Solving Systems Of Equations A Comprehensive Guide

Jul 16, 2025 by ADMIN 51 views

Solving a System of Equations: A Comprehensive Guide

In the realm of mathematics, systems of equations serve as powerful tools for modeling and solving real-world problems. This article delves into the process of solving a system of linear equations, providing a step-by-step guide to unravel the solution. We will tackle the following system:

x + 2y + 5z = 600
x - z = 0
x + y + z = 960

This system comprises three equations with three unknowns (x, y, and z). Our mission is to determine the values of these unknowns that simultaneously satisfy all three equations. To embark on this quest, we'll employ the method of augmented matrices, a systematic approach for solving linear systems.

Transforming into an Augmented Matrix

The first step involves converting the system of equations into an augmented matrix. An augmented matrix is a compact representation of the system, where the coefficients of the variables and the constants are arranged in a rectangular array. For our system, the augmented matrix takes the form:

1 2 5 | 600
1 0 -1 | 0
1 1 1 | 960

Each row of the matrix corresponds to an equation in the system, and each column represents the coefficients of a specific variable (x, y, and z) or the constant term. The vertical line separates the coefficient matrix from the constant terms.

Gaussian Elimination: The Path to Solution

Now, we employ the technique of Gaussian elimination, a pivotal method for transforming the augmented matrix into row-echelon form. Row-echelon form is a specific arrangement of the matrix that simplifies the process of solving for the unknowns. The essence of Gaussian elimination lies in performing elementary row operations, which manipulate the rows of the matrix without altering the underlying solution of the system.

The elementary row operations at our disposal are:

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

Our goal is to systematically apply these operations to transform the matrix into row-echelon form, where:

The first non-zero element (leading entry) in each row is 1.
The leading entry in each row is to the right of the leading entry in the row above it.
Rows with all zero entries are at the bottom of the matrix.

Let's embark on the Gaussian elimination process:

Step 1: Eliminate x from the second and third equations.
- Subtract row 1 from row 2: R2 = R2 - R1
- Subtract row 1 from row 3: R3 = R3 - R1
The matrix transforms to:
```
1 2 5 | 600
0 -2 -6 | -600
0 -1 -4 | 360
```
Step 2: Make the leading entry in the second row 1.
- Divide row 2 by -2: R2 = R2 / -2
The matrix becomes:
```
1 2 5 | 600
0 1 3 | 300
0 -1 -4 | 360
```
Step 3: Eliminate y from the third equation.
- Add row 2 to row 3: R3 = R3 + R2
The matrix transforms to:
```
1 2 5 | 600
0 1 3 | 300
0 0 -1 | 660
```
Step 4: Make the leading entry in the third row 1.
- Multiply row 3 by -1: R3 = R3 * -1
The matrix becomes:
```
1 2 5 | 600
0 1 3 | 300
0 0 1 | -660
```

Now, the matrix is in row-echelon form.

Back-Substitution: Unveiling the Solution

With the matrix in row-echelon form, we can employ the technique of back-substitution to solve for the unknowns. Back-substitution involves starting from the last equation and working our way upwards, solving for one variable at a time.

From the third row, we have:

z = -660

Substituting this value into the second row, we get:

y + 3(-660) = 300 y - 1980 = 300 y = 2280

Finally, substituting the values of y and z into the first row, we obtain:

x + 2(2280) + 5(-660) = 600 x + 4560 - 3300 = 600 x = -660

Therefore, the solution to the system of equations is:

x = -660, y = 2280, z = -660

Verification: Ensuring Accuracy

To ensure the accuracy of our solution, we can substitute these values back into the original equations. If the equations hold true, our solution is verified.

-660 + 2(2280) + 5(-660) = -660 + 4560 - 3300 = 600 (True)
-660 - (-660) = 0 (True)
-660 + 2280 + (-660) = 960 (True)

Since the solution satisfies all three equations, we can confidently declare it as the correct solution.

Now that we've successfully solved the system of equations, let's delve into a discussion of the solution and the techniques we employed. The solution we obtained, x = -660, y = 2280, and z = -660, represents the unique set of values that simultaneously satisfy all three equations in the system. This implies that there is only one possible combination of x, y, and z that can make all the equations true.

Gaussian elimination, the cornerstone of our solution, is a versatile method for solving systems of linear equations. Its systematic approach ensures that we can transform the augmented matrix into a form that readily reveals the solution. The elementary row operations, the building blocks of Gaussian elimination, allow us to manipulate the matrix without altering the underlying solution.
Back-substitution, the final step in our journey, provides a straightforward way to extract the values of the unknowns from the row-echelon form of the matrix. By starting from the last equation and working our way upwards, we can systematically solve for each variable.
Verification, an essential step in any mathematical endeavor, confirms the accuracy of our solution. By substituting the obtained values back into the original equations, we can ensure that they hold true, solidifying our confidence in the result.

In conclusion, solving systems of equations is a fundamental skill in mathematics with applications spanning various fields. The method of augmented matrices, coupled with Gaussian elimination and back-substitution, provides a robust framework for tackling these systems. By mastering these techniques, we equip ourselves with the tools to solve a wide range of mathematical problems and real-world scenarios.

In this article, we will meticulously explore the process of solving a system of three linear equations with three unknowns. Our focus will be on the system presented earlier:

x + 2y + 5z = 600
x - z = 0
x + y + z = 960

We will utilize the powerful technique of augmented matrices combined with Gaussian elimination and back-substitution to arrive at the solution. This method provides a systematic and organized way to solve such systems, ensuring accuracy and efficiency.

Transforming the System into an Augmented Matrix

The initial step in our approach is to represent the system of equations in the form of an augmented matrix. An augmented matrix is a compact and convenient way to represent the coefficients and constants of the equations. It consists of a rectangular array where each row corresponds to an equation, and each column represents the coefficients of a variable or the constant term. For the given system, the augmented matrix is:

1 2 5 | 600
1 0 -1 | 0
1 1 1 | 960

Here, the first three columns represent the coefficients of the variables x, y, and z, respectively, while the last column represents the constant terms on the right-hand side of the equations. The vertical line separates the coefficient matrix from the constant terms.

Gaussian Elimination: A Systematic Approach

Our next step involves employing the technique of Gaussian elimination to transform the augmented matrix into row-echelon form. Row-echelon form is a specific arrangement of the matrix that makes it easier to solve for the unknowns. The core idea behind Gaussian elimination is to perform elementary row operations on the matrix until it reaches row-echelon form.

Elementary row operations are operations that can be performed on the rows of a matrix without changing the solution of the underlying system of equations. These operations include:

Swapping two rows: This operation allows us to change the order of the equations.
Multiplying a row by a non-zero constant: This operation scales an equation by a constant factor.
Adding a multiple of one row to another row: This operation combines two equations to eliminate a variable.

The goal of Gaussian elimination is to transform the matrix into row-echelon form, where:

The first non-zero element (leading entry) in each row is 1.
The leading entry in each row is to the right of the leading entry in the row above it.
Rows with all zero entries are at the bottom of the matrix.

Let's apply Gaussian elimination to our augmented matrix:

Step 1: Eliminate x from the second and third equations.
- Subtract row 1 from row 2: R2 = R2 - R1
- Subtract row 1 from row 3: R3 = R3 - R1
This gives us the following matrix:
```
1 2 5 | 600
0 -2 -6 | -600
0 -1 -4 | 360
```
Step 2: Make the leading entry in the second row 1.
- Divide row 2 by -2: R2 = R2 / -2
The matrix becomes:
```
1 2 5 | 600
0 1 3 | 300
0 -1 -4 | 360
```
Step 3: Eliminate y from the third equation.
- Add row 2 to row 3: R3 = R3 + R2
This transforms the matrix to:
```
1 2 5 | 600
0 1 3 | 300
0 0 -1 | 660
```
Step 4: Make the leading entry in the third row 1.
- Multiply row 3 by -1: R3 = R3 * -1
The resulting matrix is in row-echelon form:
```
1 2 5 | 600
0 1 3 | 300
0 0 1 | -660
```

Back-Substitution: Unveiling the Solution

Now that the matrix is in row-echelon form, we can employ the technique of back-substitution to solve for the unknowns. Back-substitution involves starting from the last equation and working our way upwards, solving for one variable at a time.

From the third row, we have:

z = -660

Substituting this value into the second row, we get:

y + 3(-660) = 300 y - 1980 = 300 y = 2280

Finally, substituting the values of y and z into the first row, we obtain:

x + 2(2280) + 5(-660) = 600 x + 4560 - 3300 = 600 x = -660

Therefore, the solution to the system of equations is:

x = -660, y = 2280, z = -660

Understanding the underlying concepts is crucial for mastering the art of solving systems of equations. In this section, we will delve into the key concepts that govern the methods and techniques we employ.

Linear Equations and Systems

A linear equation is an equation in which the variables appear only to the first power and are not multiplied together. A system of linear equations is a set of two or more linear equations involving the same variables. The goal of solving a system of linear equations is to find the values of the variables that satisfy all the equations simultaneously.

Augmented Matrices: A Compact Representation

An augmented matrix provides a compact and efficient way to represent a system of linear equations. It consists of the coefficients of the variables and the constant terms arranged in a rectangular array. Each row of the matrix corresponds to an equation, and each column represents the coefficients of a specific variable or the constant term. The vertical line separates the coefficient matrix from the constant terms.

Gaussian Elimination: The Power of Row Operations

Gaussian elimination is a systematic method for solving systems of linear equations by transforming the augmented matrix into row-echelon form. The core idea behind Gaussian elimination is to perform elementary row operations on the matrix until it reaches row-echelon form. Elementary row operations are operations that can be performed on the rows of a matrix without changing the solution of the underlying system of equations. These operations include swapping two rows, multiplying a row by a non-zero constant, and adding a multiple of one row to another row.

Row-Echelon Form: Simplifying the Solution Process

Row-echelon form is a specific arrangement of the matrix that makes it easier to solve for the unknowns. In row-echelon form, the first non-zero element (leading entry) in each row is 1, the leading entry in each row is to the right of the leading entry in the row above it, and rows with all zero entries are at the bottom of the matrix.

Back-Substitution: A Step-by-Step Approach

Back-substitution is a technique used to solve for the unknowns after the matrix has been transformed into row-echelon form. It involves starting from the last equation and working our way upwards, solving for one variable at a time. By substituting the values of the known variables into the preceding equations, we can systematically solve for all the unknowns.

Solutions to Systems of Equations

Systems of linear equations can have three possible types of solutions:

Unique solution: The system has exactly one solution, which means there is only one set of values for the variables that satisfies all the equations simultaneously.
Infinitely many solutions: The system has an infinite number of solutions, which means there are many sets of values for the variables that satisfy all the equations. This occurs when the equations are dependent, meaning that one or more equations can be derived from the others.
No solution: The system has no solution, which means there is no set of values for the variables that satisfies all the equations. This occurs when the equations are inconsistent, meaning that they contradict each other.

Understanding these key concepts is essential for effectively solving systems of linear equations. By mastering these concepts, you will be well-equipped to tackle a wide range of mathematical problems and real-world scenarios involving systems of equations.

In this comprehensive guide, we have embarked on a journey to unravel the intricacies of solving systems of equations. We have explored the power of augmented matrices, the elegance of Gaussian elimination, and the efficiency of back-substitution. By mastering these techniques, you have equipped yourself with the tools to tackle a wide array of mathematical challenges.

Systems of equations serve as fundamental building blocks in various fields, including mathematics, physics, engineering, and economics. They provide a framework for modeling and solving real-world problems involving multiple variables and constraints.
The method of augmented matrices offers a systematic and organized approach to solving systems of equations. By representing the system in matrix form, we can leverage the power of matrix operations to transform the system into a more manageable form.
Gaussian elimination, the cornerstone of our solution strategy, provides a step-by-step procedure for transforming the augmented matrix into row-echelon form. This form simplifies the process of solving for the unknowns.
Back-substitution, the final step in our journey, allows us to efficiently extract the values of the unknowns from the row-echelon form of the matrix. By working our way upwards from the last equation, we can systematically solve for each variable.

As you continue your mathematical explorations, remember that practice is key to mastering any technique. The more you solve systems of equations, the more confident and proficient you will become. Embrace the challenges, persevere through the complexities, and relish the satisfaction of arriving at the solution.

May this guide serve as a valuable resource in your mathematical endeavors. Happy solving!