Solve the following system of equations by using Gauss- Elimination method: 3x - y + 2z = 12 x + 2y + 8z = 11 2x - 2y - z = 2

We will use the **Gauss Elimination Method** to solve the system of linear equations:

1. \( 3x - y + 2z = 12 \)
2. \( x + 2y + 8z = 11 \)
3. \( 2x - 2y - z = 2 \)

The steps are as follows:

### Step 1: Write the system of equations in augmented matrix form.
The augmented matrix for the given system is:

\[
\left[\begin{array}{ccc|c}
3 & -1 & 2 & 12 \\
1 & 2 & 8 & 11 \\
2 & -2 & -1 & 2
\end{array}\right]
\]

### Step 2: Apply row operations to convert the matrix into an upper triangular form.

#### 1st Operation: Eliminate the \(x\)-terms from the 2nd and 3rd rows.

We use \( R_1 \) as the pivot row and perform row operations:

- \( R_2 \leftarrow R_2 - \frac{1}{3}R_1 \) to eliminate \(x\) from the second row.
- \( R_3 \leftarrow R_3 - \frac{2}{3}R_1 \) to eliminate \(x\) from the third row.

\[
\left[\begin{array}{ccc|c}
3 & -1 & 2 & 12 \\
0 & \frac{7}{3} & \frac{22}{3} & 7 \\
0 & -\frac{4}{3} & -\frac{7}{3} & -6
\end{array}\right]
\]

#### 2nd Operation: Eliminate the \(y\)-term from the 3rd row.

We use \( R_2 \) as the pivot for eliminating \(y\) from the 3rd row:

- \( R_3 \leftarrow R_3 - \left(\frac{-4/3}{7/3}\right)R_2 = R_3 + \frac{4}{7}R_2 \)

\[
\left[\begin{array}{ccc|c}
3 & -1 & 2 & 12 \\
0 & \frac{7}{3} & \frac{22}{3} & 7 \\
0 & 0 & \frac{2}{7} & -\frac{10}{7}
\end{array}\right]
\]

### Step 3: Back substitution to solve for the variables.

1. From the 3rd row: 
   \[
   \frac{2}{7}z = -\frac{10}{7} \quad \Rightarrow \quad z = -5
   \]

2. Substitute \( z = -5 \) into the 2nd row equation:
   \[
   \frac{7}{3}y + \frac{22}{3}(-5) = 7 \quad \Rightarrow \quad \frac{7}{3}y - \frac{110}{3} = 7
   \]
   Multiply through by 3 to simplify:
   \[
   7y - 110 = 21 \quad \Rightarrow \quad 7y = 131 \quad \Rightarrow \quad y = 19
   \]

3. Substitute \( y = 19 \) and \( z = -5 \) into the 1st row equation:
   \[
   3x - y + 2z = 12 \quad \Rightarrow \quad 3x - 19 + 2(-5) = 12
   \]
   Simplify:
   \[
   3x - 19 - 10 = 12 \quad \Rightarrow \quad 3x = 41 \quad \Rightarrow \quad x = \frac{41}{3}
   \]

### Solution:

The solution to the system of equations is:

\[
x = \frac{41}{3}, \quad y = 19, \quad z = -5
\]

Would you like more details or have any further questions?

Here are 5 related questions:
1. How can we use Gauss-Jordan elimination for this system?
2. What is the difference between Gauss elimination and Gauss-Jordan elimination?
3. How do you determine if a system of equations has no solution or infinitely many solutions?
4. How can you solve systems of equations using matrix inversion?
5. Can the Gauss elimination method be used for non-square matrices?

**Tip**: Always check if the matrix is singular (i.e., determinant is zero) before solving using elimination, as this can indicate no unique solutions exist.

Solve the following system of equations by using Gauss-Elimination method: 3x - y + 2z = 12, x + 2y + 8z = 11, 2x - 2y - z = 2

Explore how to solve a system of three linear equations using the Gauss Elimination Method. This step-by-step solution includes converting to an augmented matrix, applying row operations, and performing back substitution. Perfect for high school students in Grades 10-12 learning linear algebra concepts.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation