3 Linear Equations With 3 Variables

Here's an in-depth exploration of solving systems of three linear equations with three variables, covering various methods and offering practical examples to solidify understanding.

Solving Systems of Three Linear Equations with Three Variables

A system of three linear equations with three variables, typically x, y, and z, represents three planes in a three-dimensional space. The solution to the system is the point (or set of points) where all three planes intersect. This intersection can be a single point, a line, or no intersection at all, indicating no solution. Mastering the techniques to solve these systems is fundamental in various fields, including engineering, physics, economics, and computer science.

Understanding the Basics

Before diving into the solution methods, let's define what constitutes a linear equation in this context:

• Form: A linear equation in three variables has the form ax + by + cz = d, where a, b, c, and d are constants, and x, y, and z are the variables.

• System: A system of three such equations looks like this:

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

The goal is to find the values of x, y, and z that satisfy all three equations simultaneously.

Methods for Solving Systems of Linear Equations

Several methods can be used to solve these systems. Here, we'll focus on the three most common:

1. Substitution
2. Elimination (or Addition)
3. Using Matrices (Gaussian Elimination & Reduced Row Echelon Form)

1. Substitution Method

The substitution method involves solving one equation for one variable and substituting that expression into the other equations. This reduces the system to two equations with two variables, which can then be solved using similar techniques.

Steps:

1. Solve for one variable: Choose one of the equations and solve it for one of the variables (e.g., solve the first equation for x). Select the equation and variable that look easiest to isolate.
2. Substitute: Substitute the expression obtained in step 1 into the other two equations. This will give you two equations with two variables.
3. Solve the reduced system: Solve the system of two equations with two variables using either substitution or elimination.
4. Back-substitute: Once you find the values of two variables, substitute them back into one of the original equations (or the equation from step 1) to find the value of the third variable.
5. Verify: Check your solution by substituting the values of x, y, and z into all three original equations to ensure they are satisfied.

Example:

Consider the following system:

x + y + z = 6  (1)
2x - y + z = 3  (2)
x + 2y - z = 2  (3)

1. Solve for one variable: From equation (1), we can easily solve for x:

   x = 6 - y - z

2. Substitute: Substitute this expression for x into equations (2) and (3):

   • Equation (2): 2(6 - y - z) - y + z = 3 => 12 - 2y - 2z - y + z = 3 => -3y - z = -9
   • Equation (3): (6 - y - z) + 2y - z = 2 => 6 - y - z + 2y - z = 2 => y - 2z = -4

   Now we have a reduced system:

   -3y - z = -9   (4)
   y - 2z = -4    (5)
   

3. Solve the reduced system: Solve equation (5) for y: y = 2z - 4. Substitute this into equation (4):

   -3(2z - 4) - z = -9 => -6z + 12 - z = -9 => -7z = -21 => z = 3

   Substitute z = 3 back into y = 2z - 4:

   y = 2(3) - 4 = 6 - 4 = 2

4. Back-substitute: Substitute y = 2 and z = 3 back into the equation x = 6 - y - z:

   x = 6 - 2 - 3 = 1

5. Verify:

   • Equation (1): 1 + 2 + 3 = 6 (True)
   • Equation (2): 2(1) - 2 + 3 = 3 (True)
   • Equation (3): 1 + 2(2) - 3 = 2 (True)

Therefore, the solution is x = 1, y = 2, z = 3, or the ordered triple (1, 2, 3).

2. Elimination (or Addition) Method

The elimination method involves adding or subtracting multiples of equations to eliminate one variable at a time. This simplifies the system until you can solve for one variable and then back-substitute to find the others.

Steps:

1. Choose a variable to eliminate: Look for a variable that has coefficients that are easy to manipulate (e.g., coefficients that are multiples of each other or have opposite signs).
2. Eliminate the variable from two equations: Multiply one or both of the equations by suitable constants so that the coefficients of the chosen variable are opposites. Then, add the equations to eliminate that variable.
3. Repeat the process: Repeat step 2 with a different pair of equations (usually one of the original equations and the third equation) to eliminate the same variable. You should now have two equations with two variables.
4. Solve the reduced system: Solve the system of two equations with two variables using either substitution or elimination.
5. Back-substitute: Once you find the values of two variables, substitute them back into one of the original equations to find the value of the third variable.
6. Verify: Check your solution by substituting the values of x, y, and z into all three original equations to ensure they are satisfied.

Example:

Using the same system as before:

x + y + z = 6  (1)
2x - y + z = 3  (2)
x + 2y - z = 2  (3)

1. Choose a variable to eliminate: y looks easiest to eliminate because equation (1) has +y and equation (2) has -y.

2. Eliminate y from equations (1) and (2): Add equations (1) and (2) directly:

   (x + y + z) + (2x - y + z) = 6 + 3 => 3x + 2z = 9 (4)

3. Eliminate y from equations (1) and (3): Multiply equation (1) by -2:

   -2(x + y + z) = -2(6) => -2x - 2y - 2z = -12

   Add this modified equation to equation (3):

   (-2x - 2y - 2z) + (x + 2y - z) = -12 + 2 => -x - 3z = -10 (5)

4. Solve the reduced system: Now we have two equations with two variables:

   3x + 2z = 9   (4)
   -x - 3z = -10  (5)
   

   Multiply equation (5) by 3:

   -3x - 9z = -30

   Add this to equation (4):

   (3x + 2z) + (-3x - 9z) = 9 + (-30) => -7z = -21 => z = 3

   Substitute z = 3 into equation (5):

   -x - 3(3) = -10 => -x - 9 = -10 => -x = -1 => x = 1

5. Back-substitute: Substitute x = 1 and z = 3 into equation (1):

   1 + y + 3 = 6 => y + 4 = 6 => y = 2

6. Verify: (Same as before)

Therefore, the solution is x = 1, y = 2, z = 3, or (1, 2, 3).

3. Using Matrices (Gaussian Elimination & Reduced Row Echelon Form)

This method leverages the power of matrices to systematically solve the system of equations. It's particularly useful for larger systems with many variables.

Key Concepts:

• Augmented Matrix: Represent the system of equations as a matrix by taking the coefficients of the variables and the constants on the right-hand side. For the system:

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

  The augmented matrix is:

  [ a1  b1  c1 | d1 ]
  [ a2  b2  c2 | d2 ]
  [ a3  b3  c3 | d3 ]
  

• Elementary Row Operations: These are operations that can be performed on the rows of a matrix without changing the solution of the corresponding system of equations:

  • Swapping two rows: Ri <-> Rj
  • Multiplying a row by a non-zero constant: k * Ri -> Ri
  • Adding a multiple of one row to another row: Ri + k * Rj -> Ri

• Gaussian Elimination: The goal is to transform the augmented matrix into row echelon form (REF). In REF, the first non-zero entry in each row (called the leading entry or pivot) is to the right of the leading entry of the row above it. All entries below a leading entry are zero.

• Reduced Row Echelon Form (RREF): Further transform the matrix from REF into RREF. In RREF, the leading entry in each non-zero row is 1, and all other entries in the column containing a leading 1 are zero. This form directly gives the solution to the system.

Steps:

1. Form the augmented matrix: Represent the system of equations as an augmented matrix.
2. Perform Gaussian Elimination: Use elementary row operations to transform the matrix into row echelon form (REF).
3. Perform further row operations (if needed): Transform the matrix from REF into reduced row echelon form (RREF).
4. Interpret the RREF: The RREF will directly give you the solution for x, y, and z.
5. Verify: Check your solution by substituting the values of x, y, and z into all three original equations to ensure they are satisfied.

Example:

Using the same system as before:

x + y + z = 6  (1)
2x - y + z = 3  (2)
x + 2y - z = 2  (3)

1. Form the augmented matrix:

   [ 1  1  1 | 6 ]
   [ 2 -1  1 | 3 ]
   [ 1  2 -1 | 2 ]
   

2. Perform Gaussian Elimination (to get to REF):

   • R2 - 2*R1 -> R2: Subtract 2 times row 1 from row 2.

     [ 1  1  1 | 6 ]
     [ 0 -3 -1 | -9 ]
     [ 1  2 -1 | 2 ]
     

   • R3 - R1 -> R3: Subtract row 1 from row 3.

     [ 1  1  1 | 6 ]
     [ 0 -3 -1 | -9 ]
     [ 0  1 -2 | -4 ]
     

   • Swap R2 and R3:

     [ 1  1  1 | 6 ]
     [ 0  1 -2 | -4 ]
     [ 0 -3 -1 | -9 ]
     

   • R3 + 3*R2 -> R3: Add 3 times row 2 to row 3.

     [ 1  1  1 | 6 ]
     [ 0  1 -2 | -4 ]
     [ 0  0 -7 | -21 ]
     

     Now the matrix is in Row Echelon Form (REF).

3. Perform further row operations (to get to RREF):

   • R3 / -7 -> R3: Divide row 3 by -7.

     [ 1  1  1 | 6 ]
     [ 0  1 -2 | -4 ]
     [ 0  0  1 | 3 ]
     

   • R2 + 2*R3 -> R2: Add 2 times row 3 to row 2.

     [ 1  1  1 | 6 ]
     [ 0  1  0 | 2 ]
     [ 0  0  1 | 3 ]
     

   • R1 - R3 -> R1: Subtract row 3 from row 1.

     [ 1  1  0 | 3 ]
     [ 0  1  0 | 2 ]
     [ 0  0  1 | 3 ]
     

   • R1 - R2 -> R1: Subtract row 2 from row 1.

     [ 1  0  0 | 1 ]
     [ 0  1  0 | 2 ]
     [ 0  0  1 | 3 ]
     

     Now the matrix is in Reduced Row Echelon Form (RREF).

4. Interpret the RREF: The RREF directly tells us that x = 1, y = 2, z = 3.

5. Verify: (Same as before)

Therefore, the solution is x = 1, y = 2, z = 3, or (1, 2, 3).

Special Cases

Not all systems of three linear equations with three variables have a unique solution. There are two special cases to consider:

• No Solution (Inconsistent System): This occurs when the planes do not intersect at any common point. In terms of the augmented matrix, this will result in a row in RREF that looks like [0 0 0 | 1] (or any non-zero number on the right-hand side), which represents the equation 0 = 1, a contradiction.
• Infinitely Many Solutions (Dependent System): This occurs when the planes intersect in a line or when all three equations represent the same plane. In terms of the augmented matrix, this will result in at least one row of zeros in RREF. You'll need to express the solution in terms of a parameter (usually t).

Example of No Solution:

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

If you try to solve this system using any of the methods above, you will eventually encounter a contradiction, indicating that there is no solution.

Example of Infinitely Many Solutions:

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

Notice that the second and third equations are just multiples of the first equation. This system represents the same plane, and there are infinitely many solutions. You can express the solution in terms of a parameter. For example, let z = t. Then, x + y + t = 1, so x + y = 1 - t. We can solve for x in terms of y and t: x = 1 - t - y. Therefore, the solution can be written as (1 - t - y, y, t), where y and t can be any real numbers. Alternatively, solving for y gives (x, 1 - t - x, t).

Practical Applications

Solving systems of linear equations is a fundamental skill with applications in various fields:

• Engineering: Analyzing circuits, determining forces in structures, and modeling fluid flow.
• Physics: Solving for unknown quantities in mechanics, electromagnetism, and thermodynamics.
• Economics: Modeling supply and demand, analyzing market equilibrium, and optimizing resource allocation.
• Computer Science: Solving linear programming problems, creating computer graphics, and developing machine learning algorithms.
• Statistics: Linear regression, solving for coefficients in statistical models.
• Cryptography: Some cryptographic algorithms rely on solving systems of equations.

Tips and Tricks

• Look for easy eliminations or substitutions: Before blindly applying a method, examine the equations for opportunities to simplify the process.
• Be organized: Keep your work neat and organized to avoid errors, especially when using the elimination method or working with matrices.
• Double-check your calculations: It's easy to make mistakes when performing algebraic manipulations. Always double-check your work to ensure accuracy.
• Use technology: For larger systems, consider using calculators or computer software that can perform matrix operations.
• Understand the geometry: Visualize the equations as planes in 3D space to gain a better understanding of the possible solution types.

Conclusion

Solving systems of three linear equations with three variables is a critical skill with broad applications. Understanding the different methods, including substitution, elimination, and matrix techniques, allows you to approach a variety of problems effectively. By mastering these techniques and practicing regularly, you can confidently solve complex systems and apply them to real-world scenarios. Remember to always check your solutions and be aware of the special cases where there may be no solution or infinitely many solutions. With practice and a solid understanding of the underlying concepts, you can become proficient in solving these important mathematical problems.