Representing the System as a Matrix

Oct 13, 2025 by sorusgentrification.com 80 views

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Introduction

In mathematics, solving systems of linear equations is a fundamental concept that has numerous applications in various fields, including physics, engineering, and computer science. One of the most efficient methods for solving systems of linear equations is the Gaussian elimination method, which involves transforming the system into upper triangular form using elementary row operations. In this article, we will explore how to solve a system of linear equations using matrices and the Gaussian elimination method with back-substitution.

The System of Equations

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

{ \begin{cases} x + 2y = 0 \\ x + 3y + z = 1 \\ -2x - y - z = -13 \end{cases} \}

This system consists of three linear equations with three variables: x, y, and z. Our goal is to find the values of x, y, and z that satisfy all three equations simultaneously.

Representing the System as a Matrix

To apply the Gaussian elimination method, we need to represent the system of equations as an augmented matrix. The augmented matrix is a matrix that combines the coefficient matrix and the constant matrix.

{ \begin{bmatrix} 1 & 2 & 0 & 0 \\ 1 & 3 & 1 & 1 \\ -2 & -1 & -1 & -13 \end{bmatrix} \}

The coefficient matrix is the matrix of coefficients of the variables, while the constant matrix is the matrix of constants on the right-hand side of the equations.

Applying Gaussian Elimination

The Gaussian elimination method involves transforming the augmented matrix into upper triangular form using elementary row operations. The goal is to eliminate the entries below the main diagonal, making it easier to solve the system.

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 the first row to the second row.

{ \begin{bmatrix} 1 & 2 & 0 & 0 \\ 0 & 5 & 1 & 1 \\ -2 & -1 & -1 & -13 \end{bmatrix} \}

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 2/5 times the second row to the third row.

{ \begin{bmatrix} 1 & 2 & 0 & 0 \\ 0 & 5 & 1 & 1 \\ 0 & 3/5 & -3/5 & -33/5 \end{bmatrix} \}

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 1/5 times the third row to the second row.

{ \begin{bmatrix} 1 & 2 & 0 & 0 \\ 0 & 5 & 4/5 & 16/5 \\ 0 & 3/5 & -3/5 & -33/5 \end{bmatrix} \}

Step 4: Eliminate the Entry Below the Main Diagonal in the Third Column (continued)

To eliminate the entry below the main diagonal in the third column, we need to add 3/5 times the third row to the second row.

{ \begin{bmatrix} 1 & 2 & 0 & 0 \\ 0 & 5 & 0 & -1 \\ 0 & 3/5 & -3/5 & -33/5 \end{bmatrix} \}

Step 5: Eliminate the Entry Below the Main Diagonal in the Third Column (continued)

To eliminate the entry below the main diagonal in the third column, we need to add 3/5 times the third row to the second row.

{ \begin{bmatrix} 1 & 2 & 0 & 0 \\ 0 & 5 & 0 & -1 \\ 0 & 0 & -3/5 & -33/5 \end{bmatrix} \}

Back-Substitution

Now that we have transformed the augmented matrix into upper triangular form, we can use back-substitution to find the values of x, y, and z.

Step 3: Solve for z

To solve for z, we can use the third equation:

{ -3/5z = -33/5 \}

Multiplying both sides by -5/3, we get:

{ z = 11 \}

Step 2: Solve for y

To solve for y, we can use the second equation:

{ 5y = -1 \}

Dividing both sides by 5, we get:

{ y = -1/5 \}

Step 1: Solve for x

To solve for x, we can use the first equation:

{ x + 2y = 0 \}

Substituting the value of y, we get:

{ x + 2(-1/5) = 0 \}

Simplifying, we get:

{ x = 2/5 \}

Conclusion

In this article, we have shown how to solve a system of linear equations using matrices and the Gaussian elimination method with back-substitution. We have transformed the augmented matrix into upper triangular form using elementary row operations and then used back-substitution to find the values of x, y, and z. The solution to the system is x = 2/5, y = -1/5, and z = 11.

Discussion

The Gaussian elimination method is a powerful tool for solving systems of linear equations. It involves transforming the augmented matrix into upper triangular form using elementary row operations and then using back-substitution to find the values of the variables. This method is particularly useful when the system has a large number of variables or when the coefficients are complex.

However, the Gaussian elimination method can be computationally intensive and may not be suitable for large systems. In such cases, other methods such as LU decomposition or QR decomposition may be more efficient.

References

[1] Strang, G. (1988). Linear Algebra and Its Applications. 3rd ed. New York: Harcourt Brace Jovanovich.
[2] Anton, H. (1994). Elementary Linear Algebra. 7th ed. New York: John Wiley & Sons.
[3] Lay, D. C. (2005). Linear Algebra and Its Applications. 3rd ed. Boston: Addison-Wesley.

Keywords

Gaussian elimination
Back-substitution
Linear equations
Matrices
Elementary row operations
Upper triangular form
System of linear equations
Linear algebra

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Introduction

In our previous article, we explored how to solve systems of linear equations using matrices and the Gaussian elimination method with back-substitution. In this article, we will answer some of the most frequently asked questions (FAQs) about this topic.

Q&A

Q: What is the Gaussian elimination method?

A: The Gaussian elimination method is a technique used to solve systems of linear equations by transforming the augmented matrix into upper triangular form using elementary row operations.

Q: What is the purpose of back-substitution in Gaussian elimination?

A: Back-substitution is used to find the values of the variables in the system of linear equations after the augmented matrix has been transformed into upper triangular form.

Q: What are the advantages of using Gaussian elimination to solve systems of linear equations?

A: The advantages of using Gaussian elimination include:

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

Q: What are the disadvantages of using Gaussian elimination to solve systems of linear equations?

A: The disadvantages of using Gaussian elimination include:

It can be computationally intensive for large systems.
It may not be suitable for systems with a large number of variables.
It may not be suitable for systems with complex coefficients.

Q: What is the difference between Gaussian elimination and LU decomposition?

A: Gaussian elimination is a technique used to transform the augmented matrix into upper triangular form, while LU decomposition is a technique used to factor the coefficient matrix into the product of two matrices, L and U.

Q: When should I use Gaussian elimination to solve a system of linear equations?

A: You should use Gaussian elimination when:

The system has a small number of variables.
The coefficients are simple.
You want a simple and efficient method for solving the system.

Q: When should I use LU decomposition to solve a system of linear equations?

A: You should use LU decomposition when:

The system has a large number of variables.
The coefficients are complex.
You want a more efficient method for solving the system.

Q: Can I use Gaussian elimination to solve a system of linear equations with complex coefficients?

A: Yes, you can use Gaussian elimination to solve a system of linear equations with complex coefficients. However, you may need to use a more complex algorithm to handle the complex arithmetic.

Q: Can I use Gaussian elimination to solve a system of linear equations with a large number of variables?

A: Yes, you can use Gaussian elimination to solve a system of linear equations with a large number of variables. However, you may need to use a more complex algorithm to handle the large number of variables.

Conclusion

In this article, we have answered some of the most frequently asked questions (FAQs) about solving systems of linear equations using matrices and the Gaussian elimination method with back-substitution. We hope that this article has been helpful in clarifying some of the concepts and techniques involved in solving systems of linear equations.

Discussion

The Gaussian elimination method is a powerful tool for solving systems of linear equations. However, it may not be suitable for all types of systems. In such cases, other methods such as LU decomposition or QR decomposition may be more efficient.

References

[1] Strang, G. (1988). Linear Algebra and Its Applications. 3rd ed. New York: Harcourt Brace Jovanovich.
[2] Anton, H. (1994). Elementary Linear Algebra. 7th ed. New York: John Wiley & Sons.
[3] Lay, D. C. (2005). Linear Algebra and Its Applications. 3rd ed. Boston: Addison-Wesley.

Keywords

Gaussian elimination
Back-substitution
Linear equations
Matrices
Elementary row operations
Upper triangular form
System of linear equations
Linear algebra
LU decomposition
QR decomposition
Complex coefficients
Large number of variables