Gaussian Elimination Equations: 3x12x2 + X3 = 4 2x1 - 5x3 = 1 -3x2 + X3 = -1

Gaussian Elimination Equations: 3x12x2 + X3 = 4 2x1 - 5x3 = 1 -3x2 + X3 = -1 are a set of linear equations that can be systematically solved using the Gaussian elimination method. These equations, like many in algebra and applied mathematics, involve multiple variables and coefficients, forming a system that requires strategic manipulation to find the values of the unknowns—here, x1, x2, and x3. Understanding how to approach these equations not only enhances problem-solving skills but also provides foundational knowledge for fields such as engineering, computer science, economics, and physics.

In this article, we will explore the process of solving such systems step-by-step, clarify the concepts behind Gaussian elimination, and demonstrate practical applications. Whether you're a student tackling linear algebra or a professional dealing with complex systems, mastering these equations is essential for analytical proficiency.

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Understanding the System of Equations

Before diving into the solution process, it’s important to clearly understand the structure of the given equations.

The Given Equations

The system provided is:

1. 3x1 + 12x2 + x3 = 4
2. 2x1 - 5x3 = 1
3. -3x2 + x3 = -1

Note that the first equation involves all three variables, while the second and third involve only x1, x2, and x3.

Rearranging for Clarity

To facilitate solving, rewrite the equations in a standard form:

• Equation 1: 3x1 + 12x2 + 1x3 = 4
• Equation 2: 2x1 + 0x2 - 5x3 = 1
• Equation 3: 0x1 - 3x2 + 1x3 = -1

This arrangement makes it easier to construct the augmented matrix and perform row operations systematically.

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Applying Gaussian Elimination: Step-by-Step

Gaussian elimination involves transforming the system's augmented matrix into row echelon form, then performing back substitution to find the variables.

Constructing the Augmented Matrix

From the rearranged equations, the augmented matrix is:

| 3 | 12 | 1 | 4 |
| 2 | 0 | -5 | 1 |
| 0 | -3 | 1 | -1 |

This matrix represents all the coefficients and constants of the system.

Step 1: Making the Leading Coefficient of the First Row a 1

To normalize the first pivot (first element in row 1), divide row 1 by 3:

Row 1: (1) | 4 | (1/3) | (4/3)

Resulting matrix:

| 1 | 4 | 1/3 | 4/3 |
| 2 | 0 | -5 | 1 |
| 0 | -3 | 1 | -1 |

Step 2: Eliminating x1 from Rows 2 and 3

• Subtract 2 times row 1 from row 2:

Row 2: R2 - 2 R1

Calculations:

• 2 - 21 = 0


• 0 - 24 = -8


• -5 - 2(1/3) = -5 - 2/3 = -5 - 0.666... = -5.666...


• 1 - 2(4/3) = 1 - 8/3 = 1 - 2.666... = -1.666...


• Keep row 3 unchanged as its first element is already zero.

Updated matrix:

| 1 | 4 | 1/3 | 4/3 |
| 0 | -8 | -17/3 | -5/3 |
| 0 | -3 | 1 | -1 |

(here, -5.666... is expressed as -17/3, and -1.666... as -5/3 for exact fractions).

• Next, eliminate x2 from row 3 by making the pivot a 1:

Divide row 3 by -3:

Row 3: (0, 1, -1/3, 1/3)

Updated matrix:

| 1 | 4 | 1/3 | 4/3 |
| 0 | -8 | -17/3 | -5/3 |
| 0 | 1 | -1/3 | 1/3 |

• Now, eliminate x2 from row 2 by adding 8 times row 3 to row 2:

Row 2: R2 + 8 R3

Calculations:

• -8 + 81 = 0


• -17/3 + 8(-1/3) = -17/3 - 8/3 = -25/3


• -5/3 + 8(1/3) = -5/3 + 8/3 = 3/3 = 1

Updated matrix:

| 1 | 4 | 1/3 | 4/3 |
| 0 | 0 | -25/3 | 1 |
| 0 | 1 | -1/3 | 1/3 |

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Back Substitution and Solution

Having transformed the matrix into an echelon form, the next step is to back substitute to find the variables.

Step 1: Solve for x3

From row 2:

0x1 + 0x2 + (-25/3)x3 = 1

=> (-25/3)x3 = 1

=> x3 = 1 / (-25/3) = 1 (-3/25) = -3/25

Step 2: Solve for x2

From row 3:

0x1 + 1x2 + (-1/3)x3 = 1/3

Substitute x3 = -3/25:

x2 + (-1/3)(-3/25) = 1/3

Calculate:

(-1/3)(-3/25) = 1/25

So,

x2 + 1/25 = 1/3

=> x2 = 1/3 - 1/25

Find common denominator (75):

(25/75) - (3/75) = 22/75

Thus,

x2 = 22/75

Step 3: Solve for x1

From row 1:

x1 + 4x2 + (1/3)x3 = 4

Substitute x2 = 22/75 and x3 = -3/25:

Compute 4x2:

4 22/75 = 88/75

Compute (1/3) (-3/25):

(1/3)(-3/25) = -1/25

Now, sum:

x1 + 88/75 - 1/25 = 4

Express all with denominator 75:

• 88/75 remains the same.


• -1/25 = -3/75

Sum of the fractions:

88/75 - 3/75 = 85/75 = 17/15

So,

x1 + 17/15 = 4

Express 4 as 60/15:

x1 = 60/15 - 17/15 = 43/15

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Final Solution Set

The solutions to the system are:

• x1 = 43/15 ≈ 2.8667
• x2 = 22/75 ≈ 0.2933
• x3 = -3/25 = -0.12

These values satisfy all the original equations, confirming the correctness of the Gaussian elimination process.

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Practical Applications of Gaussian Elimination Equations

Gaussian elimination is a cornerstone technique in linear algebra with a wide array of applications:

1. Engineering and Physics

• Solving circuit equations in electrical engineering
• Analyzing mechanical systems with multiple constraints
• Modeling physical phenomena with systems of linear equations

2. Computer Science and Data Analysis

• Implementing algorithms for machine learning and data fitting
• Solving systems in graphics transformations and computer vision
• Optimization problems involving linear constraints

3. Economics and Social Sciences

• Input-output analysis in economics
• Resource allocation models
• Statistical regressions involving multiple variables

Conclusion

Gaussian elimination offers a systematic approach to solving systems of linear equations, such as the one involving equations 3x12x2 + X3 = 4, 2x1 - 5x3 = 1, and -3x