https://r2.erweima.ai/mathbot/image/1726426906727wmm9vdqw.jpg

The system of linear equations provided in the image is:

\[
\begin{aligned}
1. &\quad x + y + z = 2 \\
2. &\quad 3x - 2y - z = 4 \\
3. &\quad -2x + y + 2z = 2 \\
\end{aligned}
\]

We will solve this system using **Gaussian elimination**.

### Step 1: Write the augmented matrix
The system can be represented as an augmented matrix:

\[
\left[
\begin{array}{ccc|c}
1 & 1 & 1 & 2 \\
3 & -2 & -1 & 4 \\
-2 & 1 & 2 & 2 \\
\end{array}
\right]
\]

### Step 2: Perform Gaussian elimination

1. **First pivot (Row 1, Column 1):** 
   The pivot element is 1 in row 1. We aim to make the first element of rows 2 and 3 zero by performing row operations:
   \[
   R_2 \to R_2 - 3R_1 \quad \text{and} \quad R_3 \to R_3 + 2R_1
   \]
   After applying the row operations, we get:
   \[
   \left[
   \begin{array}{ccc|c}
   1 & 1 & 1 & 2 \\
   0 & -5 & -4 & -2 \\
   0 & 3 & 4 & 6 \\
   \end{array}
   \right]
   \]

2. **Second pivot (Row 2, Column 2):**
   The pivot element is -5 in row 2. We now eliminate the second element of row 3:
   \[
   R_3 \to R_3 + \frac{3}{5}R_2
   \]
   After performing this operation, the augmented matrix becomes:
   \[
   \left[
   \begin{array}{ccc|c}
   1 & 1 & 1 & 2 \\
   0 & -5 & -4 & -2 \\
   0 & 0 & \frac{8}{5} & \frac{12}{5} \\
   \end{array}
   \right]
   \]

3. **Third pivot (Row 3, Column 3):**
   The pivot element is \(\frac{8}{5}\), so we simplify row 3 by dividing it by \(\frac{8}{5}\):
   \[
   R_3 \to \frac{5}{8}R_3
   \]
   which gives:
   \[
   \left[
   \begin{array}{ccc|c}
   1 & 1 & 1 & 2 \\
   0 & -5 & -4 & -2 \\
   0 & 0 & 1 & \frac{3}{2} \\
   \end{array}
   \right]
   \]

### Step 3: Back substitution
Now that we have the matrix in row-echelon form, we can solve the system using back substitution:

1. From row 3: \( z = \frac{3}{2} \)
2. From row 2: 
   \[
   -5y - 4z = -2 \implies -5y - 4\left(\frac{3}{2}\right) = -2 \implies -5y - 6 = -2 \implies y = \frac{4}{5}
   \]
3. From row 1:
   \[
   x + y + z = 2 \implies x + \frac{4}{5} + \frac{3}{2} = 2
   \]
   Solving for \(x\):
   \[
   x + \frac{8}{10} + \frac{15}{10} = 2 \implies x + \frac{23}{10} = 2 \implies x = \frac{-3}{10}
   \]

### Final Solution:
\[
x = \frac{-3}{10}, \quad y = \frac{4}{5}, \quad z = \frac{3}{2}
\]

Let me know if you'd like more details or clarifications!

---

Here are five related questions:

1. How does Gaussian elimination differ from the Gauss-Jordan method?
2. What are the potential cases where Gaussian elimination leads to no solutions or infinitely many solutions?
3. How would you solve this system using matrix inversion?
4. Can this system be solved graphically? What would the geometrical interpretation be?
5. How can partial pivoting improve the stability of the Gaussian elimination process?

**Tip**: When solving linear systems, always check for the possibility of row swapping to avoid numerical instability (this is called pivoting).

Solve the system of linear equations using the Gaussian elimination method:
1. x + y + z = 2
2. 3x - 2y - z = 4
3. -2x + y + 2z = 2

Learn how to solve a 3x3 system of linear equations step-by-step using Gaussian elimination. The example involves solving for x, y, and z using augmented matrices and back substitution. Suitable for students in Grades 10-12.

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