Solving Linear Systems In 3 Variables

Solving linear systems in three variables might seem daunting at first, but with a systematic approach, it becomes a manageable task. Understanding the underlying principles and mastering the techniques will equip you with a powerful tool for solving real-world problems in various fields, from engineering and economics to computer science and data analysis.

Introduction to Linear Systems in 3 Variables

A linear system in three variables consists of three equations, each representing a plane in three-dimensional space. The solution to the system is the point where all three planes intersect, if such a point exists. This intersection can be a single point, a line, or no intersection at all, leading to different types of solutions:

• Unique Solution: The three planes intersect at a single point. This point represents the unique values of the three variables (x, y, z) that satisfy all three equations simultaneously.
• Infinite Solutions: The three planes intersect along a line or coincide entirely. This means there are infinitely many points (x, y, z) that satisfy all three equations.
• No Solution: The three planes do not have a common intersection point. This can happen if the planes are parallel or intersect in pairs but not all together.

Before diving into the methods, let's establish the general form of a linear equation in three variables:

ax + by + cz = d

Where a, b, c, and d are constants, and x, y, and z are the variables. A system of three such equations forms a linear system in three variables.

Methods for Solving Linear Systems in 3 Variables

Several methods can be used to solve these systems, each with its own advantages and disadvantages. We will explore three common methods:

1. Substitution Method
2. Elimination Method
3. Matrix Method (using Gaussian Elimination)

1. Substitution Method

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

Steps:

1. Solve for one variable: Choose one of the three equations and solve it for one of the variables (e.g., solve for x in terms of y and z). Pick the equation and variable that appear easiest to isolate.
2. Substitute: Substitute the expression obtained in step 1 into the other two equations. This will result in two equations with two variables.
3. Solve the 2x2 system: Solve the resulting system of two equations in two variables using either substitution or elimination method.
4. Back-substitute: Once you have the values of two variables, substitute them back into the expression obtained in step 1 to find the value of the third variable.
5. Verify the solution: Substitute the values of all three variables into the original three equations to ensure they satisfy all equations.

Example:

Solve the following system of equations:

1. x + y + z = 6
2. 2x - y + z = 3
3. x + 2y - z = 2

• Step 1: Solve for one variable

  From equation (1), solve for x:

  x = 6 - y - z

• Step 2: Substitute

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

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

• Step 3: Solve the 2x2 system

  Now we have a system of two equations with two variables:

  1. -3y - z = -9
  2. y - 2z = -4

  Solve equation (5) for y: y = 2z - 4

  Substitute this expression for y into equation (4):

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

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

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

• Step 4: Back-substitute

  Substitute y = 2 and z = 3 back into the expression for x:

  x = 6 - y - z => x = 6 - 2 - 3 => x = 1

• Step 5: Verify the solution

  The solution is x = 1, y = 2, z = 3. Verify this solution by substituting these values into the original equations:

  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 to the system is (1, 2, 3).

2. Elimination Method

The elimination method involves adding or subtracting multiples of the equations to eliminate one variable at a time. This process is repeated until only one variable remains, which can then be solved. The values of the other variables are found by back-substitution.

Steps:

1. Choose a variable to eliminate: Select a variable that you want to eliminate first. Look for equations where the coefficients of that variable are either the same or easily made the same (or opposite) by multiplying the equation by a constant.
2. Eliminate the variable from two equations: Multiply one or both of the chosen equations by constants so that the coefficients of the variable you want to eliminate are opposites. Add the equations together; this will eliminate that variable.
3. Repeat the process: Repeat steps 1 and 2 with a different pair of equations (usually one of the original equations and the new equation obtained in step 2), eliminating the same variable. This will leave you with two equations in two variables.
4. Solve the 2x2 system: Solve the resulting system of two equations in two variables using either substitution or elimination method.
5. Back-substitute: Once you have the values of two variables, substitute them back into one of the original equations to find the value of the third variable.
6. Verify the solution: Substitute the values of all three variables into the original three equations to ensure they satisfy all equations.

Example:

Solve the following system of equations:

1. x + y + z = 6
2. 2x - y + z = 3
3. x + 2y - z = 2

• Step 1: Choose a variable to eliminate

  Let's eliminate y first.

• Step 2: Eliminate the variable from two equations

  Add equation (1) and equation (2) to eliminate y:

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

• Step 3: Repeat the process

  Multiply equation (2) by 2 to get 4x - 2y + 2z = 6. Then add this to equation (3):

  (4x - 2y + 2z) + (x + 2y - z) = 6 + 2 => 5x + z = 8 (5)

• Step 4: Solve the 2x2 system

  Now we have a system of two equations with two variables:

  1. 3x + 2z = 9
  2. 5x + z = 8

  Multiply equation (5) by -2 to get -10x - 2z = -16. Add this to equation (4):

  (3x + 2z) + (-10x - 2z) = 9 + (-16) => -7x = -7 => x = 1

  Substitute x = 1 back into equation (5):

  5(1) + z = 8 => 5 + z = 8 => z = 3

• Step 5: Back-substitute

  Substitute x = 1 and z = 3 back into equation (1):

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

• Step 6: Verify the solution

  The solution is x = 1, y = 2, z = 3. We already verified this solution in the substitution method example.

  Therefore, the solution to the system is (1, 2, 3).

3. Matrix Method (Gaussian Elimination)

The matrix method provides a systematic way to solve linear systems using matrices. Gaussian elimination is a common technique used to transform the augmented matrix into row-echelon form, from which the solution can be easily obtained.

Steps:

1. Write the augmented matrix: Represent the system of equations as an augmented matrix. This matrix consists of the coefficients of the variables and the constants on the right-hand side of the equations.

2. Perform row operations to get to row-echelon form: Apply elementary row operations to transform the matrix into row-echelon form. Row-echelon form is achieved when:

   • The first non-zero entry in each row (called the leading entry) is 1.
   • The leading entry in each row is to the right of the leading entry in the row above it.
   • Rows with all zero entries are at the bottom of the matrix.

   The elementary row operations are:

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

3. Back-substitution: Once the matrix is in row-echelon form, use back-substitution to find the values of the variables. Start with the last row and work your way up.

4. Verify the solution: Substitute the values of all three variables into the original three equations to ensure they satisfy all equations.

Example:

Solve the following system of equations using Gaussian elimination:

1. x + y + z = 6
2. 2x - y + z = 3
3. x + 2y - z = 2

• Step 1: Write the augmented matrix

  The augmented matrix is:

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

• Step 2: Perform row operations to get to row-echelon form

  • R2 = R2 - 2*R1: (Subtract 2 times the first row from the second row)

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

  • R3 = R3 - R1: (Subtract the first row from the third row)

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

  • R2 = R2 / -3: (Divide the second row by -3)

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

  • R3 = R3 - R2: (Subtract the second row from the third row)

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

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

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

• Step 3: Back-substitution

  The matrix is now in row-echelon form. We can write the corresponding system of equations:

  1. x + y + z = 6
  2. y + (1/3)z = 3
  3. z = 3

  From equation (3), we have z = 3.

  Substitute z = 3 into equation (2):

  y + (1/3)(3) = 3 => y + 1 = 3 => y = 2

  Substitute y = 2 and z = 3 into equation (1):

  x + 2 + 3 = 6 => x + 5 = 6 => x = 1

• Step 4: Verify the solution

  The solution is x = 1, y = 2, z = 3. We already verified this solution in the previous examples.

  Therefore, the solution to the system is (1, 2, 3).

Special Cases: Infinite Solutions and No Solution

Not all linear systems have a unique solution. Here's how to identify cases with infinite solutions or no solution when using Gaussian elimination:

• Infinite Solutions: If, during Gaussian elimination, you obtain a row of zeros (e.g., [0 0 0 | 0]), this indicates that the system has infinite solutions. This means the equations are dependent, and one or more equations can be expressed as a linear combination of the others. To express the solutions, you would typically solve for two variables in terms of the third (parameter).

• No Solution: If you obtain a row that looks like [0 0 0 | c] where c is a non-zero constant, this indicates that the system has no solution. This represents a contradiction; the equations are inconsistent and cannot be simultaneously satisfied.

Example (Infinite Solutions):

Consider the system:

1. x + y + z = 3
2. 2x + 2y + 2z = 6
3. 3x + 3y + 3z = 9

Applying Gaussian elimination, you'll quickly find that equations 2 and 3 are just multiples of equation 1. The augmented matrix would reduce to something like:

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

This means x + y + z = 3. We can solve for x in terms of y and z: x = 3 - y - z. Let y = a and z = b, where a and b are parameters. Then x = 3 - a - b. The solution set is (3 - a - b, a, b), where a and b can be any real numbers.

Example (No Solution):

Consider the system:

1. x + y + z = 1
2. x + y + z = 2
3. x - y + z = 0

Applying Gaussian elimination, you might subtract equation 1 from equation 2, resulting in:

[ 1 1 1 | 1 ] [ 0 0 0 | 1 ] [ x - y + z = 0]

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

Practical Applications

Solving linear systems in three variables has numerous practical applications across various disciplines:

• Engineering: Analyzing electrical circuits, structural mechanics, and fluid dynamics often involves solving systems of linear equations. For example, determining the currents in different branches of an electrical circuit requires solving a linear system derived from Kirchhoff's laws.

• Economics: Linear systems are used in economic modeling to analyze supply and demand, equilibrium prices, and resource allocation. For instance, input-output models use linear systems to determine the interdependencies between different sectors of an economy.

• Computer Graphics: In 3D computer graphics, linear algebra, including solving linear systems, is fundamental for transformations, projections, and rendering. Determining the intersection point of a ray of light with a triangle requires solving a linear system.

• Data Analysis: Linear regression, a common technique in data analysis, involves finding the best-fit line or plane for a set of data points. This often requires solving a system of linear equations.

• Chemistry: Balancing chemical equations involves solving a system of linear equations to determine the stoichiometric coefficients.

Tips and Tricks for Success

• Organization is Key: Keep your work organized and clearly label each step. This will help you avoid errors and make it easier to track your progress.
• Double-Check Your Work: Mistakes are common when solving linear systems. Always double-check your arithmetic and substitutions to ensure accuracy.
• Choose the Right Method: Consider the specific system of equations when choosing a method. The substitution method may be easier for systems where one variable is easily isolated, while the elimination method may be more efficient for other systems. The matrix method is generally more systematic, especially for larger systems.
• Practice Makes Perfect: The more you practice solving linear systems, the more comfortable and confident you will become. Work through a variety of examples and try different methods to develop your skills.
• Use Technology: Calculators and computer software can be helpful for solving linear systems, especially for larger or more complex systems. However, it is important to understand the underlying concepts and techniques before relying on technology. Tools like Wolfram Alpha or MATLAB can be used to verify your solutions.

Conclusion

Solving linear systems in three variables is a fundamental skill with wide-ranging applications. By understanding the underlying principles and mastering the various methods, you can confidently tackle these problems and apply them to real-world scenarios. Whether you choose the substitution method, elimination method, or matrix method, remember to stay organized, double-check your work, and practice regularly to develop your proficiency. Understanding how to handle special cases like infinite solutions and no solution is also crucial for a complete understanding.