Solving Systems Of Equations Using Matrices Row Operations A Comprehensive Guide

Jul 15, 2025 by ADMIN 81 views

In mathematics, particularly in linear algebra, systems of equations are fundamental. A system of equations is a set of two or more equations containing the same variables. The solution to a system of equations is the set of values that satisfy all equations simultaneously. One powerful method for solving these systems is using matrices and row operations. This article will delve into how to solve a system of equations using matrices and row operations, providing a step-by-step guide and addressing the scenario where a system might be inconsistent, meaning it has no solution.

Understanding Systems of Equations and Matrices

Before diving into the solution process, it's crucial to understand the basics. A system of linear equations can be represented in matrix form, which simplifies the solving process. Let’s consider a general system of three linear equations with three variables:

a₁x + b₁y + c₁z = d₁
a₂x + b₂y + c₂z = d₂
a₃x + b₃y + c₃z = d₃

This system can be represented in matrix form as AX = B, where:

A is the coefficient matrix:

| a₁  b₁  c₁ |
| a₂  b₂  c₂ |
| a₃  b₃  c₃ |

X is the variable matrix:

| x |
| y |
| z |

B is the constant matrix:

| d₁ |
| d₂ |
| d₃ |

The augmented matrix combines the coefficient matrix A and the constant matrix B, denoted as [A|B]. This is the form we will manipulate using row operations to solve the system.

The Power of Row Operations

Row operations are the key to solving systems of equations using matrices. These operations allow us to manipulate the augmented matrix without changing the solution set of the original system. There are three elementary row operations:

Swapping two rows: This operation involves interchanging the positions of two rows in the matrix. It is useful for rearranging the matrix to achieve a more desirable form.
Multiplying a row by a non-zero constant: This operation involves multiplying all elements in a row by the same non-zero constant. It is used to scale rows, making it easier to eliminate variables.
Adding a multiple of one row to another: This operation involves adding a multiple of one row to another row. This is the most powerful operation, as it allows us to eliminate variables and simplify the system.

The Goal: Row-Echelon Form and Reduced Row-Echelon Form

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

Row-Echelon Form: A matrix is in row-echelon form if:
1. All non-zero rows (rows with at least one non-zero element) are above any rows of all zeros.
2. The leading coefficient (the first non-zero number from the left, also called the pivot) of a non-zero row is always strictly to the right of the leading coefficient of the row above it.
3. All entries in a column below a leading coefficient are zeros.
Reduced Row-Echelon Form: A matrix is in reduced row-echelon form if:
1. It is in row-echelon form.
2. The leading coefficient in each non-zero row is 1.
3. Each leading 1 is the only non-zero entry in its column.

Step-by-Step Solution Using Row Operations

Now, let's apply these concepts to the given system of equations:

x - 3y + 4z = 16
2x + y + z = 4
-4x + 3y - 3z = -16

Step 1: Write the Augmented Matrix

First, we write the augmented matrix corresponding to the system:

[A|B] = 
| 1  -3   4 | 16 |
| 2   1   1 |  4 |
| -4   3  -3 | -16 |

Step 2: Apply Row Operations to Get Row-Echelon Form

Our goal is to transform this matrix into row-echelon form. We'll use row operations to eliminate the entries below the first leading coefficient (which is 1 in the first row).

Eliminate the 2 in the second row:
- Replace Row 2 with Row 2 - 2 * Row 1 (R₂ → R₂ - 2R₁):
```
| 1  -3   4 | 16 |
| 0   7  -7 | -28 |
| -4   3  -3 | -16 |
```
Eliminate the -4 in the third row:
- Replace Row 3 with Row 3 + 4 * Row 1 (R₃ → R₃ + 4R₁):
```
| 1  -3   4 | 16 |
| 0   7  -7 | -28 |
| 0  -9  13 | 48 |
```

Simplify Row 2 by dividing by 7 (R₂ → R₂ / 7):

| 1  -3   4 | 16 |
| 0   1  -1 | -4 |
| 0  -9  13 | 48 |

Eliminate the -9 in the third row:
- Replace Row 3 with Row 3 + 9 * Row 2 (R₃ → R₃ + 9R₂):
```
| 1  -3   4 | 16 |
| 0   1  -1 | -4 |
| 0   0   4 | 12 |
```

Simplify Row 3 by dividing by 4 (R₃ → R₃ / 4):

| 1  -3   4 | 16 |
| 0   1  -1 | -4 |
| 0   0   1 | 3 |

Now the matrix is in row-echelon form.

Step 3: Apply Row Operations to Get Reduced Row-Echelon Form

Next, we transform the matrix into reduced row-echelon form.

Eliminate the -1 in the second row:
- Replace Row 2 with Row 2 + Row 3 (R₂ → R₂ + R₃):
```
| 1  -3   4 | 16 |
| 0   1   0 | -1 |
| 0   0   1 | 3 |
```
Eliminate the 4 in the first row:
- Replace Row 1 with Row 1 - 4 * Row 3 (R₁ → R₁ - 4R₃):
```
| 1  -3   0 | 4 |
| 0   1   0 | -1 |
| 0   0   1 | 3 |
```
Eliminate the -3 in the first row:
- Replace Row 1 with Row 1 + 3 * Row 2 (R₁ → R₁ + 3R₂):
```
| 1   0   0 | 1 |
| 0   1   0 | -1 |
| 0   0   1 | 3 |
```

Now the matrix is in reduced row-echelon form.

Step 4: Read the Solution

The reduced row-echelon form of the matrix directly gives us the solution:

x = 1
y = -1
z = 3

Thus, the solution to the system of equations is (1, -1, 3).

Identifying Inconsistent Systems

Sometimes, a system of equations has no solution, which is called an inconsistent system. This can be identified during the row reduction process. If at any point, you obtain a row in the augmented matrix of the form:

| 0  0  0 | k |

where k is a non-zero number, this indicates an inconsistency. This row corresponds to the equation 0 = k, which is impossible.

Example of an Inconsistent System

Consider the following system:

x + y = 2
2x + 2y = 5

The augmented matrix is:

| 1  1 | 2 |
| 2  2 | 5 |

Apply the row operation R₂ → R₂ - 2R₁:

| 1  1 | 2 |
| 0  0 | 1 |

The second row represents the equation 0 = 1, which is a contradiction. Therefore, this system is inconsistent and has no solution.

Conclusion

Solving systems of equations using matrices and row operations is a powerful and systematic method. By transforming the augmented matrix into row-echelon or reduced row-echelon form, we can easily find the solution or determine if the system is inconsistent. This method is particularly valuable for systems with many equations and variables, where manual algebraic manipulation becomes cumbersome. Understanding the concepts of row operations and matrix forms is essential for anyone studying linear algebra and its applications. Whether you are dealing with simple two-variable systems or complex multi-variable problems, the matrix method provides a robust approach to finding solutions or identifying inconsistencies.

By mastering these techniques, you can confidently tackle a wide range of mathematical problems involving systems of equations, making this a crucial skill for students, engineers, and anyone working with mathematical models in various fields.

Solve the Following System of Equations Using Matrices (Row Operations). If the System Has No Solution, Say That It Is Inconsistent

Let's revisit the initial problem and recap the step-by-step solution using matrices and row operations. The system of equations we need to solve is:

x - 3y + 4z = 16
2x + y + z = 4
-4x + 3y - 3z = -16

Step 1: Form the Augmented Matrix

The first step in solving this system using matrices is to write the augmented matrix. This matrix combines the coefficients of the variables and the constants on the right side of the equations. The augmented matrix for our system is:

| 1  -3   4 | 16 |
| 2   1   1 |  4 |
| -4   3  -3 | -16 |

Step 2: Apply Row Operations to Obtain Row-Echelon Form

The next step is to use elementary row operations to transform the matrix into row-echelon form. This involves making the entries below the leading diagonal elements (pivots) zero.

Eliminate the 2 in the second row:
- Replace Row 2 with Row 2 - 2 * Row 1 (R₂ → R₂ - 2R₁):
Performing this operation, we get:
```
| 1  -3   4 | 16 |
| 0   7  -7 | -28 |
| -4   3  -3 | -16 |
```
Eliminate the -4 in the third row:
- Replace Row 3 with Row 3 + 4 * Row 1 (R₃ → R₃ + 4R₁):
This operation results in:
```
| 1  -3   4 | 16 |
| 0   7  -7 | -28 |
| 0  -9  13 | 48 |
```
Simplify Row 2 by dividing by 7 (R₂ → R₂ / 7):

Dividing Row 2 by 7, we obtain:
```
| 1  -3   4 | 16 |
| 0   1  -1 | -4 |
| 0  -9  13 | 48 |
```
Eliminate the -9 in the third row:
- Replace Row 3 with Row 3 + 9 * Row 2 (R₃ → R₃ + 9R₂):
Applying this operation, we get:
```
| 1  -3   4 | 16 |
| 0   1  -1 | -4 |
| 0   0   4 | 12 |
```
Simplify Row 3 by dividing by 4 (R₃ → R₃ / 4):

Dividing Row 3 by 4, the matrix becomes:
```
| 1  -3   4 | 16 |
| 0   1  -1 | -4 |
| 0   0   1 | 3 |
```
At this point, the matrix is in row-echelon form.

Step 3: Transform the Matrix to Reduced Row-Echelon Form

Now, we need to transform the matrix to reduced row-echelon form, where the leading coefficients are 1, and all other entries in the same column are 0.

Eliminate the -1 in the second row:
- Replace Row 2 with Row 2 + Row 3 (R₂ → R₂ + R₃):
This results in:
```
| 1  -3   4 | 16 |
| 0   1   0 | -1 |
| 0   0   1 | 3 |
```
Eliminate the 4 in the first row:
- Replace Row 1 with Row 1 - 4 * Row 3 (R₁ → R₁ - 4R₃):
The matrix becomes:
```
| 1  -3   0 | 4 |
| 0   1   0 | -1 |
| 0   0   1 | 3 |
```
Eliminate the -3 in the first row:
- Replace Row 1 with Row 1 + 3 * Row 2 (R₁ → R₁ + 3R₂):
Finally, we get:
```
| 1   0   0 | 1 |
| 0   1   0 | -1 |
| 0   0   1 | 3 |
```
The matrix is now in reduced row-echelon form.

Step 4: Read the Solution from the Reduced Row-Echelon Form

The reduced row-echelon form of the matrix directly provides the solution to the system of equations:

x = 1
y = -1
z = 3

Thus, the solution to the given system of equations is (1, -1, 3).

Conclusion

In summary, solving systems of equations using matrices and row operations is a systematic approach that involves transforming the augmented matrix into row-echelon and then reduced row-echelon form. This method is especially useful for larger systems of equations where manual algebraic methods can become cumbersome. By following the steps outlined above, you can solve a wide variety of linear systems efficiently. Additionally, understanding when a system is inconsistent, which occurs when a row of the form [0 0 ... 0 | k] (where k is non-zero) is encountered, is crucial in linear algebra. The example provided illustrates the power and versatility of matrix methods in solving linear equations.