4. Solve the system of equations using augmented matrix methods. x ₁-2x ₂ = 1 2x₁ - x₂ = 7 ==

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### Solving a System of Equations Using Augmented Matrix Methods

#### Problem Statement:

Solve the system of equations using augmented matrix methods.

\[
x_1 - 2x_2 = -1
\]
\[
2x_1 - x_2 = 7
\]

#### Explanation:

To solve this system using the augmented matrix method, follow these steps:

1. **Write the augmented matrix**: Convert the system of equations into an augmented matrix. 

   \[
   \begin{bmatrix}
   1 & -2 & | & -1 \\
   2 & -1 & | & 7 
   \end{bmatrix}
   \]

2. **Row operations**: Use row operations to transform the matrix into row-echelon form (or reduced row-echelon form).

   - First, make the first element in the first column a leading 1 (which it already is), and use it to make all elements below it zero.

   - Subtract 2 times the first row from the second row.

   \[
   \begin{bmatrix}
   1 & -2 & | & -1 \\
   0 & 3 & | & 9 
   \end{bmatrix}
   \]

3. **Back substitution**: Once the matrix is in row-echelon form, solve for the variables starting from the last row.

   - From the second row \(0x_1 + 3x_2 = 9\), solve for \(x_2\):
   
   \[
   x_2 = \frac{9}{3} = 3
   \]

   - Substitute \(x_2 = 3\) into the first equation \(x_1 - 2x_2 = -1\) to solve for \(x_1\):

   \[
   x_1 - 2(3) = -1 \\
   x_1 - 6 = -1 \\
   x_1 = 5
   \]

#### Solution:

\(x_1 = 5\) and \(x_2 = 3\).

This concludes the solution using the augmented matrix method.
Transcribed Image Text:### Solving a System of Equations Using Augmented Matrix Methods #### Problem Statement: Solve the system of equations using augmented matrix methods. \[ x_1 - 2x_2 = -1 \] \[ 2x_1 - x_2 = 7 \] #### Explanation: To solve this system using the augmented matrix method, follow these steps: 1. **Write the augmented matrix**: Convert the system of equations into an augmented matrix. \[ \begin{bmatrix} 1 & -2 & | & -1 \\ 2 & -1 & | & 7 \end{bmatrix} \] 2. **Row operations**: Use row operations to transform the matrix into row-echelon form (or reduced row-echelon form). - First, make the first element in the first column a leading 1 (which it already is), and use it to make all elements below it zero. - Subtract 2 times the first row from the second row. \[ \begin{bmatrix} 1 & -2 & | & -1 \\ 0 & 3 & | & 9 \end{bmatrix} \] 3. **Back substitution**: Once the matrix is in row-echelon form, solve for the variables starting from the last row. - From the second row \(0x_1 + 3x_2 = 9\), solve for \(x_2\): \[ x_2 = \frac{9}{3} = 3 \] - Substitute \(x_2 = 3\) into the first equation \(x_1 - 2x_2 = -1\) to solve for \(x_1\): \[ x_1 - 2(3) = -1 \\ x_1 - 6 = -1 \\ x_1 = 5 \] #### Solution: \(x_1 = 5\) and \(x_2 = 3\). This concludes the solution using the augmented matrix method.
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