Solving 3x3 Systems Of Equations A Comprehensive Guide

by Admin 55 views

Solving systems of equations can seem daunting, especially when dealing with 3x3 systems. However, with a systematic approach and a clear understanding of the underlying principles, these problems can be tackled effectively. This article aims to provide a comprehensive guide on how to solve 3x3 systems of equations, using the example provided:

2x - 3y - 2z = 4
x + 3y + 2z = -7
-4x - 4y - 2z = 10

We will explore various methods, including elimination, substitution, and matrix operations, to find the solution to this system. By the end of this guide, you will have a solid understanding of how to approach and solve 3x3 systems of equations.

Understanding 3x3 Systems of Equations

Before diving into the solution, let's first understand what a 3x3 system of equations entails. A 3x3 system consists of three linear equations, each containing three variables, typically denoted as x, y, and z. The goal is to find the values of these variables that satisfy all three equations simultaneously. Geometrically, each equation represents a plane in three-dimensional space, and the solution to the system corresponds to the point where all three planes intersect. This intersection point, if it exists, represents the unique solution to the system.

In the given system:

2x - 3y - 2z = 4
x + 3y + 2z = -7
-4x - 4y - 2z = 10

We have three equations with three unknowns. The challenge lies in manipulating these equations in a way that allows us to isolate the variables and determine their values. There are several methods to achieve this, each with its own advantages and disadvantages. We will explore the most common methods, including elimination and substitution, and demonstrate how they can be applied to solve the system effectively. Additionally, we will touch upon matrix operations, which provide a more streamlined approach for solving larger systems of equations.

Method 1: Elimination

The elimination method is a powerful technique for solving systems of equations. The core idea is to manipulate the equations in such a way that one or more variables are eliminated, thereby reducing the system to a simpler form. This is achieved by adding or subtracting multiples of the equations from each other. The goal is to create equations where the coefficients of one variable are opposites, so that when the equations are added, that variable cancels out. In our example system:

2x - 3y - 2z = 4  (1)
x + 3y + 2z = -7  (2)
-4x - 4y - 2z = 10 (3)

Notice that the coefficients of 'y' and 'z' in equations (1) and (2) are opposites. This makes them ideal candidates for elimination. Adding equations (1) and (2) directly eliminates both 'y' and 'z':

(2x - 3y - 2z) + (x + 3y + 2z) = 4 + (-7)
3x = -3
x = -1

Now that we have found the value of 'x', we can use it to simplify the system further. We can choose any two equations from the original system and substitute x = -1 into them. Let's use equations (2) and (3):

(-1) + 3y + 2z = -7  (4)
-4(-1) - 4y - 2z = 10 (5)

Simplifying these equations, we get:

3y + 2z = -6  (4)
-4y - 2z = 6   (5)

Now, we can add equations (4) and (5) to eliminate 'z':

(3y + 2z) + (-4y - 2z) = -6 + 6
-y = 0
y = 0

With 'x' and 'y' determined, we can substitute their values into any of the original equations to find 'z'. Let's use equation (2):

(-1) + 3(0) + 2z = -7
2z = -6
z = -3

Therefore, the solution to the system of equations is x = -1, y = 0, and z = -3.

Method 2: Substitution

The substitution method is another effective technique for solving systems of equations. It involves solving one equation for one variable and then substituting that expression into the other equations. This reduces the number of variables in the system, making it easier to solve. To illustrate this method, let's revisit our example system:

2x - 3y - 2z = 4  (1)
x + 3y + 2z = -7  (2)
-4x - 4y - 2z = 10 (3)

We can start by solving equation (2) for 'x':

x = -7 - 3y - 2z  (4)

Now, substitute this expression for 'x' into equations (1) and (3):

2(-7 - 3y - 2z) - 3y - 2z = 4  (5)
-4(-7 - 3y - 2z) - 4y - 2z = 10 (6)

Simplifying these equations, we get:

-14 - 6y - 4z - 3y - 2z = 4
-9y - 6z = 18  (7)

28 + 12y + 8z - 4y - 2z = 10
8y + 6z = -18 (8)

Now we have a system of two equations with two variables, 'y' and 'z'. We can multiply equation (8) by 9 and equation (7) by 8 to eliminate the variable 'y':

72y + 54z = -162
-72y - 48z = 144

Adding them up, we get:

6z = -18
z = -3

Substituting z = -3 into equation (8):

8y + 6(-3) = -18
8y - 18 = -18
8y = 0
y = 0

Finally, substituting y = 0 and z = -3 into equation (4):

x = -7 - 3(0) - 2(-3)
x = -7 + 6
x = -1

Thus, the solution obtained through the substitution method is x = -1, y = 0, and z = -3, which matches the result obtained using the elimination method. The substitution method is particularly useful when one of the equations can be easily solved for one variable in terms of the others.

Method 3: Matrix Operations

Matrix operations offer a more streamlined and efficient approach for solving systems of equations, especially for larger systems. This method involves representing the system of equations in matrix form and then using techniques such as Gaussian elimination or matrix inversion to find the solution. The matrix representation of our example system is:

| 2 -3 -2 |   | x |   | 4 |
| 1  3  2 | * | y | = | -7|
|-4 -4 -2 |   | z |   | 10|

This can be written in the form AX = B, where A is the coefficient matrix, X is the variable matrix, and B is the constant matrix.

A = | 2 -3 -2 |
    | 1  3  2 |
    |-4 -4 -2 |

X = | x |
    | y |
    | z |

B = | 4 |
    | -7|
    | 10|

Gaussian Elimination

Gaussian elimination is a method to transform the augmented matrix [A|B] into row-echelon form or reduced row-echelon form. This is done by performing elementary row operations, which include:

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

Let's form the augmented matrix for our system:

| 2 -3 -2  4 |
| 1  3  2 -7 |
|-4 -4 -2 10 |

We'll perform row operations to get this matrix into reduced row-echelon form. First, we swap row 1 and row 2:

| 1  3  2 -7 |
| 2 -3 -2  4 |
|-4 -4 -2 10 |

Next, we replace row 2 with row 2 - 2 * row 1 and row 3 with row 3 + 4 * row 1:

| 1  3  2 -7 |
| 0 -9 -6 18 |
| 0  8  6 -18|

Divide row 2 by -9:

| 1  3  2 -7 |
| 0  1 2/3 -2 |
| 0  8  6 -18|

Replace row 3 with row 3 - 8 * row 2:

| 1  3  2 -7 |
| 0  1 2/3 -2 |
| 0  0 -2/3 -2|

Multiply row 3 by -3/2:

| 1  3  2 -7 |
| 0  1 2/3 -2 |
| 0  0  1  3 |

Now, we perform back-substitution. From the last row, we have z = -3. From the second row:

y + (2/3)z = -2
y + (2/3)(-3) = -2
y - 2 = -2
y = 0

From the first row:

x + 3y + 2z = -7
x + 3(0) + 2(-3) = -7
x - 6 = -7
x = -1

So, the solution is x = -1, y = 0, and z = -3.

Matrix Inversion

If the coefficient matrix A is invertible, we can find the solution by multiplying both sides of the equation AX = B by the inverse of A:

X = A^(-1)B

For a 3x3 matrix, finding the inverse can be computationally intensive, but it's a straightforward process. However, for simplicity, we won't demonstrate the manual computation of the inverse here. Instead, we'll use the result we obtained from Gaussian elimination to confirm the solution.

The solution x = -1, y = 0, and z = -3 can be verified by substituting these values into the original equations, confirming the accuracy of our calculations using matrix operations.

Conclusion

In this article, we have thoroughly explored methods for solving a 3x3 system of equations. We demonstrated three primary techniques: elimination, substitution, and matrix operations. Each method offers a unique approach to tackling these systems, and the choice of method often depends on the specific characteristics of the equations. Whether you prefer the systematic approach of elimination, the variable isolation of substitution, or the streamlined efficiency of matrix operations, understanding these techniques empowers you to solve a wide range of mathematical problems. The solution to the example system is x = -1, y = 0, and z = -3, a result consistently obtained across all methods discussed. By mastering these methods, you'll be well-equipped to handle any 3x3 system of equations you encounter.