this does not have upper triangular matric

You're correct. The matrix transformation I performed didn't lead to a correct upper triangular matrix. Let's go through the correct steps to achieve the upper triangular form using the Gauss Elimination method.

### Step 1: Write the system of equations in standard form
The equations are already in standard form:

1. \( 3x + 4y + 5z = 18 \)
2. \( 2x - y + 8z = 13 \)
3. \( 5x - 2y + 7z = 20 \)

### Step 2: Write the augmented matrix
The augmented matrix corresponding to the system is:

\[
\begin{pmatrix}
3 & 4 & 5 & | & 18 \\
2 & -1 & 8 & | & 13 \\
5 & -2 & 7 & | & 20 \\
\end{pmatrix}
\]

### Step 3: Perform row operations to get an upper triangular matrix

#### a) Make the first element of the first column a pivot (if necessary):
The first row is fine as is, so let's eliminate the elements below it in the first column.

- **Eliminate the element in the second row, first column:**

   \[
   R2 \rightarrow R2 - \frac{2}{3}R1
   \]

   Calculation:
   \[
   \begin{pmatrix}
   3 & 4 & 5 & | & 18 \\
   0 & -\frac{11}{3} & \frac{14}{3} & | & 1 \\
   5 & -2 & 7 & | & 20 \\
   \end{pmatrix}
   \]

- **Eliminate the element in the third row, first column:**

   \[
   R3 \rightarrow R3 - \frac{5}{3}R1
   \]

   Calculation:
   \[
   \begin{pmatrix}
   3 & 4 & 5 & | & 18 \\
   0 & -\frac{11}{3} & \frac{14}{3} & | & 1 \\
   0 & -\frac{26}{3} & \frac{4}{3} & | & -10 \\
   \end{pmatrix}
   \]

#### b) Make the pivot in the second column:

- **Eliminate the element in the third row, second column:**

   We now eliminate the element below the pivot \( y \) in the third row by the following operation:

   \[
   R3 \rightarrow R3 - \frac{-26/3}{-11/3}R2
   \]

   Simplification gives:

   \[
   R3 \rightarrow R3 + 2 \times R2
   \]

   Calculation:
   \[
   \begin{pmatrix}
   3 & 4 & 5 & | & 18 \\
   0 & -\frac{11}{3} & \frac{14}{3} & | & 1 \\
   0 & 0 & -6 & | & -8 \\
   \end{pmatrix}
   \]

Now we have the upper triangular form.

### Step 4: Back substitution

1. **Solve for \( z \):**

   \[
   -6z = -8 \quad \Rightarrow \quad z = \frac{4}{3}
   \]

2. **Solve for \( y \):**

   Using the second row:

   \[
   -\frac{11}{3}y + \frac{14}{3} \times \frac{4}{3} = 1
   \]
   Simplifying:

   \[
   -\frac{11}{3}y + \frac{56}{9} = 1 \quad \Rightarrow \quad y = \frac{53}{33}
   \]

3. **Solve for \( x \):**

   Using the first row:

   \[
   3x + 4 \times \frac{53}{33} + 5 \times \frac{4}{3} = 18
   \]

   Simplifying:

   \[
   x = \frac{14}{9}
   \]

### Conclusion
The correct solutions are:

- \( x = \frac{14}{9} \)
- \( y = \frac{53}{33} \)
- \( z = \frac{4}{3} \)

Would you like more details on these steps? Here are some related questions:

1. What does it mean for a matrix to be in upper triangular form?
2. How can you check if a system of equations has a unique solution?
3. How does pivoting help in the Gauss Elimination method?
4. Can the Gauss Elimination method be used on non-square matrices?
5. What are the conditions for a system of linear equations to have no solutions?

**Tip:** Always keep track of each row operation to avoid mistakes when forming the upper triangular matrix.

Explore the Gauss Elimination method to transform a matrix into upper triangular form. Detailed steps and examples provided for solving systems of equations using matrix operations. Suitable for advanced high school students and anyone interested in linear algebra concepts.

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