x-2y+3z = 1 O Solve x +2y-z = 13 by using Gaussian Elimination. (3x + 2y - 5z = 3

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### Solving Systems of Equations Using Gaussian Elimination

---

**Problem Statement:**

Solve the system of equations using Gaussian Elimination:

1. \( x - 2y + 3z = 1 \)
2. \( x + 2y - z = 13 \)
3. \( 3x + 2y - 5z = 3 \)

**Hint:** Find the Row Echelon Form

---

**Solution Steps:**

1. **Write the system as a matrix:**

   \[
   \begin{bmatrix}
   1 & -2 & 3 & | & 1 \\
   1 & 2 & -1 & | & 13 \\
   3 & 2 & -5 & | & 3
   \end{bmatrix}
   \]

2. **Perform Gaussian Elimination to reach Row Echelon Form:**

   - Subtract the first row from the second row:
     \[
     (-x + 2y - z = 13) - (x - 2y + 3z = 1) \rightarrow 4y - 4z = 12
     \]
     Simplifying gives:
     \[
     4y - 4z = 12 \quad \Rightarrow \quad y - z = 3
     \]

   - Subtract three times the first row from the third row:
     \[
     (3x + 2y - 5z = 3) - 3(x - 2y + 3z = 1) \rightarrow -4y + 14z = 0
     \]
     Simplifying gives:
     \[
     -4y + 14z = 0 \quad \Rightarrow \quad 2y - 7z = 0 \quad
     \]

3. **Further simplification of the equations results in:**
   
   \[
   x + y + z = 1
   \]
   \[
   y - z = 2
   \]
   \[
   z = 3
   \]

4. **Final Solution:**

   \[
   \begin{cases}
   x + y + z = 1 \\
   y - z = 2 \\
   z = 3
   \end{cases}
   \]
Transcribed Image Text:### Solving Systems of Equations Using Gaussian Elimination --- **Problem Statement:** Solve the system of equations using Gaussian Elimination: 1. \( x - 2y + 3z = 1 \) 2. \( x + 2y - z = 13 \) 3. \( 3x + 2y - 5z = 3 \) **Hint:** Find the Row Echelon Form --- **Solution Steps:** 1. **Write the system as a matrix:** \[ \begin{bmatrix} 1 & -2 & 3 & | & 1 \\ 1 & 2 & -1 & | & 13 \\ 3 & 2 & -5 & | & 3 \end{bmatrix} \] 2. **Perform Gaussian Elimination to reach Row Echelon Form:** - Subtract the first row from the second row: \[ (-x + 2y - z = 13) - (x - 2y + 3z = 1) \rightarrow 4y - 4z = 12 \] Simplifying gives: \[ 4y - 4z = 12 \quad \Rightarrow \quad y - z = 3 \] - Subtract three times the first row from the third row: \[ (3x + 2y - 5z = 3) - 3(x - 2y + 3z = 1) \rightarrow -4y + 14z = 0 \] Simplifying gives: \[ -4y + 14z = 0 \quad \Rightarrow \quad 2y - 7z = 0 \quad \] 3. **Further simplification of the equations results in:** \[ x + y + z = 1 \] \[ y - z = 2 \] \[ z = 3 \] 4. **Final Solution:** \[ \begin{cases} x + y + z = 1 \\ y - z = 2 \\ z = 3 \end{cases} \]
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