Systems Of Linear Equations In 3 Variables

Let's explore the fascinating world of systems of linear equations in three variables, a cornerstone of algebra with widespread applications in various fields. We will delve into what these systems are, how to solve them, and why they are so important.

Introduction to Systems of Linear Equations in 3 Variables

A system of linear equations in three variables is a set of two or more linear equations, each containing the same three variables (typically x, y, and z). The goal is to find values for these variables that satisfy all equations simultaneously. This solution, if it exists, represents the point where all the planes represented by these equations intersect.

A single linear equation in three variables represents a plane in three-dimensional space. Therefore, solving a system of these equations involves finding the intersection of these planes. This intersection can be a single point, a line, or no intersection at all.

Understanding Linear Equations in 3 Variables

Before tackling systems, let's ensure we understand the basic building block: the linear equation in three variables.

General Form: The standard form of a linear equation in three variables is:

Ax + By + Cz = D

where:

• A, B, and C are the coefficients of the variables x, y, and z, respectively.
• D is a constant term.
• A, B, C, and D are real numbers.

Geometric Interpretation: As mentioned, a linear equation in this form represents a plane in three-dimensional space. Each point (x, y, z) that satisfies the equation lies on this plane.

Example: Consider the equation 2x + y - z = 5. This represents a plane. One point that lies on this plane is (2, 1, 0) because 2(2) + 1 - 0 = 5. There are infinitely many points that satisfy this equation, all forming the plane.

Methods for Solving Systems of Linear Equations in 3 Variables

Several methods can be used to solve systems of 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 equations.
2. Elimination Method (also known as the Addition Method): This involves adding or subtracting multiples of equations to eliminate one variable at a time.
3. Gaussian Elimination and Row Echelon Form: A systematic approach using matrix operations to transform the system into a more easily solvable form.
4. Using Matrices and Determinants (Cramer's Rule): A method utilizing determinants to directly calculate the values of the variables, applicable when the determinant of the coefficient matrix is non-zero.

Let's explore each method in detail with examples:

1. Substitution Method

The substitution method aims to reduce the system of three equations to a system of two equations by expressing one variable in terms of the others.

Steps:

1. Solve for one variable: Choose one equation 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 look easiest to isolate.
2. Substitute: Substitute the expression you found in Step 1 into the other two equations. This will eliminate the variable you solved for, leaving you with two equations in two variables.
3. Solve the new system: Solve the resulting system of two equations in two variables using any method (substitution or elimination).
4. Back-substitute: Once you have the values for two variables, substitute them back into any of the original equations (or the expression you found in Step 1) to find the value of the third variable.

Example:

Consider the following system:

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

Solution:

1. Solve for x in Equation 1: x = 6 - y - z

2. Substitute into Equation 2 and Equation 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

3. Solve the new system: We now have the system:

   • -3y - z = -9
   • y - 2z = -4

   Let's solve the second equation for y: y = 2z - 4. Substitute this into the first equation:

   • -3(2z - 4) - z = -9 => -6z + 12 - z = -9 => -7z = -21 => z = 3
   • Now, substitute z = 3 back into y = 2z - 4: y = 2(3) - 4 => y = 2

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

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

2. Elimination Method (Addition Method)

The elimination method involves strategically adding or subtracting multiples of equations to eliminate one variable at a time.

Steps:

1. Choose a variable to eliminate: Select a variable you want to eliminate. Look for equations where the coefficients of that variable are either the same or easily made the same (or opposites) by multiplying the entire equation by a constant.
2. Multiply equations (if necessary): Multiply one or both equations by a constant so that the coefficients of the chosen variable are opposites.
3. Add or subtract equations: Add the equations together. This will eliminate the chosen variable.
4. Solve the new system: You'll now have a system of two equations with two variables. Solve this system using either substitution or elimination.
5. Back-substitute: Once you have the values for two variables, substitute them back into any of the original equations to find the value of the third variable.

Example:

Using the same system as before:

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

Solution:

1. Eliminate y from Equation 1 and Equation 2: Notice that the y terms have opposite signs. Adding Equation 1 and Equation 2 directly will eliminate y:

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

2. Eliminate y from Equation 1 and Equation 3: To eliminate y, multiply Equation 1 by -2:

   • -2(x + y + z) = -2(6) => -2x - 2y - 2z = -12
   • Now add this to Equation 3: (-2x - 2y - 2z) + (x + 2y - z) = -12 + 2 => -x - 3z = -10 (Equation 5)

3. Solve the new system: We now have the system:

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

   Multiply Equation 5 by 3: -3x - 9z = -30. Now 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

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

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

3. Gaussian Elimination and Row Echelon Form

Gaussian elimination is a systematic method for solving systems of linear equations by transforming the system's augmented matrix into row echelon form or reduced row echelon form. This method is particularly useful for larger systems.

Steps:

1. Write the augmented matrix: Represent the system of equations as an augmented matrix. The augmented matrix consists of the coefficients of the variables and the constants, separated by a vertical line.

2. Perform row operations: Use elementary row operations to transform the matrix into row echelon form or reduced row echelon form. 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. Row Echelon Form: A matrix is in row echelon form if:

   • All non-zero rows are above any rows of all zeros.
   • The leading coefficient (the first non-zero number from the left) of a non-zero row is always strictly to the right of the leading coefficient of the row above it.

4. Reduced Row Echelon Form: A matrix is in reduced row echelon form if it is in row echelon form and also satisfies:

   • The leading coefficient in each non-zero row is 1.
   • Each leading 1 is the only non-zero entry in its column.

5. Solve for the variables: Once the matrix is in row echelon form or reduced row echelon form, solve for the variables using back-substitution.

Example:

Using the same system:

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

Solution:

1. Augmented Matrix:

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

2. Row Operations:

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

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

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

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

   • R2 <-> R3 (Swap row 2 and row 3):

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

   • R3 -> R3 + 3R2 (Add 3 times row 2 to row 3):

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

   • R3 -> R3 / -7 (Divide row 3 by -7):

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

   • R2 -> R2 + 2R3 (Add 2 times row 3 to row 2):

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

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

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

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

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

3. Solve: The matrix is now in reduced row echelon form. We can directly read off the solution:

   • x = 1
   • y = 2
   • z = 3

The solution is (1, 2, 3).

4. Using Matrices and Determinants (Cramer's Rule)

Cramer's Rule provides a direct way to solve for the variables using determinants. It's applicable only when the determinant of the coefficient matrix is non-zero (meaning the system has a unique solution).

Steps:

1. Form the coefficient matrix (D): Create a matrix using the coefficients of the variables in the system of equations.
2. Calculate the determinant of D (|D|): Find the determinant of the coefficient matrix. If |D| = 0, Cramer's Rule cannot be used (the system either has no solution or infinitely many solutions).
3. Form matrices Dx, Dy, and Dz: To find x, replace the first column of D with the constants from the right-hand side of the equations to create matrix Dx. Similarly, replace the second column to create Dy and the third column to create Dz.
4. Calculate the determinants |Dx|, |Dy|, and |Dz|: Find the determinants of these new matrices.
5. Solve for x, y, and z:
   • x = |Dx| / |D|
   • y = |Dy| / |D|
   • z = |Dz| / |D|

Calculating the Determinant of a 3x3 Matrix:

For a 3x3 matrix:

| a b c |
| d e f |
| g h i |

The determinant is calculated as:

• |D| = a(ei - fh) - b(di - fg) + c(dh - eg)

Example:

Using the same system:

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

Solution:

1. Coefficient Matrix (D):

   [ 1  1  1 ]
   [ 2 -1  1 ]
   [ 1  2 -1 ]
   

2. Determinant of D (|D|):

   • |D| = 1((-1)(-1) - 12) - 1(2*(-1) - 11) + 1(22 - (-1)*1)
   • |D| = 1(1 - 2) - 1(-2 - 1) + 1(4 + 1)
   • |D| = 1(-1) - 1(-3) + 1(5)
   • |D| = -1 + 3 + 5 = 7

3. Matrices Dx, Dy, Dz:

   • Dx:

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

   • Dy:

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

   • Dz:

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

4. Determinants |Dx|, |Dy|, |Dz|:

   • |Dx| = 6((-1)(-1) - 12) - 1(3*(-1) - 12) + 1(32 - (-1)*2) = 6(-1) - 1(-5) + 1(8) = -6 + 5 + 8 = 7
   • |Dy| = 1(3*(-1) - 12) - 6(2(-1) - 11) + 1(22 - 3*1) = 1(-5) - 6(-3) + 1(1) = -5 + 18 + 1 = 14
   • |Dz| = 1((-1)2 - 32) - 1(22 - 31) + 6(2*2 - (-1)*1) = 1(-8) - 1(1) + 6(5) = -8 - 1 + 30 = 21

5. Solve:

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

The solution is (1, 2, 3).

Possible Solutions: Unique, Infinite, or No Solution

When solving a system of linear equations in three variables, there are three possible outcomes:

• Unique Solution: The system has exactly one solution, represented by a single ordered triple (x, y, z). Geometrically, this means the three planes intersect at a single point.
• Infinite Solutions: The system has infinitely many solutions. This occurs when the equations are dependent, meaning one or more equations can be derived from the others. Geometrically, this means the planes intersect along a line or are coincident (the same plane).
• No Solution: The system has no solution. This occurs when the equations are inconsistent, meaning there is no set of values for x, y, and z that can satisfy all equations simultaneously. Geometrically, this means the planes do not have a common intersection point; they might be parallel or intersect in pairs without a common intersection.

Identifying the Type of Solution:

• Gaussian Elimination: In Gaussian elimination, a unique solution results in a reduced row echelon form with a leading 1 in each column corresponding to a variable. Infinite solutions result in a row of zeros, indicating a dependent system. No solution results in a row of the form [0 0 0 | c], where c is a non-zero constant, indicating an inconsistent system.
• Cramer's Rule: Cramer's Rule can only be used to find a unique solution. If the determinant of the coefficient matrix (|D|) is zero, the system either has infinite solutions or no solution.

Applications of Systems of Linear Equations in 3 Variables

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

• Engineering: Solving for currents in electrical circuits, analyzing structural stresses, and modeling fluid flow.
• Physics: Describing the motion of objects in three-dimensional space, solving problems in optics, and analyzing forces.
• Computer Graphics: Transforming objects in 3D space, rendering images, and creating animations.
• Economics: Modeling supply and demand in multiple markets.
• Statistics: Performing regression analysis and solving for parameters in statistical models.
• Chemistry: Balancing chemical equations.

Tips for Solving Systems of Linear Equations

• Stay Organized: Keep your work neat and organized to avoid errors. Label your equations and steps clearly.
• Check Your Solutions: After finding a solution, substitute the values of x, y, and z back into the original equations to verify that they are satisfied.
• Choose the Right Method: Consider the specific system of equations and choose the method that seems most efficient. Substitution is good when one variable is easily isolated. Elimination is good when coefficients are easily made opposites. Gaussian elimination is good for larger systems. Cramer's rule is good for systems with a unique solution when you only need to find the value of one variable.
• Be Careful with Signs: Pay close attention to signs when multiplying and adding equations. A small sign error can lead to an incorrect solution.
• Practice, Practice, Practice: The more you practice solving systems of linear equations, the better you will become at it.

In conclusion, understanding and mastering systems of linear equations in three variables is a valuable skill with broad applications. By understanding the different methods for solving these systems and practicing regularly, you can confidently tackle a wide range of problems in mathematics, science, and engineering.