Solving Linear Equations With Augmented Matrix Methods A Step By Step Guide

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In the realm of linear algebra, solving systems of linear equations is a fundamental task. These systems arise in various fields, including engineering, physics, economics, and computer science. One powerful and systematic approach to solving such systems is the method of augmented matrices. This article will delve into the augmented matrix method, providing a step-by-step guide to solving the system of linear equations:

-6x₁ + 4x₂ = 2
12x₁ - 8x₂ = -4

We will explore the underlying principles, demonstrate the application of the method, and discuss the different types of solutions that can arise. By the end of this article, you will have a solid understanding of how to use augmented matrices to solve systems of linear equations efficiently and accurately.

Understanding Augmented Matrices

Before diving into the solution process, it's crucial to grasp the concept of an augmented matrix. An augmented matrix is a compact representation of a system of linear equations. It combines the coefficients of the variables and the constants into a single matrix, making it easier to perform row operations and solve the system.

To form an augmented matrix, we first extract the coefficients of the variables and the constants from the equations. For the given system:

-6x₁ + 4x₂ = 2
12x₁ - 8x₂ = -4

The coefficients are -6, 4, 12, and -8, and the constants are 2 and -4. We arrange these numbers into a matrix, separating the coefficients from the constants with a vertical line. This vertical line represents the equals signs in the original equations.

The augmented matrix for the system is:

[ -6  4 |  2 ]
[ 12 -8 | -4 ]

The matrix has two rows, corresponding to the two equations, and three columns. The first two columns represent the coefficients of x₁ and x₂, respectively, and the third column represents the constants.

The power of the augmented matrix lies in its ability to represent the entire system of equations in a concise and manageable form. We can then perform elementary row operations on the matrix, which correspond to manipulating the equations in the system. These operations allow us to transform the matrix into a simpler form, making it easier to solve for the variables.

Elementary Row Operations

Elementary row operations are the cornerstone of the augmented matrix method. These operations allow us to manipulate the rows of the matrix without changing the solution set of the corresponding system of equations. There are three fundamental types of elementary row operations:

  1. Row Switching: This operation involves interchanging two rows of the matrix. It's useful for rearranging the rows to facilitate other operations.
  2. Row Multiplication: This operation involves multiplying a row by a nonzero constant. It's used to scale a row, often to create a leading 1.
  3. Row Addition: This operation involves adding a multiple of one row to another row. It's the most powerful operation for eliminating variables and simplifying the system.

Each elementary row operation corresponds to a valid manipulation of the original equations. Row switching corresponds to changing the order of the equations. Row multiplication corresponds to multiplying an equation by a constant. Row addition corresponds to adding a multiple of one equation to another equation.

The goal of applying elementary row operations is to transform the augmented matrix into row-echelon form or reduced row-echelon form. These forms make it easy to read off the solution to the system.

Solving the System Using Augmented Matrix Methods

Now, let's apply the augmented matrix method to solve the given system of linear equations:

-6x₁ + 4x₂ = 2
12x₁ - 8x₂ = -4

We've already formed the augmented matrix:

[ -6  4 |  2 ]
[ 12 -8 | -4 ]

Our goal is to transform this matrix into row-echelon form or reduced row-echelon form. Row-echelon form is a matrix where:

  • The first nonzero entry in each row (the leading entry) is 1.
  • Each leading entry is in a column to the right of the leading entry in the row above it.
  • Rows with all zero entries are at the bottom of the matrix.

Reduced row-echelon form is a stricter form where, in addition to the above conditions:

  • Each leading entry is the only nonzero entry in its column.

We'll aim for reduced row-echelon form, as it directly reveals the solution.

Here are the steps we'll take:

  1. Get a leading 1 in the first row: We can achieve this by dividing the first row by -6:

    R₁ → R₁ / -6
[ 1  -2/3 | -1/3 ]
[ 12 -8   | -4   ]

  2. Eliminate the 12 in the second row: We can do this by adding -12 times the first row to the second row:

    R₂ → R₂ + (-12)R₁
[ 1  -2/3 | -1/3 ]
[ 0   0    |  0    ]

Now, the matrix is in row-echelon form and reduced row-echelon form. The matrix represents the following system of equations:

x₁ - (2/3)x₂ = -1/3
0 = 0

The second equation, 0 = 0, is always true and doesn't provide any additional information. The first equation tells us that x₁ can be expressed in terms of x₂:

x₁ = (2/3)x₂ - 1/3

This indicates that the system has infinitely many solutions. We can express the solution set in parametric form. Let x₂ = t, where t is a parameter. Then,

x₁ = (2/3)t - 1/3

Therefore, the solution set is:

(x₁, x₂) = ((2/3)t - 1/3, t)

where t is any real number.

Interpreting the Solution

The solution we obtained, (x₁, x₂) = ((2/3)t - 1/3, t), represents a line in the x₁x₂-plane. This means that there are infinitely many points (x₁, x₂) that satisfy the system of equations. Each value of the parameter t corresponds to a different point on this line.

The fact that we obtained infinitely many solutions is a consequence of the two original equations being linearly dependent. If we multiply the first equation by -2, we get the second equation. This means that the two equations represent the same line, and any point on that line is a solution to the system.

If we had obtained a unique solution, the matrix in reduced row-echelon form would have had a leading 1 in each column corresponding to a variable, and the constants would have directly given the values of the variables. If we had obtained a contradiction (e.g., 0 = 1), the system would have had no solutions.

Different Types of Solutions

When solving systems of linear equations, there are three possible types of solutions:

  1. Unique Solution: The system has exactly one solution, represented by a single point in the variable space. This occurs when the lines (in 2D) or planes (in 3D) intersect at a single point.
  2. Infinitely Many Solutions: The system has an infinite number of solutions, represented by a line or a plane (or higher-dimensional analogues) in the variable space. This occurs when the equations are linearly dependent, meaning they represent the same geometric object.
  3. No Solution: The system has no solutions, meaning there is no point that satisfies all equations simultaneously. This occurs when the lines or planes are parallel and do not intersect.

The augmented matrix method provides a systematic way to determine which type of solution a system has and to find the solution(s) if they exist. The final form of the matrix in row-echelon or reduced row-echelon form reveals the nature of the solution.

Advantages of Augmented Matrix Methods

The augmented matrix method offers several advantages over other methods for solving systems of linear equations:

  • Systematic Approach: It provides a clear and systematic procedure for solving systems, reducing the chance of errors.
  • Efficiency: It's often more efficient than substitution or elimination methods, especially for larger systems.
  • Universality: It works for systems with any number of equations and variables.
  • Solution Type Determination: It allows us to determine whether a system has a unique solution, infinitely many solutions, or no solution.

These advantages make the augmented matrix method a valuable tool for anyone working with linear systems.

Conclusion

In this article, we've explored the augmented matrix method for solving systems of linear equations. We've seen how to form an augmented matrix, perform elementary row operations, and interpret the resulting matrix to find the solution. We've also discussed the different types of solutions that can arise and the advantages of using augmented matrix methods.

The augmented matrix method is a powerful and versatile technique that can be applied to a wide range of problems involving linear systems. By mastering this method, you'll gain a valuable skill for solving mathematical problems in various fields.

In the specific case of the system:

-6x₁ + 4x₂ = 2
12x₁ - 8x₂ = -4

We found that the system has infinitely many solutions, which can be expressed in parametric form as (x₁, x₂) = ((2/3)t - 1/3, t), where t is any real number.