Multiply a row by a constant

Oct 13, 2025 by leolavanderia.com.br 122 views

Introduction

In mathematics, a system of linear equations is a set of two or more linear equations that involve the same set of variables. Solving a system of linear equations means finding the values of the variables that satisfy all the equations in the system. In this article, we will focus on solving a system of three linear equations with three variables.

The System of Equations

The system of equations we will be solving is:

\begin{array}{l} 4x + 3y + 6z = 3 \\ 5x + 5y + 6z = 5 \\ 6x + 3y + 6z = 3 \end{array}

Method of Solution

There are several methods to solve a system of linear equations, including substitution, elimination, and matrix methods. In this article, we will use the elimination method to solve the system of equations.

Step 1: Write the Augmented Matrix

To solve the system of equations using the elimination method, we first need to write the augmented matrix. The augmented matrix is a matrix that includes the coefficients of the variables and the constants on the right-hand side of the equations.

4	3	6	3
5	5	6	5
6	3	6	3

Step 2: Perform Row Operations

To solve the system of equations, we need to perform row operations on the augmented matrix. The goal is to transform the matrix into row-echelon form, where all the entries below the leading entry in each row are zeros.

First, we will multiply the first row by 5 and the second row by 4 to make the entries in the first column equal.

20	15	30	15
20	20	24	20
6	3	6	3

Next, we will subtract the first row from the second row to eliminate the entry in the first column.

20	15	30	15
0	5	-6	5
6	3	6	3

Step 3: Continue Performing Row Operations

We will continue performing row operations to transform the matrix into row-echelon form.

First, we will multiply the second row by 6 and the third row by 5 to make the entries in the second column equal.

20	15	30	15
0	30	-36	30
30	15	30	15

Next, we will subtract the second row from the third row to eliminate the entry in the second column.

20	15	30	15
0	30	-36	30
0	-15	36	-15

Step 4: Solve for the Variables

Now that the matrix is in row-echelon form, we can solve for the variables.

From the third row, we can see that $-15y + 36z = -15$ . We can solve for $y$ in terms of $z$ by multiplying both sides by $-1$ and then dividing both sides by $15$ .

y = \frac{36}{15}z + 1

From the second row, we can see that $30y - 36z = 30$ . We can substitute the expression for $y$ in terms of $z$ into this equation.

30\left(\frac{36}{15}z + 1\right) - 36z = 30

Simplifying this equation, we get:

72z + 30 - 36z = 30

Combine like terms:

36z + 30 = 30

Subtract 30 from both sides:

36z = 0

Divide both sides by 36:

z = 0

Now that we have found the value of $z$ , we can substitute this value into the expression for $y$ in terms of $z$ .

y = \frac{36}{15}(0) + 1

Simplify:

y = 1

Finally, we can substitute the values of $y$ and $z$ into one of the original equations to solve for $x$ .

From the first equation, we have:

4x + 3y + 6z = 3

Substitute $y = 1$ and $z = 0$ :

4x + 3(1) + 6(0) = 3

Simplify:

4x + 3 = 3

Subtract 3 from both sides:

4x = 0

Divide both sides by 4:

x = 0

Therefore, the solution to the system of equations is $(x, y, z) = (0, 1, 0)$ .

Conclusion

In this article, we have solved a system of three linear equations with three variables using the elimination method. We have shown that the solution to the system of equations is $(x, y, z) = (0, 1, 0)$ . This solution satisfies all three equations in the system.

Discussion

The system of equations we have solved is a simple example of a system of linear equations. In general, systems of linear equations can be much more complex and may involve many more variables. However, the methods we have used in this article can be applied to solve systems of linear equations of any size.

Answer

The correct answer is:

a. $(x=1, y=0, z=-1)$ is not the correct solution.
b. $(x=-1, y=2, z=1)$ is not the correct solution.
c. $(x=0, y=1, z=0)$ is the correct solution.

References

[1] "Linear Algebra and Its Applications" by Gilbert Strang
[2] "Introduction to Linear Algebra" by Jim Hefferon
Solving a System of Linear Equations: Q&A =====================================

Introduction

In our previous article, we solved a system of three linear equations with three variables using the elimination method. In this article, we will answer some frequently asked questions about solving systems of linear equations.

Q: What is a system of linear equations?

A: A system of linear equations is a set of two or more linear equations that involve the same set of variables. Solving a system of linear equations means finding the values of the variables that satisfy all the equations in the system.

Q: What are the different methods for solving a system of linear equations?

A: There are several methods for solving a system of linear equations, including:

Substitution method: This method involves solving one equation for one variable and then substituting that expression into the other equations.
Elimination method: This method involves adding or subtracting equations to eliminate variables.
Matrix method: This method involves using matrices to represent the system of equations and then solving for the variables.

Q: What is the augmented matrix?

A: The augmented matrix is a matrix that includes the coefficients of the variables and the constants on the right-hand side of the equations. It is used to represent the system of equations in a compact form.

Q: How do I perform row operations on the augmented matrix?

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

Swap two rows: Swap two rows of the matrix to change the order of the equations.
Multiply a row by a constant: Multiply a row of the matrix by a constant to change the coefficients of the variables.
Add a multiple of one row to another row: Add a multiple of one row to another row to change the coefficients of the variables.

Q: What is row-echelon form?

A: Row-echelon form is a form of the augmented matrix where all the entries below the leading entry in each row are zeros. It is used to represent the system of equations in a simplified form.

Q: How do I solve for the variables in row-echelon form?

A: To solve for the variables in row-echelon form, you need to follow these steps:

Back-substitute: Back-substitute the values of the variables into the equations to find the values of the variables.
Solve for the variables: Solve for the variables using the equations.

Q: What are some common mistakes to avoid when solving a system of linear equations?

A: Some common mistakes to avoid when solving a system of linear equations include:

Not following the order of operations: Not following the order of operations can lead to incorrect solutions.
Not checking for consistency: Not checking for consistency can lead to incorrect solutions.
Not using the correct method: Not using the correct method can lead to incorrect solutions.

Conclusion

In this article, we have answered some frequently asked questions about solving systems of linear equations. We have covered topics such as the different methods for solving a system of linear equations, the augmented matrix, row operations, row-echelon form, and solving for the variables. We hope that this article has been helpful in answering your questions about solving systems of linear equations.

References

[1] "Linear Algebra and Its Applications" by Gilbert Strang
[2] "Introduction to Linear Algebra" by Jim Hefferon