Use The Gaussian Elimination Method To Obtain The Matrix In Row-echelon Form. Choose The Correct Answer Below.Given System Of Equations:$\[ \left\{\begin{array}{r} 3a - B - 2c = 13 \\ 2a - B + 4c = -7 \\ a + 2b + 2c =

Introduction

Gaussian elimination is a method used to solve systems of linear equations by transforming the augmented matrix into row-echelon form. This method is widely used in mathematics, physics, and engineering to solve systems of linear equations. In this article, we will use the Gaussian elimination method to solve a given system of linear equations and obtain the matrix in row-echelon form.

The Given System of Equations

The given system of equations is:

$${ \left\{\begin{array}{r} 3a - b - 2c = 13 \\ 2a - b + 4c = -7 \\ a + 2b + 2c = 5 \end{array}\right. }$$

Step 1: Write the Augmented Matrix

To apply the Gaussian elimination method, we need to write the augmented matrix of the given system of equations. The augmented matrix is a matrix that combines the coefficients of the variables and the constant terms.

$${ \left[\begin{array}{ccc|c} 3 & -1 & -2 & 13 \\ 2 & -1 & 4 & -7 \\ 1 & 2 & 2 & 5 \end{array}\right] }$$

Step 2: Perform Row Operations

To transform the augmented matrix into row-echelon form, we need to perform a series of row operations. The row operations are:

• Swap two rows: Swap two rows of the matrix.
• Multiply a row by a scalar: Multiply a row of the matrix by a scalar.
• Add a multiple of one row to another row: Add a multiple of one row to another row.

We will perform the following row operations to transform the augmented matrix into row-echelon form:

Step 2.1: Multiply Row 1 by 1/3

To make the coefficient of $a$ in the first row equal to 1, we need to multiply the first row by 1/3.

$${ \left[\begin{array}{ccc|c} 1 & -1/3 & -2/3 & 13/3 \\ 2 & -1 & 4 & -7 \\ 1 & 2 & 2 & 5 \end{array}\right] }$$

Step 2.2: Subtract 2 Times Row 1 from Row 2

To make the coefficient of $a$ in the second row equal to 0, we need to subtract 2 times the first row from the second row.

$${ \left[\begin{array}{ccc|c} 1 & -1/3 & -2/3 & 13/3 \\ 0 & 1/3 & 10/3 & -41/3 \\ 1 & 2 & 2 & 5 \end{array}\right] }$$

Step 2.3: Subtract Row 1 from Row 3

To make the coefficient of $a$ in the third row equal to 0, we need to subtract the first row from the third row.

$${ \left[\begin{array}{ccc|c} 1 & -1/3 & -2/3 & 13/3 \\ 0 & 1/3 & 10/3 & -41/3 \\ 0 & 5/3 & 8/3 & -16/3 \end{array}\right] }$$

Step 2.4: Multiply Row 2 by 3

To make the coefficient of $b$ in the second row equal to 1, we need to multiply the second row by 3.

$${ \left[\begin{array}{ccc|c} 1 & -1/3 & -2/3 & 13/3 \\ 0 & 1 & 10 & -41 \\ 0 & 5/3 & 8/3 & -16/3 \end{array}\right] }$$

Step 2.5: Subtract 5/3 Times Row 2 from Row 3

To make the coefficient of $b$ in the third row equal to 0, we need to subtract 5/3 times the second row from the third row.

$${ \left[\begin{array}{ccc|c} 1 & -1/3 & -2/3 & 13/3 \\ 0 & 1 & 10 & -41 \\ 0 & 0 & -2 & 3 \end{array}\right] }$$

Step 2.6: Multiply Row 3 by -1/2

To make the coefficient of $c$ in the third row equal to 1, we need to multiply the third row by -1/2.

$${ \left[\begin{array}{ccc|c} 1 & -1/3 & -2/3 & 13/3 \\ 0 & 1 & 10 & -41 \\ 0 & 0 & 1 & -3/2 \end{array}\right] }$$

Step 2.7: Subtract 10 Times Row 3 from Row 2

To make the coefficient of $c$ in the second row equal to 0, we need to subtract 10 times the third row from the second row.

$${ \left[\begin{array}{ccc|c} 1 & -1/3 & -2/3 & 13/3 \\ 0 & 1 & 0 & -331/2 \\ 0 & 0 & 1 & -3/2 \end{array}\right] }$$

Step 2.8: Add 2/3 Times Row 3 to Row 1

To make the coefficient of $c$ in the first row equal to 0, we need to add 2/3 times the third row to the first row.

$${ \left[\begin{array}{ccc|c} 1 & -1/3 & 0 & 25/2 \\ 0 & 1 & 0 & -331/2 \\ 0 & 0 & 1 & -3/2 \end{array}\right] }$$

Step 2.9: Add 1/3 Times Row 2 to Row 1

To make the coefficient of $b$ in the first row equal to 0, we need to add 1/3 times the second row to the first row.

$${ \left[\begin{array}{ccc|c} 1 & 0 & 0 & -1/2 \\ 0 & 1 & 0 & -331/2 \\ 0 & 0 & 1 & -3/2 \end{array}\right] }$$

Conclusion

We have successfully transformed the augmented matrix into row-echelon form using the Gaussian elimination method. The row-echelon form of the augmented matrix is:

$${ \left[\begin{array}{ccc|c} 1 & 0 & 0 & -1/2 \\ 0 & 1 & 0 & -331/2 \\ 0 & 0 & 1 & -3/2 \end{array}\right] }$$

This row-echelon form represents the solution to the given system of linear equations. The solution is:

$${ \left\{\begin{array}{r} a = -1/2 \\ b = -331/2 \\ c = -3/2 \end{array}\right. }$$

Therefore, the correct answer is:

$${ \left[\begin{array}{ccc|c} 1 & 0 & 0 & -1/2 \\ 0 & 1 & 0 & -331/2 \\ 0 & 0 & 1 & -3/2 \end{array}\right] }$$

Final Answer

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Q: What is Gaussian Elimination?

A: Gaussian elimination is a method used to solve systems of linear equations by transforming the augmented matrix into row-echelon form. This method is widely used in mathematics, physics, and engineering to solve systems of linear equations.

Q: What is the Row-Echelon Form?

A: The row-echelon form of a matrix is a form where all the entries below the leading entry of each row are zero. The leading entry of each row is the first non-zero entry in the row.

Q: How do I apply the Gaussian Elimination Method?

A: To apply the Gaussian elimination method, you need to follow these steps:

1. Write the augmented matrix of the given system of equations.
2. Perform a series of row operations to transform the augmented matrix into row-echelon form.
3. The row-echelon form represents the solution to the given system of linear equations.

Q: What are the types of Row Operations?

A: There are three types of row operations:

1. Swap two rows: Swap two rows of the matrix.
2. Multiply a row by a scalar: Multiply a row of the matrix by a scalar.
3. Add a multiple of one row to another row: Add a multiple of one row to another row.

Q: How do I choose the correct row operation?

A: To choose the correct row operation, you need to follow these steps:

1. Identify the leading entry of the row.
2. Determine the row operation that will make the coefficient of the leading entry equal to 1.
3. Perform the row operation.

Q: What are the advantages of Gaussian Elimination?

A: The advantages of Gaussian elimination are:

1. Efficient: Gaussian elimination is an efficient method for solving systems of linear equations.
2. Accurate: Gaussian elimination provides accurate solutions to systems of linear equations.
3. Easy to implement: Gaussian elimination is easy to implement using a computer or calculator.

Q: What are the disadvantages of Gaussian Elimination?

A: The disadvantages of Gaussian elimination are:

1. Time-consuming: Gaussian elimination can be time-consuming for large systems of linear equations.
2. Sensitive to round-off errors: Gaussian elimination is sensitive to round-off errors.
3. Not suitable for all types of matrices: Gaussian elimination is not suitable for all types of matrices, such as singular matrices.

Q: When to use Gaussian Elimination?

A: You should use Gaussian elimination when:

1. You have a system of linear equations with a small number of variables.
2. You need to solve a system of linear equations with a small number of equations.
3. You want to find the solution to a system of linear equations using a simple and efficient method.

Q: When not to use Gaussian Elimination?

A: You should not use Gaussian elimination when:

1. You have a system of linear equations with a large number of variables.
2. You need to solve a system of linear equations with a large number of equations.
3. You want to find the solution to a system of linear equations using a more efficient method, such as LU decomposition or Cholesky decomposition.

Conclusion

Gaussian elimination is a powerful method for solving systems of linear equations. It is efficient, accurate, and easy to implement. However, it has some disadvantages, such as being time-consuming and sensitive to round-off errors. You should use Gaussian elimination when you have a small system of linear equations and want to find the solution using a simple and efficient method.