Solving And Classifying Linear Equation Systems A Comprehensive Guide

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In mathematics, solving systems of linear equations is a fundamental skill with widespread applications in various fields such as engineering, physics, economics, and computer science. A system of linear equations consists of two or more linear equations involving the same variables. The solution to a system of linear equations is the set of values for the variables that satisfy all equations simultaneously. This article delves into the process of solving and classifying different systems of linear equations, providing a comprehensive guide for students and professionals alike. We will explore various techniques, including Gaussian elimination, matrix methods, and discuss the nature of solutions—whether a system has a unique solution, infinitely many solutions, or no solution at all.

Understanding Linear Equation Systems

A linear equation system is a collection of two or more linear equations that share the same set of variables. A linear equation is one in which the highest power of any variable is 1. For example, ax + by + cz = d is a linear equation, where x, y, and z are variables, and a, b, c, and d are constants. A solution to a system of linear equations is a set of values for the variables that makes all the equations in the system true simultaneously. A system of linear equations can have one unique solution, infinitely many solutions, or no solution. The nature of the solutions depends on the relationships between the equations in the system.

Methods for Solving Linear Equation Systems

Several methods can be used to solve systems of linear equations. Some of the most common methods include:

  1. Substitution: This method involves solving one equation for one variable and then substituting that expression into the other equations.
  2. Elimination: Also known as the addition method, this involves adding or subtracting multiples of the equations to eliminate one variable at a time.
  3. Gaussian Elimination: A systematic method that uses row operations to transform the system's augmented matrix into row-echelon form or reduced row-echelon form.
  4. Matrix Methods: Using matrix operations, such as finding the inverse of a matrix or using Cramer's rule, to solve the system.

Classifying Linear Equation Systems

Linear equation systems can be classified based on their solutions:

  • Consistent System: A system that has at least one solution. This includes systems with a unique solution and systems with infinitely many solutions.
  • Inconsistent System: A system that has no solution.
  • Independent System: A system with a unique solution.
  • Dependent System: A system with infinitely many solutions. In a dependent system, the equations are essentially multiples of each other, or there is some linear combination of the equations that results in another equation in the system.

A) Solving the System: x + y - z = 1, 2x - 3y - 3z = 3, x - 3y - 3z = 2

Let's dive into solving the first system of linear equations:

x + y - z = 1
2x - 3y - 3z = 3
x - 3y - 3z = 2

To effectively solve and classify this system, we will employ the Gaussian elimination method. Gaussian elimination is a powerful technique that systematically transforms a system of linear equations into an equivalent upper triangular form, making it easier to solve. This method involves performing elementary row operations on the augmented matrix of the system.

Step 1: Construct the Augmented Matrix

The first step in Gaussian elimination is to represent the system of equations as an augmented matrix. The augmented matrix consists of the coefficients of the variables and the constants on the right-hand side of the equations. For this system, the augmented matrix is:

[ 1  1 -1 | 1 ]
[ 2 -3 -3 | 3 ]
[ 1 -3 -3 | 2 ]

Step 2: Perform Row Operations

The goal is to transform the matrix into an upper triangular form, where all entries below the main diagonal are zero. This is achieved through elementary row operations, which include:

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

First, we want to eliminate the x term in the second and third rows. To do this, we perform the following operations:

  • Replace Row 2 with Row 2 - 2 * Row 1.
  • Replace Row 3 with Row 3 - Row 1.

This gives us the new matrix:

[  1   1  -1 |  1 ]
[  0  -5  -1 |  1 ]
[  0  -4  -2 |  1 ]

Next, we want to eliminate the y term in the third row. To do this, we can multiply Row 2 by -4 and Row 3 by 5, and then add the results:

  • Replace Row 3 with 5 * Row 3 - 4 * Row 2
[  1   1  -1 |  1 ]
[  0  -5  -1 |  1 ]
[  0   0  -6 |  1 ]

Step 3: Back-Substitution

Now that we have the matrix in upper triangular form, we can use back-substitution to solve for the variables. The matrix represents the following system of equations:

x +   y -  z = 1
   -5y -  z = 1
      -6z = 1

From the third equation, we can directly solve for z:

z = -1/6

Substituting z = -1/6 into the second equation, we can solve for y:

-5y - (-1/6) = 1
-5y + 1/6 = 1
-5y = 5/6
y = -1/6

Finally, substituting y = -1/6 and z = -1/6 into the first equation, we can solve for x:

x + (-1/6) - (-1/6) = 1
x = 1

Step 4: Classify the System

We have found a unique solution for the system: x = 1, y = -1/6, and z = -1/6. Therefore, the system is consistent and independent.

  • Consistent: Because a solution exists.
  • Independent: Because there is a unique solution.

B) Solving the System: -x + y - z = 4, x - y + z = 0, x - y = 2

Let's proceed to the next system:

-x + y - z = 4
x - y + z = 0
x - y = 2

Step 1: Construct the Augmented Matrix

The augmented matrix for this system is:

[ -1  1 -1 | 4 ]
[  1 -1  1 | 0 ]
[  1 -1  0 | 2 ]

Step 2: Perform Row Operations

We aim to transform the matrix into upper triangular form. First, let's eliminate the x term in the second and third rows:

  • Replace Row 2 with Row 2 + Row 1
  • Replace Row 3 with Row 3 + Row 1

This yields:

[ -1  1 -1 | 4 ]
[  0  0  0 | 4 ]
[  0  0 -1 | 6 ]

Step 3: Analyze the System

Observe the second row of the matrix. It represents the equation:

0x + 0y + 0z = 4

This simplifies to 0 = 4, which is a contradiction. This means that the system of equations is inconsistent and has no solution.

Step 4: Classify the System

  • Inconsistent: Because there is no solution.

C) Solving the System: 2x + 2y + 3z = 5, x + y + 2z = 3, 3x + 4y + 2z = 0

Now, let's tackle the third system:

2x + 2y + 3z = 5
x +   y + 2z = 3
3x + 4y + 2z = 0

Step 1: Construct the Augmented Matrix

The augmented matrix for this system is:

[ 2  2  3 | 5 ]
[ 1  1  2 | 3 ]
[ 3  4  2 | 0 ]

Step 2: Perform Row Operations

To begin, let's swap Row 1 and Row 2 to get a smaller number in the first row:

  • Swap Row 1 and Row 2
[ 1  1  2 | 3 ]
[ 2  2  3 | 5 ]
[ 3  4  2 | 0 ]

Next, eliminate the x terms in the second and third rows:

  • Replace Row 2 with Row 2 - 2 * Row 1
  • Replace Row 3 with Row 3 - 3 * Row 1
[  1  1  2 |  3 ]
[  0  0 -1 | -1 ]
[  0  1 -4 | -9 ]

Swap Row 2 and Row 3:

[  1  1  2 |  3 ]
[  0  1 -4 | -9 ]
[  0  0 -1 | -1 ]

Step 3: Back-Substitution

The matrix now represents the following system:

x + y + 2z = 3
    y - 4z = -9
       -z = -1

From the third equation, we have z = 1. Substituting z = 1 into the second equation:

y - 4(1) = -9
y = -5

Substituting y = -5 and z = 1 into the first equation:

x + (-5) + 2(1) = 3
x - 5 + 2 = 3
x = 6

Step 4: Classify the System

We have found a unique solution: x = 6, y = -5, and z = 1. The system is consistent and independent.

  • Consistent: Because a solution exists.
  • Independent: Because there is a unique solution.

D) Solving the System: 2x + 6y + z = 11, x + 3y - z = 9, 4x + 9y - 7z = 38

Let's move on to the fourth system:

2x + 6y +  z = 11
x + 3y -  z = 9
4x + 9y - 7z = 38

Step 1: Construct the Augmented Matrix

The augmented matrix for this system is:

[  2  6  1 | 11 ]
[  1  3 -1 |  9 ]
[  4  9 -7 | 38 ]

Step 2: Perform Row Operations

Swap Row 1 and Row 2:

[  1  3 -1 |  9 ]
[  2  6  1 | 11 ]
[  4  9 -7 | 38 ]

Eliminate the x terms in the second and third rows:

  • Replace Row 2 with Row 2 - 2 * Row 1
  • Replace Row 3 with Row 3 - 4 * Row 1
[  1  3 -1 |  9 ]
[  0  0  3 | -7 ]
[  0 -3 -3 |  2 ]

Swap Row 2 and Row 3:

[  1  3 -1 |  9 ]
[  0 -3 -3 |  2 ]
[  0  0  3 | -7 ]

Step 3: Back-Substitution

The matrix now represents the following system:

x + 3y - z = 9
   -3y - 3z = 2
        3z = -7

From the third equation, z = -7/3. Substituting z into the second equation:

-3y - 3(-7/3) = 2
-3y + 7 = 2
-3y = -5
y = 5/3

Substituting y and z into the first equation:

x + 3(5/3) - (-7/3) = 9
x + 5 + 7/3 = 9
x = 9 - 5 - 7/3
x = 5/3

Step 4: Classify the System

We have a unique solution: x = 5/3, y = 5/3, and z = -7/3. The system is consistent and independent.

  • Consistent: Because a solution exists.
  • Independent: Because there is a unique solution.

E) Solving the System: x + y + 3z = -5, 3x - 2y + 4z = 0, 2x - 3y + z = 5

Let's solve the fifth system:

x + y + 3z = -5
3x - 2y + 4z = 0
2x - 3y +  z = 5

Step 1: Construct the Augmented Matrix

The augmented matrix for this system is:

[ 1  1  3 | -5 ]
[ 3 -2  4 |  0 ]
[ 2 -3  1 |  5 ]

Step 2: Perform Row Operations

Eliminate the x terms in the second and third rows:

  • Replace Row 2 with Row 2 - 3 * Row 1
  • Replace Row 3 with Row 3 - 2 * Row 1
[  1  1  3 | -5 ]
[  0 -5 -5 | 15 ]
[  0 -5 -5 | 15 ]

Notice that Row 2 and Row 3 are identical. Let's simplify Row 2 by dividing by -5:

[  1  1  3 | -5 ]
[  0  1  1 | -3 ]
[  0 -5 -5 | 15 ]

Now, eliminate the y term in the third row:

  • Replace Row 3 with Row 3 + 5 * Row 2
[  1  1  3 | -5 ]
[  0  1  1 | -3 ]
[  0  0  0 |  0 ]

Step 3: Back-Substitution

The matrix now represents the following system:

x + y + 3z = -5
y + z = -3
0 = 0

Since the last row is all zeros, this indicates that the system has infinitely many solutions. We can express the solutions in terms of a parameter. Let z = t, where t is any real number. From the second equation:

y + t = -3
y = -3 - t

Substituting y and z into the first equation:

x + (-3 - t) + 3t = -5
x - 3 - t + 3t = -5
x + 2t = -2
x = -2 - 2t

Step 4: Classify the System

The system has infinitely many solutions, which can be expressed as x = -2 - 2t, y = -3 - t, and z = t. The system is consistent and dependent.

  • Consistent: Because solutions exist.
  • Dependent: Because there are infinitely many solutions.

F) Solving the System: 2x - 3y + 5z = 0, x - y + z = 0, 3x + 2y - 12z = 0

Finally, let's address the last system:

2x - 3y +  5z = 0
x -  y +  z = 0
3x + 2y - 12z = 0

Step 1: Construct the Augmented Matrix

The augmented matrix for this system is:

[  2 -3  5 | 0 ]
[  1 -1  1 | 0 ]
[  3  2 -12 | 0 ]

Step 2: Perform Row Operations

Swap Row 1 and Row 2:

[  1 -1  1 | 0 ]
[  2 -3  5 | 0 ]
[  3  2 -12 | 0 ]

Eliminate the x terms in the second and third rows:

  • Replace Row 2 with Row 2 - 2 * Row 1
  • Replace Row 3 with Row 3 - 3 * Row 1
[  1 -1  1 | 0 ]
[  0 -1  3 | 0 ]
[  0  5 -15 | 0 ]

Multiply Row 2 by -1:

[  1 -1  1 | 0 ]
[  0  1 -3 | 0 ]
[  0  5 -15 | 0 ]

Eliminate the y term in the third row:

  • Replace Row 3 with Row 3 - 5 * Row 2
[  1 -1  1 | 0 ]
[  0  1 -3 | 0 ]
[  0  0  0 | 0 ]

Step 3: Back-Substitution

The matrix represents the following system:

x - y + z = 0
y - 3z = 0
0 = 0

Again, the last row being all zeros indicates infinitely many solutions. Let z = t. From the second equation:

y - 3t = 0
y = 3t

Substituting y and z into the first equation:

x - 3t + t = 0
x - 2t = 0
x = 2t

Step 4: Classify the System

The system has infinitely many solutions: x = 2t, y = 3t, and z = t. The system is consistent and dependent.

  • Consistent: Because solutions exist.
  • Dependent: Because there are infinitely many solutions.

Conclusion

In summary, we have systematically solved and classified six different systems of linear equations using Gaussian elimination and back-substitution. We identified systems with unique solutions (consistent and independent), systems with no solutions (inconsistent), and systems with infinitely many solutions (consistent and dependent). Mastering these techniques is crucial for various applications in mathematics, science, and engineering. The ability to analyze and solve linear equation systems provides a powerful tool for modeling and understanding real-world phenomena.

By understanding these methods, you can confidently approach and solve a wide range of linear equation systems, making informed decisions based on the nature of their solutions. Whether you are a student tackling homework problems or a professional working on complex models, these skills are indispensable.