Solve The System Of Equations Using Matrices. Use The Gaussian Elimination Method With Back-substitution.$\[ \begin{cases} x - 5y = Z - 7 \\ x = 5 + Y - Z \\ x + Y + 5z = 23 \end{cases} \\]

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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, economics, and computer science. One of the most efficient methods for solving systems of linear equations is the Gaussian elimination method, which involves transforming the augmented matrix 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 given system of equations is:

{ \begin{cases} x - 5y = z - 7 \\ x = 5 + y - z \\ x + y + 5z = 23 \end{cases} \}$ $ To solve this system, we will first represent it in matrix form as an augmented matrix. ## Representing the System as an Augmented Matrix ------------------------------------------------ The augmented matrix for the given system is: ${ \begin{bmatrix} 1 & -5 & 1 & -7 \\ 1 & 1 & -1 & 5 \\ 1 & 1 & 5 & 23 \\ \end{bmatrix} \}$ $ ## Gaussian Elimination Method --------------------------- 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 will perform the following row operation: ${ \begin{bmatrix} 1 & -5 & 1 & -7 \\ 1 & 1 & -1 & 5 \\ 1 & 1 & 5 & 23 \\ \end{bmatrix} \sim \begin{bmatrix} 1 & -5 & 1 & -7 \\ 0 & 6 & -2 & 12 \\ 0 & 6 & 4 & 30 \\ \end{bmatrix} \}$ $ ### 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 will perform the following row operation: ${ \begin{bmatrix} 1 & -5 & 1 & -7 \\ 0 & 6 & -2 & 12 \\ 0 & 6 & 4 & 30 \\ \end{bmatrix} \sim \begin{bmatrix} 1 & -5 & 1 & -7 \\ 0 & 6 & -2 & 12 \\ 0 & 0 & 6 & 18 \\ \end{bmatrix} \}$ $ ### 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 will perform the following row operation: ${ \begin{bmatrix} 1 & -5 & 1 & -7 \\ 0 & 6 & -2 & 12 \\ 0 & 0 & 6 & 18 \\ \end{bmatrix} \sim \begin{bmatrix} 1 & -5 & 1 & -7 \\ 0 & 6 & -2 & 12 \\ 0 & 0 & 1 & 3 \\ \end{bmatrix} \}$ $ ## Back-Substitution Method ------------------------- After transforming the augmented matrix into upper triangular form, we can use the back-substitution method to solve the system. The back-substitution method involves solving the system from the bottom up, starting with the last equation and working our way up. ### Step 1: Solve the Last Equation The last equation is: ${ x + y + 5z = 23 \}$ $ Solving for $x$, we get: ${ x = 23 - y - 5z \}$ $ ### Step 2: Solve the Second-Last Equation The second-last equation is: ${ x = 5 + y - z \}$ $ Substituting the expression for $x$ from the last equation, we get: ${ 23 - y - 5z = 5 + y - z \}$ $ Simplifying, we get: ${ 18 = 6y - 4z \}$ $ Solving for $y$, we get: ${ y = 3 + \frac{2}{3}z \}$ $ ### Step 3: Solve the First Equation The first equation is: ${ x - 5y = z - 7 \}$ $ Substituting the expressions for $x$ and $y$ from the previous equations, we get: ${ 23 - y - 5z - 5(3 + \frac{2}{3}z) = z - 7 \}$ $ Simplifying, we get: ${ -2 - 5z - 10 - \frac{10}{3}z = z - 7 \}$ $ Solving for $z$, we get: ${ z = 3 \}$ $ ### Step 4: Find the Values of $y$ and $x$ Now that we have found the value of $z$, we can substitute it into the expressions for $y$ and $x$ from the previous equations. ${ y = 3 + \frac{2}{3}(3) = 5 \}$ $ ${ x = 23 - y - 5z = 23 - 5 - 15 = 3 \}$ $ ## 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 the back-substitution method to solve the system. The final solution is $x = 3$, $y = 5$, and $z = 3$. This method is a powerful tool for solving systems of linear equations and has numerous applications in various fields.<br/> # Frequently Asked Questions (FAQs) About Solving Systems of Linear Equations Using Matrices and Gaussian Elimination =========================================================== ## Introduction --------------- In our previous article, we explored how to solve a system of linear equations using matrices and the Gaussian elimination method with back-substitution. In this article, we will answer some frequently asked questions (FAQs) about this topic. ## Q: What is the Gaussian elimination method? ----------------------------------------- A: The Gaussian elimination method is a step-by-step process for solving systems of linear equations by transforming the augmented matrix into upper triangular form using elementary row operations. ## Q: What is the purpose of the augmented matrix? --------------------------------------------- A: The augmented matrix is a matrix that combines the coefficients of the variables and the constant terms of the system of linear equations. It is used to represent the system in a compact and efficient way. ## Q: What are elementary row operations? ----------------------------------------- A: Elementary row operations are basic operations that can be performed on the rows of a matrix, such as swapping two rows, multiplying a row by a non-zero constant, or adding a multiple of one row to another row. ## Q: How do I know when to stop performing elementary row operations? ---------------------------------------------------------------- A: You can stop performing elementary row operations when the augmented matrix has been transformed into upper triangular form, meaning that all the entries below the main diagonal are zero. ## Q: What is back-substitution? --------------------------- A: Back-substitution is a method for solving a system of linear equations by solving the system from the bottom up, starting with the last equation and working our way up. ## Q: Why is the Gaussian elimination method useful? ------------------------------------------------ A: The Gaussian elimination method is useful because it provides a systematic and efficient way to solve systems of linear equations, even for large systems. ## Q: Can the Gaussian elimination method be used to solve non-linear systems of equations? -------------------------------------------------------------------------------- A: No, the Gaussian elimination method can only be used to solve systems of linear equations. Non-linear systems of equations require different methods, such as numerical methods or algebraic methods. ## Q: How do I choose the order of the elementary row operations? ---------------------------------------------------------------- A: The order of the elementary row operations is not important, as long as the augmented matrix is transformed into upper triangular form. ## Q: Can the Gaussian elimination method be used to solve systems of equations with complex coefficients? ----------------------------------------------------------------------------------------- A: Yes, the Gaussian elimination method can be used to solve systems of equations with complex coefficients. ## Q: How do I interpret the results of the Gaussian elimination method? ------------------------------------------------------------------- A: The results of the Gaussian elimination method are the values of the variables that satisfy the system of linear equations. ## Q: Can the Gaussian elimination method be used to solve systems of equations with more than three variables? ----------------------------------------------------------------------------------------- A: Yes, the Gaussian elimination method can be used to solve systems of equations with more than three variables. ## Q: How do I know if the system of linear equations has a unique solution? ------------------------------------------------------------------- A: If the augmented matrix has been transformed into upper triangular form and all the entries on the main diagonal are non-zero, then the system of linear equations has a unique solution. ## Q: Can the Gaussian elimination method be used to solve systems of equations with infinitely many solutions? ----------------------------------------------------------------------------------------- A: Yes, the Gaussian elimination method can be used to solve systems of equations with infinitely many solutions. ## Q: How do I know if the system of linear equations has no solution? ---------------------------------------------------------------- A: If the augmented matrix has been transformed into upper triangular form and there is a row with all zeros except for the last entry, then the system of linear equations has no solution. ## Conclusion ---------- In this article, we have answered some 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 provided you with a better understanding of this topic and has helped you to solve systems of linear equations more efficiently.