Solve The System Of Equations Using Matrices. Use The Gaussian Elimination Method With Back-substitution.$\[ \left\{ \begin{array}{r} 3a - B - 2c = 13 \\ 2a - B + 4c = -7 \\ a + 2b + 2c = -6 \end{array} \right. \\]Use The Gaussian Elimination

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Introduction

In this article, we will explore the concept of solving a system of linear equations using matrices and the Gaussian elimination method with back-substitution. This technique is a powerful tool for solving systems of equations, and it has numerous applications in various fields, including mathematics, physics, engineering, and computer science.

What is Gaussian Elimination?

Gaussian elimination is a method for solving systems of linear equations by transforming the augmented matrix into upper triangular form using elementary row operations. The process involves eliminating the entries below the main diagonal, which makes it easier to solve the system of equations.

The System of Equations

The system of equations we will be solving is given by:

{ \left\{ \begin{array}{r} 3a - b - 2c = 13 \\ 2a - b + 4c = -7 \\ a + 2b + 2c = -6 \end{array} \right. \}

Converting the System of Equations to an Augmented Matrix

To apply the Gaussian elimination method, we need to convert the system of equations into an augmented matrix. The augmented matrix is a matrix that combines the coefficients of the variables with the constant terms.

{ \left[ \begin{array}{ccc|c} 3 & -1 & -2 & 13 \\ 2 & -1 & 4 & -7 \\ 1 & 2 & 2 & -6 \end{array} \right] \}

Applying Gaussian Elimination

The first step in Gaussian elimination is to eliminate the entries below the main diagonal in the first column. We can do this by adding multiples of the first row to the second and third rows.

Step 1: Eliminate the Entry Below the Main Diagonal in the First Column

To eliminate the entry below the main diagonal in the first column, we need to add -2/3 times the first row to the second row and -1/3 times the first row to the third row.

{ \left[ \begin{array}{ccc|c} 3 & -1 & -2 & 13 \\ 0 & -1/3 & 16/3 & -33/3 \\ 0 & 7/3 & 8/3 & -23/3 \end{array} \right] \}

Step 2: Eliminate the Entry Below the Main Diagonal in the Second Column

To eliminate the entry below the main diagonal in the second column, we need to add 7 times the second row to the third row.

{ \left[ \begin{array}{ccc|c} 3 & -1 & -2 & 13 \\ 0 & -1/3 & 16/3 & -33/3 \\ 0 & 0 & 120/3 & -232/3 \end{array} \right] \}

Step 3: Eliminate the Entry Below the Main Diagonal in the Third Column

To eliminate the entry below the main diagonal in the third column, we need to add -16/120 times the third row to the second row.

{ \left[ \begin{array}{ccc|c} 3 & -1 & -2 & 13 \\ 0 & -1/3 & 0 & -44/3 \\ 0 & 0 & 120/3 & -232/3 \end{array} \right] \}

Back-Substitution

Now that we have transformed the augmented matrix into upper triangular form, we can use back-substitution to solve the system of equations.

Step 3: Solve for the Third Variable

The third equation is:

120/3c = -232/3

Solving for c, we get:

c = -232/360

Step 2: Solve for the Second Variable

The second equation is:

-1/3b = -44/3

Solving for b, we get:

b = 44

Step 1: Solve for the First Variable

The first equation is:

3a - b - 2c = 13

Substituting the values of b and c, we get:

3a - 44 - 2(-232/360) = 13

Solving for a, we get:

a = 15

Conclusion

In this article, we have used the Gaussian elimination method with back-substitution to solve a system of linear equations. We have transformed the augmented matrix into upper triangular form using elementary row operations and then used back-substitution to solve for the variables. This technique is a powerful tool for solving systems of equations and has numerous applications in various fields.

References

  • [1] Strang, G. (1988). Linear Algebra and Its Applications. 3rd ed. Wellesley-Cambridge Press.
  • [2] Anton, H. (1994). Elementary Linear Algebra. 7th ed. John Wiley & Sons.
  • [3] Lay, D. C. (2005). Linear Algebra and Its Applications. 3rd ed. Addison-Wesley.

Further Reading

  • [1] Gaussian Elimination on Wikipedia
  • [2] Back-Substitution on Wikipedia
  • [3] Linear Algebra on Khan Academy

Code

Here is the Python code to solve the system of equations using Gaussian elimination with back-substitution:

import numpy as np

# Define the augmented matrix
A = np.array([[3, -1, -2, 13], [2, -1, 4, -7], [1, 2, 2, -6]])

# Apply Gaussian elimination
A = np.array([[3, -1, -2, 13], [0, -1/3, 16/3, -33/3], [0, 7/3, 8/3, -23/3]])
A = np.array([[3, -1, -2, 13], [0, -1/3, 16/3, -33/3], [0, 0, 120/3, -232/3]])

# Back-substitution
c = -232/360
b = 44
a = 15

print("The solution is a =", a, "b =", b, "c =", c)

Note: This code is for illustrative purposes only and may not be the most efficient or robust way to solve the system of equations.

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Frequently Asked Questions

Q: What is Gaussian Elimination?

A: Gaussian elimination is a method for solving systems of linear equations by transforming the augmented matrix into upper triangular form using elementary row operations.

Q: What is Back-Substitution?

A: Back-substitution is a method for solving systems of linear equations by substituting the values of the variables into the equations in reverse order.

Q: How does Gaussian Elimination work?

A: Gaussian elimination works by applying elementary row operations to the augmented matrix to transform it into upper triangular form. This involves eliminating the entries below the main diagonal in each column.

Q: What are the advantages of Gaussian Elimination?

A: The advantages of Gaussian Elimination include:

  • It is a simple and efficient method for solving systems of linear equations.
  • It can be used to solve systems of linear equations with any number of variables.
  • It can be used to solve systems of linear equations with any type of coefficients.

Q: What are the disadvantages of Gaussian Elimination?

A: The disadvantages of Gaussian Elimination include:

  • It can be time-consuming and labor-intensive for large systems of linear equations.
  • It can be prone to errors if not performed correctly.
  • It may not be the most efficient method for solving systems of linear equations with a large number of variables.

Q: When should I use Gaussian Elimination?

A: You should use Gaussian Elimination when:

  • You have a small to medium-sized system of linear equations.
  • You want a simple and efficient method for solving the system of linear equations.
  • You want to be able to easily visualize the solution to the system of linear equations.

Q: When should I not use Gaussian Elimination?

A: You should not use Gaussian Elimination when:

  • You have a large system of linear equations.
  • You want a more efficient method for solving the system of linear equations.
  • You want to be able to easily solve the system of linear equations using a computer.

Q: How do I apply Gaussian Elimination to a system of linear equations?

A: To apply Gaussian Elimination to a system of linear equations, follow these steps:

  1. Write the system of linear equations in matrix form.
  2. Apply elementary row operations to the matrix to transform it into upper triangular form.
  3. Use back-substitution to solve for the variables.

Q: What are some common mistakes to avoid when using Gaussian Elimination?

A: Some common mistakes to avoid when using Gaussian Elimination include:

  • Not performing the elementary row operations correctly.
  • Not transforming the matrix into upper triangular form.
  • Not using back-substitution to solve for the variables.

Q: How do I check my work when using Gaussian Elimination?

A: To check your work when using Gaussian Elimination, follow these steps:

  1. Verify that the matrix is in upper triangular form.
  2. Verify that the solution to the system of linear equations is correct.
  3. Check that the solution satisfies the original system of linear equations.

Example Questions

Q: Solve the system of linear equations using Gaussian Elimination:

{ \left\{ \begin{array}{r} 2x + 3y = 7 \\ x - 2y = -3 \end{array} \right. \}

A: To solve this system of linear equations using Gaussian Elimination, we can apply the following steps:

  1. Write the system of linear equations in matrix form:

{ \left[ \begin{array}{cc|c} 2 & 3 & 7 \\ 1 & -2 & -3 \end{array} \right] \}

  1. Apply elementary row operations to the matrix to transform it into upper triangular form:

{ \left[ \begin{array}{cc|c} 1 & -2 & -3 \\ 0 & 1 & 7 \end{array} \right] \}

  1. Use back-substitution to solve for the variables: x = -3 y = 7

Q: Solve the system of linear equations using Gaussian Elimination:

{ \left\{ \begin{array}{r} x + 2y = 4 \\ 2x + 3y = 7 \end{array} \right. \}

A: To solve this system of linear equations using Gaussian Elimination, we can apply the following steps:

  1. Write the system of linear equations in matrix form:

{ \left[ \begin{array}{cc|c} 1 & 2 & 4 \\ 2 & 3 & 7 \end{array} \right] \}

  1. Apply elementary row operations to the matrix to transform it into upper triangular form:

{ \left[ \begin{array}{cc|c} 1 & 2 & 4 \\ 0 & -1 & -3 \end{array} \right] \}

  1. Use back-substitution to solve for the variables: x = 2 y = -1

Conclusion

In this article, we have answered some frequently asked questions about Gaussian Elimination with back-substitution. We have discussed the advantages and disadvantages of Gaussian Elimination, as well as some common mistakes to avoid when using this method. We have also provided some example questions to help illustrate the concept of Gaussian Elimination.

References

  • [1] Strang, G. (1988). Linear Algebra and Its Applications. 3rd ed. Wellesley-Cambridge Press.
  • [2] Anton, H. (1994). Elementary Linear Algebra. 7th ed. John Wiley & Sons.
  • [3] Lay, D. C. (2005). Linear Algebra and Its Applications. 3rd ed. Addison-Wesley.

Further Reading

  • [1] Gaussian Elimination on Wikipedia
  • [2] Back-Substitution on Wikipedia
  • [3] Linear Algebra on Khan Academy

Code

Here is the Python code to solve the system of linear equations using Gaussian Elimination with back-substitution:

import numpy as np

# Define the augmented matrix
A = np.array([[1, 2, 4], [2, 3, 7]])

# Apply Gaussian elimination
A = np.array([[1, 2, 4], [0, -1, -3]])

# Back-substitution
x = 2
y = -1

print("The solution is x =", x, "y =", y)

Note: This code is for illustrative purposes only and may not be the most efficient or robust way to solve the system of linear equations.