Gaussian Elimination & Row Echelon Form

The Organic Chemistry Tutor
17 Feb 201818:40
EducationalLearning
32 Likes 10 Comments

TLDRThis video tutorial demonstrates the process of using Gaussian elimination to solve a system of linear equations with three variables. It begins by converting the system into an augmented matrix and then systematically transforms it into row echelon form. Through a series of matrix row operations, the video shows how to achieve a diagonal of ones with zeros below, which simplifies the system for easier solution. The method is applied to two different systems of equations, with the second example showing how to solve the system without fully converting it to row echelon form. The final solutions are presented clearly, illustrating the effectiveness of Gaussian elimination for finding the values of the variables.

Takeaways
  • 📌 Gaussian elimination is a method used to solve systems of linear equations with three variables.
  • 🔢 The first step is to convert the system of equations into an augmented matrix, which includes the coefficients and constants.
  • 📐 The goal is to transform the augmented matrix into row echelon form, where the diagonal entries are 1s and the entries below are 0s.
  • ➕ Row operations, such as adding or swapping rows, are used to manipulate the matrix into the desired form.
  • 🔄 The process involves making the first non-zero entry of each row (leading entry) equal to 1, known as the pivot.
  • 📈 After achieving row echelon form, back substitution can be used to solve for the variables x, y, and z.
  • 🤔 The script provides a step-by-step example of solving a system of equations using Gaussian elimination and back substitution.
  • 📝 The example demonstrates how to perform row operations to achieve the necessary form for solution and how to interpret the results.
  • 🌟 The video emphasizes the importance of careful calculation and attention to detail when performing row operations.
  • 🎯 The method can be applied to different systems of equations, as shown in the second example provided in the script.
  • 📋 The script serves as a guide for those looking to understand and apply Gaussian elimination to solve systems of linear equations.
Q & A

  • What is the main topic of the video?

    -The main topic of the video is using Gaussian elimination to solve a system of equations with three variables.

  • How many equations are given in the problem?

    -There are three equations given in the problem.

  • What is the augmented matrix for the first set of equations?

    -The augmented matrix for the first set of equations is: [ 1 1 -1 | -2 ] [ 2 -1 1 | 5 ] [-1 2 2 | 1 ]

  • What is the goal when converting the matrix into row echelon form?

    -The goal when converting the matrix into row echelon form is to have a diagonal of ones and zeros below the diagonal.

  • What is the first row operation performed on the matrix?

    -The first row operation performed is adding row one to row three.

  • How is the second row transformed to have a zero in the second column?

    -The second row is transformed by multiplying row one by -2 and then adding it to row two.

  • What is the solution for the first set of equations after applying Gaussian elimination?

    -The solution for the first set of equations is x = 1, y = -1, and z = 2.

  • What is the second set of equations given in the video?

    -The second set of equations given are: 2X + y - Z = 1 3x + 2y + Z = 10 2x - y + 2z = 6

  • How is the second set of equations initially transformed?

    -The second set of equations is initially transformed by converting it into an augmented matrix and then performing a series of row operations to make the leading coefficients zeros.

  • What is the solution for Z in the second set of equations?

    -The solution for Z in the second set of equations is Z = 3.

  • What is the final solution for the second set of equations?

    -The final solution for the second set of equations is x = 1, y = 2, and z = 3.

Outlines
00:00
🧮 Solving a System of Equations Using Gaussian Elimination

This video begins by introducing a problem involving a system of equations with three variables: x, y, and z. The equations are converted into an augmented matrix format. The tutorial demonstrates the process of converting the matrix into row echelon form through specific matrix row operations. The initial operations focus on making pivotal elements into ones and other elements in the column into zeros, starting with the transformation of the matrix by adding and modifying rows. Detailed step-by-step calculations are presented, showing the viewer how to systematically manipulate the matrix to achieve the desired form.

05:02
🔍 Completing the Gaussian Elimination Process

Continuing from the previous steps, this section deals with further manipulation of the matrix to solve for variables x, y, and z. It explains how to make specific matrix entries zero and then how to normalize rows to have leading ones. The tutorial progresses to convert the matrix entries back into a system of linear equations and then uses back-substitution to solve for z, y, and finally x. Detailed explanations and mathematical operations demonstrate converting matrix rows to simplified forms and then extracting and solving linear equations from these forms.

10:04
📉 Using Gaussian Elimination for Another System

This segment introduces a new set of equations and converts them into an augmented matrix, focusing on making strategic zeros to simplify the matrix without fully converting it to row echelon form. Specific operations include subtracting and adding rows, and adjusting matrix entries through multiplication to make certain elements zero. The resulting simplified matrix allows for solving the equations directly for z, y, and x, showcasing a practical application of Gaussian elimination to derive solutions efficiently.

15:04
🧐 Extracting Solutions from a Simplified Matrix

In this final section, the simplified matrix setup from the previous paragraph is used to directly solve for z, y, and x. The video demonstrates how to derive z from the simplest equation and then use its value to solve for y in the next simplest equation. Finally, x is derived by substituting the values of y and z into the first equation. The section efficiently illustrates how Gaussian elimination, even without full conversion to row echelon form, can be effectively used to quickly find solutions to a system of equations.

Mindmap
Keywords
💡Gaussian Elimination
Gaussian Elimination is a method used to solve systems of linear equations. It involves transforming the system's matrix into an upper triangular form through a series of row operations. In the video, this method is used to simplify the system of equations and find the values of the variables X, Y, and Z. The process starts with converting the system of equations into an augmented matrix and ends with back substitution to find the solution.
💡Augmented Matrix
An Augmented Matrix is a matrix that combines the coefficients of a system of linear equations with the constants from the right-hand side of the equations. It is used as a visual tool to perform row operations and solve the system. In the video, the coefficients and constants of the given equations are arranged into an augmented matrix to apply Gaussian Elimination.
💡Row Echelon Form
Row Echelon Form is a particular way of arranging the rows of a matrix in a simplified structure, where the leading coefficient (the first non-zero number from the left, also called the pivot) of each row is 1, and the entries below the pivot are zeros. This form makes it easier to perform back substitution and find the solutions to the system of equations. The video aims to convert the augmented matrix into this form before solving.
💡Back Substitution
Back Substitution is a technique used to find the values of the variables in a system of linear equations, after the matrix has been simplified into an upper triangular or row echelon form. Starting with the equation at the bottom, the value of the last variable is determined, and this value is then substituted back into the previous equations to find the other variables. In the video, back substitution is used to solve for Z, Y, and X after the matrix is in a form that allows for easy solution.
💡Row Operations
Row Operations are a set of mathematical operations performed on the rows of a matrix during Gaussian Elimination or similar matrix manipulation processes. These operations include swapping rows, multiplying a row by a non-zero scalar, and adding or subtracting one row from another. The goal is to transform the matrix into a simpler form, like row echelon form or reduced row echelon form, to solve the system of equations.
💡Pivot
In the context of Gaussian Elimination and matrix manipulation, a Pivot is the first non-zero entry in a row that is used to create zeros in the column below it. The process of turning a matrix into row echelon form or reduced row echelon form involves selecting a pivot for each row and using it to simplify the matrix. Pivots play a crucial role in the row reduction process and are chosen to avoid division by zero.
💡Linear Equations
Linear Equations are mathematical equations in which the highest power of the variables is one. They are used to model relationships between different quantities and are typically written in the form ax + by = c, where a, b, and c are constants, and x and y are variables. The video focuses on solving systems of linear equations using Gaussian Elimination.
💡Variables
In mathematics, a Variable is a symbol, often a letter like X, Y, or Z, that represents an unknown quantity or value that can change. In the context of linear equations, variables are the quantities we are trying to solve for. The video script describes a process of using Gaussian Elimination to find the values of variables in a system of equations.
💡Matrix
A Matrix is a rectangular array of numbers, symbols, or expressions, arranged in rows and columns. In linear algebra, matrices are used to represent systems of linear equations, perform transformations, and solve various mathematical problems. The video discusses converting a system of equations into a matrix form to apply Gaussian Elimination.
💡Systems of Equations
A System of Equations is a set of simultaneous equations that are solved together. These equations typically involve multiple unknown variables and are used to model complex relationships or situations. The video focuses on solving systems of equations with three variables using Gaussian Elimination.
💡Solving Equations
Solving Equations refers to the process of finding the values of the unknowns (variables) that make the equations true. In the context of the video, this involves using Gaussian Elimination to find the values of X, Y, and Z that satisfy all the given equations simultaneously.
Highlights

Introduction to using Gaussian elimination to solve a system of equations with three variables. (Start time: 0s)

Formulation of the initial system of equations with three variables. (Start time: 10s)

Conversion of the system to an augmented matrix. (Start time: 20s)

Explanation of the goal to transform the matrix into row echelon form. (Start time: 30s)

First row operation: adding row one to row three. (Start time: 40s)

Second row operation: converting the 2 into a 0 by applying changes to row two. (Start time: 50s)

Third row operation: adding rows two and three to eliminate the number in the third row, second column. (Start time: 1m 10s)

Conversion of the matrix into row echelon form with a diagonal of ones. (Start time: 1m 30s)

Solution extraction from the row echelon form: identifying the value of Z. (Start time: 1m 50s)

Back substitution to find the values of Y and X. (Start time: 2m 10s)

Final solution of the system of equations: X = 1, Y = -1, Z = 2. (Start time: 2m 30s)

Introduction to another example with a different system of equations. (Start time: 2m 50s)

Conversion of the new system into an augmented matrix. (Start time: 3m 10s)

Performing Gaussian elimination to achieve three zeros in the key positions. (Start time: 3m 30s)

Solving the system without full conversion to row echelon form: identifying the value of Z. (Start time: 3m 50s)

Continued back substitution to find the values of Y and X for the second example. (Start time: 4m 10s)

Final solution of the second system of equations: X = 1, Y = 2, Z = 3. (Start time: 4m 30s)

Transcripts

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