Which Of The Following Is The Solution Set Of $ \begin{cases} x - Y - 2z = 0 \ -2x + 6z = 0 \ -x + 2y + Z = 1 \end{cases} }$A. { {(1, 0, 0.5)}$}$B. { {(3t + 1, 0, T) T \in \mathbb{R }$}$C.

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Introduction

In mathematics, a system of linear equations is a set of two or more linear equations that are solved simultaneously to find the values of the variables. In this article, we will focus on solving a system of three linear equations with three variables. We will use the method of substitution and elimination to find the solution set of the given system.

The System of Linear Equations

The given system of linear equations is:

{x−y−2z=0−2x+6z=0−x+2y+z=1\begin{cases} x - y - 2z = 0 \\ -2x + 6z = 0 \\ -x + 2y + z = 1 \end{cases}

Step 1: Write Down the Augmented Matrix

To solve the system of linear equations, we will first write down the augmented matrix. The augmented matrix is a matrix that includes the coefficients of the variables and the constant terms.

[1−1−20−2060−1211]\begin{bmatrix} 1 & -1 & -2 & 0 \\ -2 & 0 & 6 & 0 \\ -1 & 2 & 1 & 1 \end{bmatrix}

Step 2: Perform Row Operations

To solve the system of linear equations, we will perform row operations on the augmented matrix. The goal is to transform the augmented matrix into row-echelon form.

Step 2.1: Multiply Row 1 by 2 and Add to Row 2

We will multiply row 1 by 2 and add it to row 2.

[1−1−200220−1211]\begin{bmatrix} 1 & -1 & -2 & 0 \\ 0 & 2 & 2 & 0 \\ -1 & 2 & 1 & 1 \end{bmatrix}

Step 2.2: Multiply Row 1 by 1 and Add to Row 3

We will multiply row 1 by 1 and add it to row 3.

[1−1−20022001−11]\begin{bmatrix} 1 & -1 & -2 & 0 \\ 0 & 2 & 2 & 0 \\ 0 & 1 & -1 & 1 \end{bmatrix}

Step 2.3: Multiply Row 2 by 1/2

We will multiply row 2 by 1/2.

[1−1−20011001−11]\begin{bmatrix} 1 & -1 & -2 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 1 & -1 & 1 \end{bmatrix}

Step 2.4: Subtract Row 2 from Row 3

We will subtract row 2 from row 3.

[1−1−20011000−21]\begin{bmatrix} 1 & -1 & -2 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & -2 & 1 \end{bmatrix}

Step 2.5: Multiply Row 3 by -1/2

We will multiply row 3 by -1/2.

[1−1−200110001−1/2]\begin{bmatrix} 1 & -1 & -2 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & -1/2 \end{bmatrix}

Step 2.6: Subtract 2 Times Row 3 from Row 1

We will subtract 2 times row 3 from row 1.

[1−1010110001−1/2]\begin{bmatrix} 1 & -1 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & -1/2 \end{bmatrix}

Step 2.7: Add Row 3 to Row 2

We will add row 3 to row 2.

[1−101012−1/2001−1/2]\begin{bmatrix} 1 & -1 & 0 & 1 \\ 0 & 1 & 2 & -1/2 \\ 0 & 0 & 1 & -1/2 \end{bmatrix}

Step 2.8: Subtract 2 Times Row 3 from Row 2

We will subtract 2 times row 3 from row 2.

[1−1010100001−1/2]\begin{bmatrix} 1 & -1 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & -1/2 \end{bmatrix}

Step 2.9: Add Row 3 to Row 1

We will add row 3 to row 1.

[1−111/20100001−1/2]\begin{bmatrix} 1 & -1 & 1 & 1/2 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & -1/2 \end{bmatrix}

Step 2.10: Subtract Row 3 from Row 1

We will subtract row 3 from row 1.

[1−1010100001−1/2]\begin{bmatrix} 1 & -1 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & -1/2 \end{bmatrix}

Step 2.11: Add Row 2 to Row 1

We will add row 2 to row 1.

[10010100001−1/2]\begin{bmatrix} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & -1/2 \end{bmatrix}

Step 3: Write Down the Solution Set

The solution set of the system of linear equations is the set of all possible solutions. We can write down the solution set by using the values of the variables.

Let x=1x = 1, y=0y = 0, and z=0.5z = 0.5. Then, the solution set is:

{(1,0,0.5)}\{(1, 0, 0.5)\}

Conclusion

In this article, we have solved a system of three linear equations with three variables using the method of substitution and elimination. We have written down the augmented matrix and performed row operations to transform it into row-echelon form. Finally, we have written down the solution set of the system of linear equations.

Answer

The solution set of the system of linear equations is:

{(1,0,0.5)}\{(1, 0, 0.5)\}

Q: What is a system of linear equations?

A: A system of linear equations is a set of two or more linear equations that are solved simultaneously to find the values of the variables.

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

A: There are several methods for solving systems 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 one or more variables.
  • Graphical method: This method involves graphing the equations on a coordinate plane and finding the point of intersection.
  • Matrix method: This method involves using matrices to represent the system of equations and then performing row operations to solve the system.

Q: What is the augmented matrix?

A: The augmented matrix is a matrix that includes the coefficients of the variables and the constant terms. It is used to represent the system of linear equations in a compact form.

Q: What are row operations?

A: Row operations are the steps taken to transform the augmented matrix into row-echelon form. These operations include:

  • Multiplying a row by a non-zero constant
  • Adding a multiple of one row to another row
  • Interchanging two rows

Q: What is row-echelon form?

A: Row-echelon form is a form of the augmented matrix where each row has a leading entry (a non-zero entry) that is to the right of the leading entry of the row above it.

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

A: To determine the solution set of a system of linear equations, you need to:

  1. Write down the augmented matrix.
  2. Perform row operations to transform the augmented matrix into row-echelon form.
  3. Identify the leading entries in each row.
  4. Write down the solution set using the values of the variables.

Q: What is the solution set of a system of linear equations?

A: The solution set of a system of linear equations is the set of all possible solutions. It is a set of ordered pairs (x, y, z) that satisfy all the equations in the system.

Q: How do I check if a solution is valid?

A: To check if a solution is valid, you need to:

  1. Plug the values of the variables into each equation.
  2. Check if the equation is true.
  3. If all the equations are true, then the solution is valid.

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

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

  • Not following the order of operations
  • Not checking for valid solutions
  • Not using the correct method for solving the system
  • Not writing down the solution set correctly

Q: How do I choose the best method for solving a system of linear equations?

A: To choose the best method for solving a system of linear equations, you need to:

  1. Look at the equations and determine the best method to use.
  2. Consider the number of variables and the complexity of the equations.
  3. Choose the method that is most efficient and easiest to understand.

Q: What are some real-world applications of solving systems of linear equations?

A: Some real-world applications of solving systems of linear equations include:

  • Physics and engineering: Solving systems of linear equations is used to model real-world problems, such as motion and forces.
  • Computer science: Solving systems of linear equations is used in computer graphics and game development.
  • Economics: Solving systems of linear equations is used to model economic systems and make predictions about the economy.
  • Biology: Solving systems of linear equations is used to model population growth and disease spread.