A) = \begin{vmatrix}1 & 1 & 1 \\ 1 & 2 & 0 \\ 1 & 0 & 2\end{vmatrix}$$

Expanding the determinant along the first row:

$$det(A) = 1 \begin{vmatrix}2 & 0 \\ 0 & 2\end{vmatrix} - 1 \begin{vmatrix}1 & 0 \\ 1 & 2\end{vmatrix} + 1 \begin{vmatrix}1 & 2 \\ 1 & 0\end{vmatrix}$$

$$det(A) = 1(2 \cdot 2 - 0 \cdot 0) - 1(1 \cdot 2 - 0 \cdot 1) + 1(1 \cdot 0 - 2 \cdot 1)$$

$$det(A) = 1(4) - 1(2) + 1(-2) = 4 - 2 - 2 = 0$$

Since the determinant is zero, the system has infinitely many solutions.

3. Rank of the Matrix

The rank of a matrix is the maximum number of linearly independent rows or columns in the matrix. The rank can be determined by reducing the matrix to row-echelon form and counting the number of non-zero rows. The rank of the matrix is related to the number of solutions of the system.

For the given matrix $A$, we have already reduced it to row-echelon form during Gaussian elimination:

$$\begin{bmatrix}1 & 1 & 1 \\ 0 & 1 & -1 \\ 0 & 0 & 0\end{bmatrix}$$

There are two non-zero rows, so the rank of matrix $A$ is 2. The number of unknowns is 3. According to the rank-nullity theorem, the dimension of the null space (the space of solutions) is the number of unknowns minus the rank of the matrix:

$$Dimension of null space = Number of unknowns - Rank(A) = 3 - 2 = 1$$

This confirms that the system has infinitely many solutions, and the solution space is one-dimensional, which aligns with our result from Gaussian elimination.

To provide a more detailed solution, let's revisit the system of equations derived from the matrix equation:

$$\begin{cases} v_1 + v_2 + v_3 = 0 \\ v_1 + 2v_2 = 0 \\ v_1 + 2v_3 = 0 \end{cases}$$

From the second equation, we can express $v_1$ in terms of $v_2$:

$$v_1 = -2v_2$$

From the third equation, we can express $v_1$ in terms of $v_3$:

$$v_1 = -2v_3$$

Equating these two expressions for $v_1$, we get:

$$-2v_2 = -2v_3 \Rightarrow v_2 = v_3$$

Now, substitute $v_1 = -2v_2$ and $v_3 = v_2$ into the first equation:

$$(-2v_2) + v_2 + v_2 = 0$$

$$0 = 0$$

This confirms that the system has infinitely many solutions. Let $v_2 = t$, where $t$ is a parameter. Then, $v_3 = t$ and $v_1 = -2t$. The solution vector is:

$$\begin{bmatrix}v_1 \\ v_2 \\ v_3\end{bmatrix} = \begin{bmatrix}-2t \\ t \\ t\end{bmatrix} = t\begin{bmatrix}-2 \\ 1 \\ 1\end{bmatrix}$$

This is the same solution we obtained using Gaussian elimination. The solutions form a line in 3D space, parameterized by $t$. For example, when $t = 1$, the solution is $\begin{bmatrix}-2 \ 1 \ 1\end{bmatrix}$, and when $t = 0$, we have the trivial solution $\begin{bmatrix}0 \ 0 \ 0\end{bmatrix}$.

In this article, we have thoroughly explored the methods for solving a system of linear equations represented in matrix form. We started by understanding the matrix equation and its components. We then applied various methods, including Gaussian elimination, calculating the determinant, and analyzing the rank of the matrix, to find the solutions. The detailed step-by-step solution demonstrated that the system has infinitely many solutions, which are multiples of the vector $\begin{bmatrix}-2 \ 1 \ 1\end{bmatrix}$.

Understanding how to solve systems of linear equations is crucial in various fields, and the techniques discussed here provide a solid foundation for tackling more complex problems. The combination of algebraic manipulation and matrix analysis allows for a comprehensive understanding of the solution space. The insights gained from this analysis can be applied to a wide range of applications, from engineering and physics to economics and computer science. The ability to determine the nature of the solutions, whether unique, infinite, or non-existent, is a powerful tool in mathematical problem-solving and practical applications.

By using methods like Gaussian elimination, determinant calculation, and rank analysis, we can effectively solve and interpret systems of linear equations. This knowledge is invaluable for anyone working in mathematical or scientific disciplines, providing a robust framework for analyzing and solving real-world problems. The principles and techniques outlined in this article serve as a strong foundation for further exploration of linear algebra and its applications.