Rewrite The Matrix Operation To Complete The Solution:$[ \begin{aligned} \left[\begin{array}{rrr|r} 1 & -1 & -3 & -15 \ 2 & 1 & -4 & -16 \ 3 & 4 & -3 & -7 \end{array}\right] & \rightarrow \left[\begin{array}{rrr|r} 1 & -1 & -3 & -15 \

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Introduction

In linear algebra, matrices are used to represent systems of equations and perform various operations such as addition, subtraction, multiplication, and inversion. Matrix operations are essential in solving systems of linear equations, finding the inverse of a matrix, and calculating determinants. In this article, we will focus on rewriting a matrix operation to complete the solution of a given system of linear equations.

The Given Matrix

The given matrix is:

[1−1−3−1521−4−1634−3−7]{ \begin{aligned} \left[\begin{array}{rrr|r} 1 & -1 & -3 & -15 \\ 2 & 1 & -4 & -16 \\ 3 & 4 & -3 & -7 \end{array}\right] \end{aligned} }

This matrix represents a system of three linear equations with three variables. The goal is to rewrite this matrix operation to complete the solution of the system.

Step 1: Perform Row Operations

To rewrite the matrix operation, we need to perform row operations to transform the matrix into row echelon form. The row echelon form of a matrix is a form where all the entries below the leading entry of each row are zeros.

Let's start by performing the following row operations:

  • Swap rows 1 and 2
  • Multiply row 1 by -1
  • Add 3 times row 1 to row 3

Performing Row Operations

Performing the above row operations, we get:

[21−4−161−1−3−1534−3−7]→[−2−14161−1−3−1534−3−7]→[1−1−3−15−2−141634−3−7]→[1−1−3−1501−2−207622]{ \begin{aligned} \left[\begin{array}{rrr|r} 2 & 1 & -4 & -16 \\ 1 & -1 & -3 & -15 \\ 3 & 4 & -3 & -7 \end{array}\right] & \rightarrow \left[\begin{array}{rrr|r} -2 & -1 & 4 & 16 \\ 1 & -1 & -3 & -15 \\ 3 & 4 & -3 & -7 \end{array}\right] & \rightarrow \left[\begin{array}{rrr|r} 1 & -1 & -3 & -15 \\ -2 & -1 & 4 & 16 \\ 3 & 4 & -3 & -7 \end{array}\right] & \rightarrow \left[\begin{array}{rrr|r} 1 & -1 & -3 & -15 \\ 0 & 1 & -2 & -2 \\ 0 & 7 & 6 & 22 \end{array}\right] \end{aligned} }

Step 2: Perform Further Row Operations

Now, we need to perform further row operations to transform the matrix into row echelon form.

Let's perform the following row operations:

  • Multiply row 2 by -1
  • Add 7 times row 2 to row 3

Performing Further Row Operations

Performing the above row operations, we get:

[1−1−3−1501−2−207622]→[1−1−3−150−12207622]→[1−1−3−1501−2−2002020]{ \begin{aligned} \left[\begin{array}{rrr|r} 1 & -1 & -3 & -15 \\ 0 & 1 & -2 & -2 \\ 0 & 7 & 6 & 22 \end{array}\right] & \rightarrow \left[\begin{array}{rrr|r} 1 & -1 & -3 & -15 \\ 0 & -1 & 2 & 2 \\ 0 & 7 & 6 & 22 \end{array}\right] & \rightarrow \left[\begin{array}{rrr|r} 1 & -1 & -3 & -15 \\ 0 & 1 & -2 & -2 \\ 0 & 0 & 20 & 20 \end{array}\right] \end{aligned} }

Step 3: Perform Final Row Operations

Now, we need to perform final row operations to transform the matrix into row echelon form.

Let's perform the following row operations:

  • Multiply row 3 by 1/20

Performing Final Row Operations

Performing the above row operation, we get:

[1−1−3−1501−2−2002020]→[1−1−3−1501−2−20011]{ \begin{aligned} \left[\begin{array}{rrr|r} 1 & -1 & -3 & -15 \\ 0 & 1 & -2 & -2 \\ 0 & 0 & 20 & 20 \end{array}\right] & \rightarrow \left[\begin{array}{rrr|r} 1 & -1 & -3 & -15 \\ 0 & 1 & -2 & -2 \\ 0 & 0 & 1 & 1 \end{array}\right] \end{aligned} }

Conclusion

In this article, we have rewritten the matrix operation to complete the solution of a given system of linear equations. We have performed row operations to transform the matrix into row echelon form. The final row echelon form of the matrix is:

[1−1−3−1501−2−20011]{ \begin{aligned} \left[\begin{array}{rrr|r} 1 & -1 & -3 & -15 \\ 0 & 1 & -2 & -2 \\ 0 & 0 & 1 & 1 \end{array}\right] \end{aligned} }

This matrix represents the solution of the system of linear equations. The solution is:

x1−x2−3x3=−15x2−2x3=−2x3=1{ \begin{aligned} x_1 - x_2 - 3x_3 & = -15 \\ x_2 - 2x_3 & = -2 \\ x_3 & = 1 \end{aligned} }

Substituting the value of x3x_3 into the second equation, we get:

x2−2(1)=−2x2=0{ \begin{aligned} x_2 - 2(1) & = -2 \\ x_2 & = 0 \end{aligned} }

Substituting the values of x2x_2 and x3x_3 into the first equation, we get:

x1−0−3(1)=−15x1=−12{ \begin{aligned} x_1 - 0 - 3(1) & = -15 \\ x_1 & = -12 \end{aligned} }

Therefore, the solution of the system of linear equations is:

x1=−12x2=0x3=1{ \begin{aligned} x_1 & = -12 \\ x_2 & = 0 \\ x_3 & = 1 \end{aligned} }

References

Further Reading

Conclusion

In conclusion, rewriting the matrix operation to complete the solution of a given system of linear equations involves performing row operations to transform the matrix into row echelon form. The final row echelon form of the matrix represents the solution of the system of linear equations. The solution is obtained by substituting the values of the variables into the equations.

Introduction

In our previous article, we discussed how to rewrite the matrix operation to complete the solution of a given system of linear equations. We performed row operations to transform the matrix into row echelon form and obtained the solution of the system. In this article, we will answer some frequently asked questions related to matrix operations and solving systems of linear equations.

Q: What is the purpose of row operations in matrix operations?

A: Row operations are used to transform a matrix into row echelon form, which is a form where all the entries below the leading entry of each row are zeros. This form is useful for solving systems of linear equations.

Q: How do I perform row operations on a matrix?

A: To perform row operations on a matrix, you need to follow these steps:

  1. Identify the leading entry of each row.
  2. Perform row swaps to move the leading entry of each row to the top.
  3. Perform row multiplications to make the leading entry of each row equal to 1.
  4. Perform row additions to make the entries below the leading entry of each row equal to 0.

Q: What is the difference between row echelon form and reduced row echelon form?

A: Row echelon form is a form where all the entries below the leading entry of each row are zeros. Reduced row echelon form is a form where all the entries above the leading entry of each row are zeros, and the leading entry of each row is equal to 1.

Q: How do I determine the solution of a system of linear equations using a matrix?

A: To determine the solution of a system of linear equations using a matrix, you need to follow these steps:

  1. Write the system of linear equations as a matrix equation.
  2. Perform row operations to transform the matrix into row echelon form.
  3. Identify the leading entry of each row and the corresponding variable.
  4. Substitute the values of the variables into the equations to obtain the solution.

Q: What is the significance of the leading entry in a matrix?

A: The leading entry in a matrix is the entry that is used to determine the solution of a system of linear equations. It is the entry that is used to perform row operations and transform the matrix into row echelon form.

Q: Can I use matrix operations to solve systems of linear equations with more than three variables?

A: Yes, you can use matrix operations to solve systems of linear equations with more than three variables. However, the process may be more complex and require more row operations.

Q: Are there any limitations to using matrix operations to solve systems of linear equations?

A: Yes, there are limitations to using matrix operations to solve systems of linear equations. For example, if the matrix is singular (i.e., its determinant is zero), then the system of linear equations may not have a unique solution.

Conclusion

In conclusion, matrix operations are a powerful tool for solving systems of linear equations. By performing row operations, we can transform a matrix into row echelon form and obtain the solution of the system. However, there are limitations to using matrix operations, and we need to be aware of these limitations when solving systems of linear equations.

References

Further Reading

Conclusion

In conclusion, matrix operations are a powerful tool for solving systems of linear equations. By performing row operations, we can transform a matrix into row echelon form and obtain the solution of the system. However, there are limitations to using matrix operations, and we need to be aware of these limitations when solving systems of linear equations.