Solve the following set of linear equations using gaussian elimination method:

x¹ + 3x² + 2x³ = 3
2x¹ - x² - 3x³ = -8
5x¹ + 2x² + x³ = 9

To solve the system of linear equations using Gaussian elimination, let's follow these steps:

The equations are:
\[
\begin{cases}
x_1 + 3x_2 + 2x_3 = 3 \\
2x_1 - x_2 - 3x_3 = -8 \\
5x_1 + 2x_2 + x_3 = 9
\end{cases}
\]

This system can be written as an augmented matrix:
\[
\begin{bmatrix}
1 & 3 & 2 & | & 3 \\
2 & -1 & -3 & | & -8 \\
5 & 2 & 1 & | & 9
\end{bmatrix}
\]

### Step 1: Make the leading coefficient of the first row (Row 1) equal to 1.
Row 1 already has a leading 1 in the \(x_1\) position.

### Step 2: Make the leading coefficient of \(x_1\) in Row 2 and Row 3 equal to zero.
To eliminate \(x_1\) from Row 2 and Row 3:
- **Row 2**: \( R_2 - 2R_1 \)
- **Row 3**: \( R_3 - 5R_1 \)

#### Perform these operations:
1. Row 2: \( [2, -1, -3, |, -8] - 2 \times [1, 3, 2, |, 3] = [0, -7, -7, |, -14] \)
2. Row 3: \( [5, 2, 1, |, 9] - 5 \times [1, 3, 2, |, 3] = [0, -13, -9, |, -6] \)

The matrix now is:
\[
\begin{bmatrix}
1 & 3 & 2 & | & 3 \\
0 & -7 & -7 & | & -14 \\
0 & -13 & -9 & | & -6
\end{bmatrix}
\]

### Step 3: Make the leading coefficient of \(x_2\) in Row 2 equal to 1.
Divide Row 2 by \(-7\):
\[
\begin{bmatrix}
1 & 3 & 2 & | & 3 \\
0 & 1 & 1 & | & 2 \\
0 & -13 & -9 & | & -6
\end{bmatrix}
\]

### Step 4: Make the leading coefficient of \(x_2\) in Row 1 and Row 3 equal to zero.
To eliminate \(x_2\) from Row 1 and Row 3:
- **Row 1**: \( R_1 - 3R_2 \)
- **Row 3**: \( R_3 + 13R_2 \)

#### Perform these operations:
1. Row 1: \( [1, 3, 2, |, 3] - 3 \times [0, 1, 1, |, 2] = [1, 0, -1, |, -3] \)
2. Row 3: \( [0, -13, -9, |, -6] + 13 \times [0, 1, 1, |, 2] = [0, 0, 4, |, 20] \)

The matrix now is:
\[
\begin{bmatrix}
1 & 0 & -1 & | & -3 \\
0 & 1 & 1 & | & 2 \\
0 & 0 & 4 & | & 20
\end{bmatrix}
\]

### Step 5: Make the leading coefficient of \(x_3\) in Row 3 equal to 1.
Divide Row 3 by 4:
\[
\begin{bmatrix}
1 & 0 & -1 & | & -3 \\
0 & 1 & 1 & | & 2 \\
0 & 0 & 1 & | & 5
\end{bmatrix}
\]

### Step 6: Make the leading coefficient of \(x_3\) in Row 1 and Row 2 equal to zero.
To eliminate \(x_3\) from Row 1 and Row 2:
- **Row 1**: \( R_1 + R_3 \)
- **Row 2**: \( R_2 - R_3 \)

#### Perform these operations:
1. Row 1: \( [1, 0, -1, |, -3] + [0, 0, 1, |, 5] = [1, 0, 0, |, 2] \)
2. Row 2: \( [0, 1, 1, |, 2] - [0, 0, 1, |, 5] = [0, 1, 0, |, -3] \)

The matrix now is in reduced row echelon form:
\[
\begin{bmatrix}
1 & 0 & 0 & | & 2 \\
0 & 1 & 0 & | & -3 \\
0 & 0 & 1 & | & 5
\end{bmatrix}
\]

### Solution:
From this, we can see:
\[
x_1 = 2, \quad x_2 = -3, \quad x_3 = 5
\]

Thus, the solution to the system is:
\[
(x_1, x_2, x_3) = (2, -3, 5)
\]

Would you like more details on any step, or have questions about this solution?

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#### Related Questions:
1. How is Gaussian elimination different from Gauss-Jordan elimination?
2. How can Gaussian elimination be applied to solve systems with infinitely many solutions?
3. What are the requirements for a matrix to have a unique solution using Gaussian elimination?
4. How does Gaussian elimination handle inconsistent systems?
5. What is the computational complexity of Gaussian elimination?

#### Tip:
Gaussian elimination is a systematic method to reduce matrices to row echelon form, making it easier to back-substitute and solve systems of equations.

Explore a step-by-step solution using Gaussian elimination to solve the system of equations x¹ + 3x² + 2x³ = 3, 2x¹ - x² - 3x³ = -8, and 5x¹ + 2x² + x³ = 9. This guide covers essential linear algebra concepts for high school students in Grades 10-12.

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