Solve The Given System Of Equations. If The System Has No Solution, State That It Is Inconsistent.$\[ \begin{aligned} x - Y &= 5 \\ 3x - 3z &= 24 \\ 3y + Z &= 9 \end{aligned} \\]

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 solve a system of three linear equations with three variables.

The System of Equations

The given system of equations is:

$$\begin{aligned} x - y &= 5 \\ 3x - 3z &= 24 \\ 3y + z &= 9 \end{aligned}$$

Step 1: Write the System of Equations in Matrix Form

To solve the system of equations, we can write it in matrix form as follows:

$$\begin{bmatrix} 1 & -1 & 0 \\ 3 & 0 & -3 \\ 0 & 3 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 5 \\ 24 \\ 9 \end{bmatrix}$$

Step 2: Find the Augmented Matrix

The augmented matrix is obtained by adding the constant terms to the coefficient matrix.

$$\begin{bmatrix} 1 & -1 & 0 & 5 \\ 3 & 0 & -3 & 24 \\ 0 & 3 & 1 & 9 \end{bmatrix}$$

Step 3: Perform Row Operations

To solve the system of equations, we need to perform row operations on the augmented matrix. We will use the following row operations:

• Multiply a row by a non-zero constant
• Add a multiple of one row to another row
• Interchange two rows

Step 3.1: Multiply Row 1 by 3 and Add to Row 2

$$\begin{bmatrix} 1 & -1 & 0 & 5 \\ 3 & 0 & -3 & 24 \\ 0 & 3 & 1 & 9 \end{bmatrix} \sim \begin{bmatrix} 1 & -1 & 0 & 5 \\ 0 & 3 & -3 & 9 \\ 0 & 3 & 1 & 9 \end{bmatrix}$$

Step 3.2: Subtract Row 2 from Row 3

$$\begin{bmatrix} 1 & -1 & 0 & 5 \\ 0 & 3 & -3 & 9 \\ 0 & 3 & 1 & 9 \end{bmatrix} \sim \begin{bmatrix} 1 & -1 & 0 & 5 \\ 0 & 3 & -3 & 9 \\ 0 & 0 & 4 & 0 \end{bmatrix}$$

Step 3.3: Divide Row 3 by 4

$$\begin{bmatrix} 1 & -1 & 0 & 5 \\ 0 & 3 & -3 & 9 \\ 0 & 0 & 4 & 0 \end{bmatrix} \sim \begin{bmatrix} 1 & -1 & 0 & 5 \\ 0 & 3 & -3 & 9 \\ 0 & 0 & 1 & 0 \end{bmatrix}$$

Step 3.4: Add 3 Times Row 3 to Row 2

$$\begin{bmatrix} 1 & -1 & 0 & 5 \\ 0 & 3 & -3 & 9 \\ 0 & 0 & 1 & 0 \end{bmatrix} \sim \begin{bmatrix} 1 & -1 & 0 & 5 \\ 0 & 3 & 0 & 27 \\ 0 & 0 & 1 & 0 \end{bmatrix}$$

Step 3.5: Divide Row 2 by 3

$$\begin{bmatrix} 1 & -1 & 0 & 5 \\ 0 & 3 & 0 & 27 \\ 0 & 0 & 1 & 0 \end{bmatrix} \sim \begin{bmatrix} 1 & -1 & 0 & 5 \\ 0 & 1 & 0 & 9 \\ 0 & 0 & 1 & 0 \end{bmatrix}$$

Step 3.6: Add Row 2 to Row 1

$$\begin{bmatrix} 1 & -1 & 0 & 5 \\ 0 & 1 & 0 & 9 \\ 0 & 0 & 1 & 0 \end{bmatrix} \sim \begin{bmatrix} 1 & 0 & 0 & 14 \\ 0 & 1 & 0 & 9 \\ 0 & 0 & 1 & 0 \end{bmatrix}$$

Conclusion

The system of equations has been solved using row operations on the augmented matrix. The solution is:

$$\begin{aligned} x &= 14 \\ y &= 9 \\ z &= 0 \end{aligned}$$

Therefore, the system of equations has a unique solution.

Inconsistent System

If the system of equations has no solution, it is called an inconsistent system. In this case, the augmented matrix will have a row of the form:

$$\begin{bmatrix} a & b & c & d \end{bmatrix}$$

where $a$, $b$, $c$, and $d$ are constants, and $a \neq 0$. If the row is of the form:

$$\begin{bmatrix} 0 & 0 & 0 & d \end{bmatrix}$$

where $d \neq 0$, then the system of equations is inconsistent.

Example

Consider the system of equations:

$$\begin{aligned} x + y &= 2 \\ 2x + 2y &= 4 \end{aligned}$$

The augmented matrix is:

$$\begin{bmatrix} 1 & 1 & 2 \\ 2 & 2 & 4 \end{bmatrix}$$

Performing row operations, we get:

$$\begin{bmatrix} 1 & 1 & 2 \\ 0 & 0 & 0 \end{bmatrix}$$

Since the last row is of the form:

$$\begin{bmatrix} 0 & 0 & 0 & d \end{bmatrix}$$

where $d = 0$, the system of equations is inconsistent.

Conclusion

In this article, we have discussed how to solve a system of linear equations using row operations on the augmented matrix. We have also seen how to determine if a system of equations is inconsistent. The solution to a system of equations is unique if the augmented matrix has a row of the form:

$$\begin{bmatrix} a & b & c & d \end{bmatrix}$$

where $a$, $b$, $c$, and $d$ are constants, and $a \neq 0$. If the row is of the form:

$$\begin{bmatrix} 0 & 0 & 0 & d \end{bmatrix}$$

$$

Introduction

In our previous article, we discussed how to solve a system of linear equations using row operations on the augmented matrix. 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 system of linear equations is a set of two or more linear equations that involve the same set of variables. For example:

$$\begin{aligned} x + y &= 2 \\ 2x + 2y &= 4 \end{aligned}$$

Q: How do I know if a system of equations has a solution?

To determine if a system of equations has a solution, you need to check if the augmented matrix has a row of the form:

$$\begin{bmatrix} a & b & c & d \end{bmatrix}$$

where $a$, $b$, $c$, and $d$ are constants, and $a \neq 0$. If the row is of the form:

$$\begin{bmatrix} 0 & 0 & 0 & d \end{bmatrix}$$

where $d \neq 0$, then the system of equations is inconsistent and has no solution.

Q: How do I solve a system of linear equations using row operations?

To solve a system of linear equations using row operations, follow these steps:

1. Write the system of equations in matrix form as an augmented matrix.
2. Perform row operations to transform the augmented matrix into row echelon form.
3. Use the row echelon form to find the solution to the system of equations.

Q: What is row echelon form?

Row echelon form is a matrix that has the following properties:

• All the rows of the matrix are non-zero.
• Each row has a leading entry (a non-zero entry) that is to the right of the leading entry of the row above it.
• All the entries below the leading entry of each row are zero.

Q: How do I perform row operations?

To perform row operations, follow these steps:

1. Multiply a row by a non-zero constant.
2. Add a multiple of one row to another row.
3. Interchange two rows.

Q: What is the difference between a consistent and inconsistent system of equations?

A consistent system of equations is a system that has a solution. An inconsistent system of equations is a system that has no solution.

Q: How do I determine if a system of equations is consistent or inconsistent?

To determine if a system of equations is consistent or inconsistent, you need to check if the augmented matrix has a row of the form:

$$\begin{bmatrix} a & b & c & d \end{bmatrix}$$

where $a$, $b$, $c$, and $d$ are constants, and $a \neq 0$. If the row is of the form:

$$\begin{bmatrix} 0 & 0 & 0 & d \end{bmatrix}$$

where $d \neq 0$, then the system of equations is inconsistent. If the row is of the form:

$$\begin{bmatrix} a & b & c & d \end{bmatrix}$$

where $a \neq 0$, then the system of equations is consistent.

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

The solution to a system of linear equations is a set of values for the variables that satisfy all the equations in the system.

Q: How do I find the solution to a system of linear equations?

To find the solution to a system of linear equations, follow these steps:

1. Write the system of equations in matrix form as an augmented matrix.
2. Perform row operations to transform the augmented matrix into row echelon form.
3. Use the row echelon form to find the solution to the system of equations.

Conclusion

In this article, we have answered some frequently asked questions about solving systems of linear equations. We have discussed how to determine if a system of equations is consistent or inconsistent, how to perform row operations, and how to find the solution to a system of linear equations.