Solving A 3x3 System Of Linear Equations

Solving a 3x3 system of linear equations might seem daunting at first, but with a systematic approach, it becomes a manageable task. This comprehensive guide will walk you through several methods, including substitution, elimination, and using matrices, to effectively solve these systems. Understanding these techniques equips you with valuable problem-solving skills applicable in various fields, from engineering to economics.

Understanding 3x3 Systems of Linear Equations

A 3x3 system of linear equations consists of three equations, each containing three variables (typically x, y, and z). The goal is to find values for these variables that satisfy all three equations simultaneously. A general form of such a system looks like this:

• 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.

Before diving into the solution methods, it's important to understand the possible outcomes:

• Unique Solution: The system has exactly one solution, meaning there is only one set of values for x, y, and z that satisfies all three equations.
• No Solution: The system is inconsistent, meaning there is no set of values for x, y, and z that can satisfy all three equations simultaneously. This often indicates parallel planes.
• Infinite Solutions: The system is dependent, meaning there are infinitely many sets of values for x, y, and z that satisfy all three equations. This often indicates that the equations represent the same plane or intersecting planes along a line.

Method 1: Solving by Substitution

The substitution method involves solving one equation for one variable and then substituting that expression into the other equations. This process reduces the system to a smaller set of equations with fewer variables, making it easier to solve.

Steps:

1. Solve for one variable: Choose one of the equations and solve it for one of the variables. Look for an equation where a variable has a coefficient of 1 or -1, as this will simplify the algebra. For example, if you have the equation x + 2y - z = 5, you could easily solve for x as x = 5 - 2y + z.

2. Substitute: Substitute the expression you found in step 1 into the other two equations. This will eliminate one variable from those two equations, leaving you with two equations in two variables.

3. Solve the 2x2 system: You now have a 2x2 system of linear equations. Solve this system using either substitution or elimination (as described below). This will give you the values for two of the variables.

4. Back-substitute: Substitute the values you found in step 3 back into the expression you found in step 1 to find the value of the third variable.

Example:

Consider the following system of equations:

• x + y + z = 6
• 2x - y + z = 3
• x + 2y - z = 2

1. Solve for x in the first equation: x = 6 - y - z

2. Substitute into the second and third equations:

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

3. Solve the 2x2 system:

   • -3y - z = -9
   • y - 2z = -4

   Solve the second equation for y: y = 2z - 4

   Substitute into the first equation: -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 = 2

4. Back-substitute: Substitute y = 2 and z = 3 back into x = 6 - y - z: x = 6 - 2 - 3 = 1

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

Method 2: Solving by Elimination

The elimination method involves adding or subtracting multiples of the equations to eliminate one variable at a time. This process also reduces the system to a smaller set of equations with fewer variables.

Steps:

1. Choose a variable to eliminate: Look for a variable that has coefficients that are easy to make opposites of each other by multiplying one or both equations by a constant.

2. Eliminate the variable from two equations: Multiply one or both of the first two equations by a constant so that the coefficients of the chosen variable are opposites. Then add the two equations together. This will eliminate the chosen variable from the resulting equation.

3. Eliminate the same variable from another pair of equations: Repeat step 2 using a different pair of equations (usually the first and third, or the second and third) to eliminate the same variable.

4. Solve the 2x2 system: You now have a 2x2 system of linear equations in two variables. Solve this system using either substitution or elimination.

5. Back-substitute: Substitute the values you found in step 4 back into one of the original equations to find the value of the third variable.

Example:

Consider the same system of equations:

• x + y + z = 6
• 2x - y + z = 3
• x + 2y - z = 2

1. Choose to eliminate y: Notice that the first and second equations already have y and -y, making them easy to eliminate.

2. Eliminate y from the first and second equations: Add the first and second equations together:

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

3. Eliminate y from the first and third equations: Multiply the first equation by -2: -2(x + y + z) = -2(6) => -2x - 2y - 2z = -12

   Add this to the third equation: (-2x - 2y - 2z) + (x + 2y - z) = -12 + 2 => -x - 3z = -10

4. Solve the 2x2 system:

   • 3x + 2z = 9
   • -x - 3z = -10

   Multiply the second equation by 3: -3x - 9z = -30

   Add this to the first equation: (3x + 2z) + (-3x - 9z) = 9 - 30 => -7z = -21 => z = 3

   Substitute z = 3 back into -x - 3z = -10: -x - 3(3) = -10 => -x - 9 = -10 => -x = -1 => x = 1

5. Back-substitute: Substitute x = 1 and z = 3 back into x + y + z = 6: 1 + y + 3 = 6 => y + 4 = 6 => y = 2

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

Method 3: Solving using Matrices (Gaussian Elimination and Row Echelon Form)

Matrices provide a concise and organized way to represent and solve systems of linear equations. Gaussian elimination is a systematic method for transforming a matrix into row echelon form, which then allows you to easily solve for the variables.

Steps:

1. Write the augmented matrix: Represent the system of equations as an augmented matrix. The coefficients of the variables form the main part of the matrix, and the constants on the right-hand side of the equations form the last column.

2. Perform row operations to get to row echelon form: Row echelon form is a matrix where:

   • All rows consisting entirely of zeros are at the bottom of the matrix.
   • The first non-zero entry (leading entry) in each row is a 1.
   • The leading entry in each row is to the right of the leading entry in the row above it.

   The allowed row operations are:

   • 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 using back-substitution: Once the matrix is in row echelon form, you can easily solve for the variables by back-substitution, starting from the last row.

Example:

Consider the same system of equations:

• x + y + z = 6
• 2x - y + z = 3
• x + 2y - z = 2

1. Write the augmented matrix:

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

2. Perform row operations to get to row echelon form:

   • Subtract 2 times row 1 from row 2 (R2 = R2 - 2R1):

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

   • Subtract row 1 from row 3 (R3 = R3 - R1):

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

   • Multiply row 2 by -1/3 (R2 = R2 * -1/3):

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

   • Subtract row 2 from row 3 (R3 = R3 - R2):

     [ 1  1  1 | 6 ]
     [ 0  1  1/3 | 3 ]
     [ 0  0 -7/3 | -7 ]
     

   • Multiply row 3 by -3/7 (R3 = R3 * -3/7):

     [ 1  1  1 | 6 ]
     [ 0  1  1/3 | 3 ]
     [ 0  0  1 | 3 ]
     

   • Subtract 1/3 times row 3 from row 2 (R2 = R2 - (1/3)R3):

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

   • Subtract row 3 from row 1 (R1 = R1 - R3):

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

   • Subtract row 2 from row 1 (R1 = R1 - R2):

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

3. Solve for the variables using back-substitution: The matrix is now in reduced row echelon form. We can directly read off the solution:

   • x = 1
   • y = 2
   • z = 3

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

Method 4: Cramer's Rule

Cramer's Rule is another method for solving systems of linear equations using determinants. While it can be elegant, it becomes computationally intensive for larger systems.

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, Dy, and Dz: Replace the first column of the coefficient matrix with the constants on the right-hand side of the equations to get the matrix for Dx. Similarly, replace the second and third columns to get the matrices for Dy and Dz, respectively. Then, calculate the determinants of these matrices.

3. Solve for the variables:

   • x = Dx / D
   • y = Dy / D
   • z = Dz / D

Important Note: Cramer's Rule only works if the determinant of the coefficient matrix (D) is not equal to zero. If D = 0, the system either has no solution or infinitely many solutions.

Example:

Consider the same system of equations:

• x + y + z = 6
• 2x - y + z = 3
• x + 2y - z = 2

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

   D = | 1  1  1 |
       | 2 -1  1 |
       | 1  2 -1 |
   
   D = 1((-1)*(-1) - (1*2)) - 1((2*(-1) - (1*1)) + 1((2*2) - (-1*1))
   D = 1(1 - 2) - 1(-2 - 1) + 1(4 + 1)
   D = -1 + 3 + 5
   D = 7
   

2. Calculate the determinants Dx, Dy, and Dz:

   Dx = | 6  1  1 |
        | 3 -1  1 |
        | 2  2 -1 |
   
   Dx = 6((-1)*(-1) - (1*2)) - 1((3*(-1) - (1*2)) + 1((3*2) - (-1*2))
   Dx = 6(1 - 2) - 1(-3 - 2) + 1(6 + 2)
   Dx = -6 + 5 + 8
   Dx = 7
   
   Dy = | 1  6  1 |
        | 2  3  1 |
        | 1  2 -1 |
   
   Dy = 1((3)*(-1) - (1*2)) - 6((2*(-1) - (1*1)) + 1((2*2) - (3*1))
   Dy = 1(-3 - 2) - 6(-2 - 1) + 1(4 - 3)
   Dy = -5 + 18 + 1
   Dy = 14
   
   Dz = | 1  1  6 |
        | 2 -1  3 |
        | 1  2  2 |
   
   Dz = 1((-1)*(2) - (3*2)) - 1((2*(2) - (3*1)) + 6((2*2) - (-1*1))
   Dz = 1(-2 - 6) - 1(4 - 3) + 6(4 + 1)
   Dz = -8 - 1 + 30
   Dz = 21
   

3. Solve for the variables:

   • x = Dx / D = 7 / 7 = 1
   • y = Dy / D = 14 / 7 = 2
   • z = Dz / D = 21 / 7 = 3

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

Practical Applications

Solving 3x3 systems of linear equations has numerous applications across various disciplines:

• Engineering: Analyzing electrical circuits, structural mechanics, and fluid dynamics often involves solving systems of equations.
• Economics: Modeling supply and demand, analyzing market equilibrium, and optimizing resource allocation.
• Computer Graphics: Transforming and manipulating 3D objects in computer graphics relies on linear algebra and solving systems of equations.
• Statistics: Linear regression and other statistical models often require solving systems of equations to estimate parameters.
• Operations Research: Optimizing logistics, scheduling, and resource management.

Tips and Tricks

• Look for Simplifications: Before applying any method, examine the equations for any obvious simplifications or relationships between the variables.
• Choose the Easiest Method: The best method depends on the specific system of equations. If one variable is easily isolated, substitution might be the best choice. If the coefficients align well for elimination, that method might be more efficient.
• Check Your Solution: Always substitute your solution back into the original equations to verify that it satisfies all three equations. This helps catch any errors made during the solution process.
• Be Organized: Keep your work organized and clearly label each step. This will make it easier to find and correct any errors.
• Practice: The more you practice solving systems of linear equations, the more comfortable and efficient you will become.

Conclusion

Solving a 3x3 system of linear equations is a fundamental skill in mathematics and various applied fields. By mastering the substitution, elimination, Gaussian elimination, and Cramer's Rule methods, you can effectively tackle these problems. Remember to choose the most appropriate method for each specific system, stay organized, and always check your solution. With practice, you'll become proficient in solving these systems and applying them to real-world problems. Understanding the underlying principles and practicing regularly will significantly enhance your problem-solving abilities and open doors to more advanced concepts in mathematics and its applications.