System Of Three Linear Equations In Three Variables

Delving into the realm of linear algebra, solving systems of three linear equations in three variables is a fundamental skill with applications ranging from engineering to economics. This article provides a comprehensive guide to understanding and solving such systems.

Introduction to Systems of Three Linear Equations

A system of three linear equations in three variables, typically x, y, and z, is a set of three equations where each equation represents a plane in three-dimensional space. The solution to this system is the point (if it exists) where all three planes intersect. This intersection point satisfies all three equations simultaneously.

A general form of such a system is:

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

Where a₁, b₁, c₁, a₂, b₂, c₂, a₃, b₃, c₃, d₁, d₂, and d₃ are constants. Solving these systems involves finding values for x, y, and z that make all three equations true. There are several methods to achieve this, each with its advantages and disadvantages.

Methods for Solving Systems of Three Linear Equations

Several methods can be employed to solve systems of three linear equations in three variables. The most common are:

1. Substitution Method: This involves solving one equation for one variable and substituting that expression into the other two equations, reducing the system to two equations in two variables.
2. Elimination Method (also known as the Addition Method): This involves adding or subtracting multiples of the equations to eliminate one variable, again reducing the system to two equations in two variables.
3. Matrix Method (using Gaussian Elimination or Gauss-Jordan Elimination): This method represents the system as an augmented matrix and uses row operations to solve for the variables.
4. Cramer's Rule: This method uses determinants to find the solution. It is efficient when the coefficient matrix is invertible.

Let's explore each method in detail with examples.

1. Substitution Method

The substitution method is particularly useful when one of the equations can easily be solved for one variable.

Steps:

1. Solve one equation for one variable: Choose the easiest equation and solve for one of the variables. For example, solve for x in terms of y and z.
2. Substitute: Substitute the expression obtained in step 1 into the other two equations. This will result in a system of two equations in two variables.
3. Solve the resulting system: Use any method (substitution or elimination) to solve this system of two equations.
4. Back-substitute: Once you have the values of two variables, substitute them back into one of the original equations (or the expression from step 1) to find the value of the third variable.

Example:

Solve the following system:

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

Solution:

1. Solve equation (1) for x:

   x = 6 - y - z (4)

2. Substitute (4) into equations (2) and (3):

   • 2(6 - y - z) - y + z = 3 => 12 - 2y - 2z - y + z = 3 => -3y - z = -9 (5)
   • (6 - y - z) + 2y - z = 2 => 6 - y - z + 2y - z = 2 => y - 2z = -4 (6)

3. Solve the resulting system of equations (5) and (6):

   We can use the elimination method on (5) and (6):

   Multiply equation (6) by 3: 3y - 6z = -12 (7)

   Add equation (5) and (7):

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

4. Back-substitute:

   • Substitute z = 3 into equation (6): y - 2(3) = -4 => y - 6 = -4 => y = 2
   • Substitute y = 2 and z = 3 into equation (4): x = 6 - 2 - 3 => x = 1

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

2. Elimination Method (Addition Method)

The elimination method is a powerful technique that involves strategically adding or subtracting multiples of equations to eliminate variables.

Steps:

1. Choose a variable to eliminate: Look for equations where the coefficients of one variable are the same or easily made the same through multiplication.
2. Multiply equations: Multiply one or more equations by constants so that the coefficients of the chosen variable are opposites in two equations.
3. Add or subtract equations: Add or subtract the modified equations to eliminate the chosen variable. This will result in a new equation with fewer variables.
4. Repeat: Repeat steps 1-3 until you have a system of two equations in two variables.
5. Solve the resulting system: Solve the system of two equations using any method.
6. Back-substitute: Substitute the values found in step 5 back into one of the original equations to find the value of the remaining variable.

Example:

Solve the following system:

• x + 2y - z = 5 (1)
• 2x - y + z = -1 (2)
• 3x + 2y + z = 4 (3)

Solution:

1. Eliminate z from equations (1) and (2):

   Add equation (1) and (2): (x + 2y - z) + (2x - y + z) = 5 + (-1) => 3x + y = 4 (4)

2. Eliminate z from equations (1) and (3):

   Multiply equation (1) by -1: -x - 2y + z = -5

   Add the modified equation (1) to equation (3): (-x - 2y + z) + (3x + 2y + z) = -5 + 4 => 2x + 2z = -1 => 2x + 0y + 2z = -1

   Now, let's rewrite equation (3): 3x + 2y + z = 4. Multiply this equation by -2 to eliminate 'y': -6x - 4y - 2z = -8

Multiply equation (1) by 2: 2x + 4y - 2z = 10

Add the modified equation (3) to the modified equation (1): (-6x - 4y - 2z) + (2x + 4y - 2z) = -8 + 10 => -4x - 4z = 2 => -2x - 2z = 1 (5) 3. Solve the resulting system of equations (4) and (5):

*   3x + y = 4 (4)
*   -2x - 2z = 1 (5)

We need another equation with 'x' and 'y' to solve for these variables. Notice that when adding equation (1) and (3), we almost eliminated y: (x+2y-z) + (3x+2y+z) = 5 + 4 = > 4x + 4y = 9 => y = (9-4x)/4

Plug this into equation (4): 3x + (9-4x)/4 = 4 => 12x + 9 - 4x = 16 => 8x = 7 => x = 7/8

Substitute the value of x into y = (9-4x)/4 => y = (9-4*(7/8))/4 = (9-7/2)/4 = (11/2)/4 = 11/8

4. Back-substitute:

   Substitute x = 7/8 and y = 11/8 into equation (1): (7/8) + 2(11/8) - z = 5 => 7/8 + 22/8 - z = 5 => 29/8 - z = 5 => z = 29/8 - 40/8 => z = -11/8

Therefore, the solution is x = 7/8, y = 11/8, and z = -11/8, or (7/8, 11/8, -11/8).

3. Matrix Method (Gaussian Elimination or Gauss-Jordan Elimination)

The matrix method is a systematic approach that leverages the power of linear algebra to solve systems of equations. It is particularly useful for larger systems.

Steps:

1. Write the augmented matrix: Represent the system of equations as an augmented matrix.
2. Perform row operations: Use elementary row operations to transform the matrix into row-echelon form (for Gaussian elimination) or reduced row-echelon form (for Gauss-Jordan elimination).
   • Swapping two rows.
   • Multiplying a row by a non-zero constant.
   • Adding a multiple of one row to another row.
3. Solve for the variables:
   • Gaussian Elimination: Use back-substitution to solve for the variables.
   • Gauss-Jordan Elimination: The solution is directly readable from the matrix.

Example:

Solve the following system:

• x - y + z = 4 (1)
• 2x + y - 3z = 0 (2)
• x + y + z = 2 (3)

Solution:

1. Write the augmented matrix:

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

2. Perform row operations (Gaussian Elimination):

   • R₂ = R₂ - 2R₁:

     [ 1  -1   1 |  4 ]
     [ 0   3  -5 | -8 ]
     [ 1   1   1 |  2 ]
     

   • R₃ = R₃ - R₁:

     [ 1  -1   1 |  4 ]
     [ 0   3  -5 | -8 ]
     [ 0   2   0 | -2 ]
     

   • R₃ = R₃ / 2:

     [ 1  -1   1 |  4 ]
     [ 0   3  -5 | -8 ]
     [ 0   1   0 | -1 ]
     

   • Swap R₂ and R₃:

     [ 1  -1   1 |  4 ]
     [ 0   1   0 | -1 ]
     [ 0   3  -5 | -8 ]
     

   • R₃ = R₃ - 3R₂:

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

   • R₃ = R₃ / -5:

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

3. Back-substitute:

   From the matrix, we have:

   • z = 1
   • y = -1
   • x - y + z = 4 => x - (-1) + 1 = 4 => x + 2 = 4 => x = 2

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

Gauss-Jordan Elimination

To perform Gauss-Jordan Elimination, we continue from the row-echelon form to get the reduced row-echelon form.

Continuing from the Gaussian elimination example:

• Add R2 to R1:

  [ 1   0   1 |  3 ]
  [ 0   1   0 | -1 ]
  [ 0   0   1 |  1 ]
  

• Subtract R3 from R1:

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

From the reduced row-echelon form: x = 2, y = -1, z = 1

4. Cramer's Rule

Cramer's Rule is a method for solving systems of linear equations using determinants. It is particularly useful when you need to find the value of only one variable.

Steps:

1. Calculate the determinant of the coefficient matrix (D): This is the matrix formed by the coefficients of the variables.

2. Calculate the determinants Dₓ, Dᵧ, and D₂: Replace the corresponding column of the coefficient matrix with the constant terms to find these determinants.

3. Solve for the variables:

   • x = Dₓ / D
   • y = Dᵧ / D
   • z = D₂ / D

Example:

Solve the following system:

• x + 2y + z = 6 (1)
• 2x - y + z = 2 (2)
• x + y - z = 0 (3)

Solution:

1. Calculate the determinant of the coefficient matrix (D):

   D = | 1  2  1 |
       | 2 -1  1 |
       | 1  1 -1 |
   

   D = 1((-1)(-1) - 11) - 2(2*(-1) - 11) + 1(21 - (-1)*1) D = 1(1 - 1) - 2(-2 - 1) + 1(2 + 1) D = 0 - 2(-3) + 1(3) D = 0 + 6 + 3 D = 9

2. Calculate Dₓ, Dᵧ, and D₂:

   • Dₓ:

     Dₓ = | 6  2  1 |
          | 2 -1  1 |
          | 0  1 -1 |
     

     Dₓ = 6((-1)(-1) - 11) - 2(2*(-1) - 10) + 1(21 - (-1)*0) Dₓ = 6(1 - 1) - 2(-2 - 0) + 1(2 - 0) Dₓ = 0 - 2(-2) + 1(2) Dₓ = 0 + 4 + 2 Dₓ = 6

   • Dᵧ:

     Dᵧ = | 1  6  1 |
          | 2  2  1 |
          | 1  0 -1 |
     

     Dᵧ = 1(2*(-1) - 10) - 6(2(-1) - 11) + 1(20 - 2*1) Dᵧ = 1(-2 - 0) - 6(-2 - 1) + 1(0 - 2) Dᵧ = -2 - 6(-3) + (-2) Dᵧ = -2 + 18 - 2 Dᵧ = 14

   • D₂:

     D₂ = | 1  2  6 |
          | 2 -1  2 |
          | 1  1  0 |
     

     D₂ = 1((-1)0 - 21) - 2(20 - 21) + 6(2*1 - (-1)*1) D₂ = 1(0 - 2) - 2(0 - 2) + 6(2 + 1) D₂ = -2 - 2(-2) + 6(3) D₂ = -2 + 4 + 18 D₂ = 20

3. Solve for the variables:

   • x = Dₓ / D = 6 / 9 = 2/3
   • y = Dᵧ / D = 14 / 9
   • z = D₂ / D = 20 / 9

Therefore, the solution is x = 2/3, y = 14/9, and z = 20/9, or (2/3, 14/9, 20/9).

Possible Outcomes of Solving Systems of Three Linear Equations

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

1. Unique Solution: The system has one and only one solution, which is the point where all three planes intersect.
2. No Solution: The system has no solution. This can occur when the planes are parallel or when they intersect in such a way that there is no common point.
3. Infinitely Many Solutions: The system has infinitely many solutions. This occurs when the planes intersect in a line or when all three equations represent the same plane.

Identifying Different Outcomes

• Unique Solution: When using algebraic methods (substitution or elimination), you will arrive at unique values for x, y, and z. When using matrix methods, the reduced row-echelon form will have a leading 1 in each row and column corresponding to a variable.
• No Solution: When using algebraic methods, you will arrive at a contradiction, such as 0 = 1. When using matrix methods, you will encounter a row in the reduced row-echelon form that looks like [0 0 0 | b], where b is a non-zero number.
• Infinitely Many Solutions: When using algebraic methods, one or more variables can be expressed in terms of another, indicating a line of intersection or a common plane. When using matrix methods, you will have fewer leading 1s than variables, indicating free variables.

Applications of Systems of Three Linear Equations

Systems of three linear equations have numerous applications in various fields, including:

• Engineering: Solving for currents in electrical circuits, determining forces in structural analysis.
• Physics: Modeling motion, solving problems in thermodynamics.
• Economics: Modeling supply and demand, optimizing resource allocation.
• Computer Graphics: Determining the intersection of planes in 3D modeling.
• Statistics: Linear regression with multiple variables.

Tips and Tricks for Solving Systems of Equations

• Choose the easiest method: Look at the equations and choose the method that seems easiest for the given system.
• Check your work: After finding a solution, substitute the values back into the original equations to make sure they are satisfied.
• Be organized: Keep your work neat and organized to avoid making mistakes.
• Use technology: Use calculators or computer software to help with complex calculations.

Conclusion

Solving systems of three linear equations in three variables is a crucial skill in many areas of mathematics, science, and engineering. By mastering the various methods—substitution, elimination, matrix methods, and Cramer's Rule—and understanding the possible outcomes, you can effectively tackle these problems and apply them to real-world situations. Remember to practice regularly and stay organized to improve your problem-solving skills. The journey through linear algebra begins with understanding these fundamental concepts, paving the way for more advanced topics and applications.