Solve the system of linear equations using elementary row operations on an augmented matrix. This problem will have you enter intermediate steps of the solution.

{
−
x
−
5
y
=
−
9
−
7
x
−
27
y
=
−
31
 

(a) Fill the corresponding augmented matrix. The rightmost column represents the constant values from the linear equations.

 	
-1
-5
-9
-7
-27
-31
 
 
 Correct
(b) Use elementary row operations on the augmented matrix in part (a) to make a matrix of the form

[
1
a
b
0
1
c
]
 ,  
[
1
a
b
0
0
1
]
 , or  
[
1
a
b
0
0
0
]
 .*

 	
1
0
0
1
 
(c) Using the matrix form in part (b), determine the solution of the corresponding system of linear equations. If the system has no solutions, enter DNE for all answers. If there are infinite number of solutions, enter  
y
  for  
y
  and find  
x
  in terms of  
y
 .

x
=
  
 and  
y
=
  


Let's work through the steps of solving this system of linear equations using elementary row operations.

We are given the system of equations:
\[
- x - 5 y = -9
\]
\[
- 7x - 27 y = -31
\]

### Step (a): Fill the augmented matrix

We form the augmented matrix based on the coefficients of \(x\), \(y\), and the constants:

\[
\begin{bmatrix}
-1 & -5 & -9 \\
-7 & -27 & -31
\end{bmatrix}
\]

### Step (b): Perform elementary row operations

We want to reduce this matrix to either row echelon form or reduced row echelon form.

1. **First operation**: Make the leading coefficient of the first row 1 (this can be done by multiplying the first row by -1):

\[
R_1 \rightarrow (-1)R_1
\]

\[
\begin{bmatrix}
1 & 5 & 9 \\
-7 & -27 & -31
\end{bmatrix}
\]

2. **Second operation**: Eliminate the \(x\)-term in the second row (we can add 7 times the first row to the second row):

\[
R_2 \rightarrow R_2 + 7R_1
\]

\[
\begin{bmatrix}
1 & 5 & 9 \\
0 & 8 & 32
\end{bmatrix}
\]

3. **Third operation**: Make the leading coefficient of the second row 1 (we divide the second row by 8):

\[
R_2 \rightarrow \frac{1}{8} R_2
\]

\[
\begin{bmatrix}
1 & 5 & 9 \\
0 & 1 & 4
\end{bmatrix}
\]

4. **Final operation**: Eliminate the \(y\)-term in the first row (we subtract 5 times the second row from the first row):

\[
R_1 \rightarrow R_1 - 5R_2
\]

\[
\begin{bmatrix}
1 & 0 & -11 \\
0 & 1 & 4
\end{bmatrix}
\]

### Step (c): Solve the system

The final matrix corresponds to the system of equations:
\[
x = -11
\]
\[
y = 4
\]

Thus, the solution to the system is:
\[
x = -11, \quad y = 4
\]

### Summary:
The solution is \(x = -11\) and \(y = 4\).

Would you like more details or have any questions?

Here are 5 related questions you can explore:
1. What are the elementary row operations used in Gaussian elimination?
2. How can you identify if a system of equations has infinite solutions?
3. How does the augmented matrix method compare to substitution or elimination methods?
4. What is reduced row echelon form, and how does it simplify finding solutions?
5. How can you handle cases where the system has no solution?

**Tip:** Always aim to make the leading coefficient of the pivot column 1 for easier row operations when solving systems of equations using matrices.

Solve the system of linear equations using elementary row operations on an augmented matrix. This problem will have you enter intermediate steps of the solution. 

-x - 5y = -9
-7x - 27y = -31

Learn how to solve a system of linear equations with two variables using elementary row operations on an augmented matrix. Follow a step-by-step approach to transform the matrix into row echelon form and find the solution. This problem covers core linear algebra concepts, ideal for students in Grades 9-12.

Math Problem Statement

{ − x − 5 y

− 9 − 7 x − 27 y

x

and
y

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation