Three Linear Equations With Three Variables

Let's delve into the world of linear equations, specifically focusing on systems involving three equations and three variables. This concept is fundamental in algebra and finds applications across various fields, from engineering and economics to computer graphics and data analysis. Understanding how to solve these systems is a crucial skill for anyone seeking to solve complex problems involving multiple interconnected factors.

Navigating the Realm of Three Linear Equations with Three Variables

Imagine a scenario where you need to determine the price of three different items based on three separate transactions. Each transaction provides a total cost and the quantity of each item purchased. This is just one example of a real-world problem that can be modeled and solved using a system of three linear equations with three variables. These systems consist of three equations, each containing three unknown variables (typically represented as x, y, and z). The goal is to find the values for these variables that simultaneously satisfy all three equations.

Comprehensive Overview: What Makes Them Tick?

A linear equation is an algebraic equation in which each term is either a constant or the product of a constant and a single variable. No variable is raised to a power greater than one, and there are no products of variables. In the context of three variables, a linear equation can be represented as:

ax + by + cz = d

Where a, b, c, and d are constants, and x, y, and z are the variables.

A system of three linear equations with three variables takes the form:

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

Here, the subscripts simply differentiate the coefficients and constants in each equation.

Why are they important?

These systems are important because they allow us to model and solve problems with multiple unknowns and constraints. They represent a set of conditions that must be simultaneously met. Finding the solution to the system provides the values for the unknowns that satisfy all the conditions.

Methods for Solving Systems of Three Linear Equations

Several methods exist for solving these systems, each with its own strengths and weaknesses. The most common methods include:

Substitution: This 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 with two variables, which can then be solved using similar techniques.
Elimination (or Addition): This method involves manipulating the equations (by multiplying them by constants) so that when two equations are added together, one of the variables is eliminated. This process is repeated until only one equation with one variable remains.
Gaussian Elimination: This is a systematic approach to elimination that uses elementary row operations to transform the system's augmented matrix into row-echelon form or reduced row-echelon form. From this form, the solution can be easily determined.
Matrix Methods (Using Inverses or Cramer's Rule): These methods involve representing the system as a matrix equation and using matrix algebra to solve for the variables. Cramer's Rule is a formula that uses determinants to find the solution, while finding the inverse of the coefficient matrix is another approach.

Let's explore each of these methods in detail with examples.

1. Substitution Method

Step 1: Solve for one variable in one equation. Choose the equation and variable that looks easiest to isolate. For example, consider the system:
- x + y + z = 6
- 2x - y + z = 3
- x + 2y - z = 2
In the first equation, x is easy to isolate: x = 6 - y - z
Step 2: Substitute this expression into the other two equations. Replace x in the second and third equations with (6 - y - z):
- 2(6 - y - z) - y + z = 3 => 12 - 2y - 2z - y + z = 3 => -3y - z = -9
- (6 - y - z) + 2y - z = 2 => 6 + y - 2z = 2 => y - 2z = -4
Step 3: Solve the resulting system of two equations with two variables. Now we have:
- -3y - z = -9
- y - 2z = -4
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
Step 4: Back-substitute to find the other variables. Substitute z = 3 back into the equation for y:

y = 2(3) - 4 = 2

Now substitute y = 2 and z = 3 back into the equation for x:

x = 6 - 2 - 3 = 1
Step 5: Check your solution. Substitute x = 1, y = 2, and z = 3 into all three original equations to verify that the solution is correct:
- 1 + 2 + 3 = 6 (Correct)
- 2(1) - 2 + 3 = 3 (Correct)
- 1 + 2(2) - 3 = 2 (Correct)
Therefore, the solution is x = 1, y = 2, and z = 3.

2. Elimination (or Addition) Method

Step 1: Choose a variable to eliminate. Look for equations where the coefficients of one variable are either the same or negatives of each other (or can easily be made that way by multiplying one or both equations by a constant). Using the same system as before:
- x + y + z = 6
- 2x - y + z = 3
- x + 2y - z = 2
Notice that the y terms in the first and second equations have coefficients of +1 and -1, respectively.
Step 2: Eliminate the chosen variable from two equations. Add the first and second equations together:

(x + y + z) + (2x - y + z) = 6 + 3 => 3x + 2z = 9
Step 3: Eliminate the same variable from a different pair of equations. To eliminate y again, we need to combine the first and third equations in a way that cancels the y terms. Multiply the first equation by -2:

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

Now add this to the third equation:

(-2x - 2y - 2z) + (x + 2y - z) = -12 + 2 => -x - 3z = -10
Step 4: Solve the resulting system of two equations with two variables. We now have:
- 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
Step 5: Back-substitute to find the other variables. Substitute z = 3 into either of the two-variable equations. Let's use 3x + 2z = 9:

3x + 2(3) = 9 => 3x + 6 = 9 => 3x = 3 => x = 1

Now substitute x = 1 and z = 3 into any of the original equations. Let's use x + y + z = 6:

1 + y + 3 = 6 => y = 2
Step 6: Check your solution. (As in the substitution method, verify the solution in all three original equations).

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

3. Gaussian Elimination

Gaussian elimination is a more systematic approach, especially useful for larger systems. It uses matrix operations, so let's briefly explain that first.

Augmented Matrix: The system of equations is represented as an augmented matrix. For example:
- x + y + z = 6
- 2x - y + z = 3
- x + 2y - z = 2
The augmented matrix is:
```
[ 1  1  1 | 6 ]
[ 2 -1  1 | 3 ]
[ 1  2 -1 | 2 ]
```
Elementary Row Operations: These are operations that don't change the solution of the system:
1. Swapping two rows.
2. Multiplying a row by a non-zero constant.
3. Adding a multiple of one row to another row.
Row-Echelon Form: The goal is to transform the matrix into row-echelon form, where:
- All non-zero rows are above any rows of all zeros.
- The leading coefficient (the first non-zero number from the left, also called the pivot) of a non-zero row is always strictly to the right of the leading coefficient of the row above it.
- All entries in a column below a leading coefficient are zeros.
Reduced Row-Echelon Form: A more restrictive form where, in addition to the conditions for row-echelon form:
- The leading coefficient in each non-zero row is 1.
- Each leading coefficient is the only non-zero entry in its column.
Steps for Gaussian Elimination:
1. Write the augmented matrix. (As shown above)
2. Use elementary row operations to transform the matrix into row-echelon form. We want to get zeros below the '1' in the first column.
  - Subtract 2 times the first row from the second row (R2 = R2 - 2R1):
```
[ 1  1  1 | 6 ]
[ 0 -3 -1 | -9 ]
[ 1  2 -1 | 2 ]
```
  - Subtract the first row from the third row (R3 = R3 - R1):
```
[ 1  1  1 | 6 ]
[ 0 -3 -1 | -9 ]
[ 0  1 -2 | -4 ]
```
  - Swap the second and third rows (R2 <-> R3):
```
[ 1  1  1 | 6 ]
[ 0  1 -2 | -4 ]
[ 0 -3 -1 | -9 ]
```
  - Add 3 times the second row to the third row (R3 = R3 + 3R2):
```
[ 1  1  1 | 6 ]
[ 0  1 -2 | -4 ]
[ 0  0 -7 | -21 ]
```
3. Transform the matrix into reduced row-echelon form (optional, but makes the final step easier).
  - Divide the third row by -7 (R3 = R3 / -7):
```
[ 1  1  1 | 6 ]
[ 0  1 -2 | -4 ]
[ 0  0  1 | 3 ]
```
  - Add 2 times the third row to the second row (R2 = R2 + 2R3):
```
[ 1  1  1 | 6 ]
[ 0  1  0 | 2 ]
[ 0  0  1 | 3 ]
```
  - Subtract the third row from the first row (R1 = R1 - R3):
```
[ 1  1  0 | 3 ]
[ 0  1  0 | 2 ]
[ 0  0  1 | 3 ]
```
  - Subtract the second row from the first row (R1 = R1 - R2):
```
[ 1  0  0 | 1 ]
[ 0  1  0 | 2 ]
[ 0  0  1 | 3 ]
```
4. Read the solution from the matrix. The matrix is now in reduced row-echelon form. The solution is directly readable: x = 1, y = 2, z = 3.

4. Matrix Methods (Using Inverses or Cramer's Rule)

Matrix Representation: The system can be written in matrix form as AX = B, where:

A is the coefficient matrix:
```
[ a1 b1 c1 ]
[ a2 b2 c2 ]
[ a3 b3 c3 ]
```
X is the variable matrix:
```
[ x ]
[ y ]
[ z ]
```
B is the constant matrix:
```
[ d1 ]
[ d2 ]
[ d3 ]
```

For our example:

A = [ 1  1  1 ]
    [ 2 -1  1 ]
    [ 1  2 -1 ]

X = [ x ]
    [ y ]
    [ z ]

B = [ 6 ]
    [ 3 ]
    [ 2 ]

Using the Inverse Matrix: If the determinant of A is not zero, then A has an inverse, denoted as A⁻¹. The solution is then:

X = A⁻¹B

Finding the inverse of a 3x3 matrix can be computationally intensive, but it's a standard procedure in linear algebra. Software like MATLAB, Mathematica, or Python with libraries like NumPy can easily compute matrix inverses.
Cramer's Rule: Cramer's Rule provides a formula to find the solution directly using determinants. Let D be the determinant of the coefficient matrix A. Let Dₓ be the determinant of the matrix formed by replacing the first column of A with B. Similarly, let Dᵧ and D₂ be the determinants of the matrices formed by replacing the second and third columns of A with B, respectively. Then:
- x = Dₓ / D
- y = Dᵧ / D
- z = D₂ / D
Again, calculating determinants for 3x3 matrices can be tedious by hand, but software can easily handle it.

Trends & Recent Developments

The core methods for solving linear equations have been established for centuries. However, the computational power available today has significantly impacted how these systems are solved in practice.

Software Solutions: Symbolic mathematics software and numerical analysis libraries in programming languages make solving large systems of linear equations routine. These tools often use variations of Gaussian elimination or LU decomposition, optimized for speed and accuracy.
Applications in Machine Learning: Solving systems of linear equations is a fundamental operation in many machine learning algorithms. For example, linear regression involves solving a system of equations to find the best-fit line or hyperplane.
Sparse Matrices: Many real-world problems lead to systems of equations where the coefficient matrix is sparse (mostly zeros). Specialized algorithms and data structures are used to efficiently solve these sparse systems.

Tips & Expert Advice

Choose the Right Method: For small systems (3x3 or smaller), substitution or elimination can be quite efficient. For larger systems, Gaussian elimination or matrix methods are generally preferred.
Look for Simplifications: Before diving into a complex solution method, see if there are any obvious simplifications. For example, can you easily eliminate a variable by adding or subtracting two equations?
Check Your Work: It's crucial to check your solution by substituting the values back into the original equations. This will catch any arithmetic errors.
Be Organized: Keep your work organized, especially when using Gaussian elimination. Clearly label each row operation to avoid mistakes.
Use Technology: Don't be afraid to use calculators or software to help with calculations, especially when dealing with large or complex systems.

FAQ (Frequently Asked Questions)

Q: Can a system of three linear equations have no solution?
- A: Yes, if the equations are inconsistent (represent planes that do not intersect at a common point).
Q: Can a system of three linear equations have infinitely many solutions?
- A: Yes, if the equations are dependent (represent planes that intersect in a line or are the same plane).
Q: Is Gaussian elimination always the best method?
- A: Not necessarily. For small systems, substitution or elimination might be faster. However, Gaussian elimination is a more systematic and robust method for larger systems.
Q: What does it mean geometrically when a system has a unique solution?
- A: Geometrically, it means that the three equations represent three planes that intersect at a single point in 3D space.
Q: What is the determinant of a matrix?
- A: The determinant is a scalar value that can be computed from the elements of a square matrix and encodes certain properties of the linear transformation described by the matrix. A non-zero determinant indicates that the matrix is invertible.

Conclusion

Solving systems of three linear equations with three variables is a fundamental skill with widespread applications. Understanding the different methods – substitution, elimination, Gaussian elimination, and matrix methods – allows you to choose the most efficient approach for a given problem. With the aid of technology, these systems can be solved quickly and accurately, enabling you to tackle complex problems in various fields. Remember to always check your solutions and look for simplifications to make the process easier.

How do you feel about tackling a system of three equations now? Are you ready to put these methods into practice?

Three Linear Equations With Three Variables

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