Solve the system of linear equations by gaussian elimination method by finding an equivalent 

system in row – echelon form x - 2y + 4z = 0, x - y + 2z = 5 and 3x - 5y + 13z = 32.

To solve the given system of linear equations using the Gaussian elimination method, we will follow these steps:

### Given System of Equations:
\[
\begin{aligned}
1) & \quad x - 2y + 4z = 0 \\
2) & \quad x - y + 2z = 5 \\
3) & \quad 3x - 5y + 13z = 32 \\
\end{aligned}
\]

### Step 1: Write the system as an augmented matrix
The augmented matrix for the system is:

\[
\begin{pmatrix}
1 & -2 & 4 & | & 0 \\
1 & -1 & 2 & | & 5 \\
3 & -5 & 13 & | & 32 \\
\end{pmatrix}
\]

### Step 2: Perform row operations to obtain row-echelon form

#### **Operation 1:** Subtract Row 1 from Row 2 to eliminate the first entry in the second row:

\[
R_2 \rightarrow R_2 - R_1
\]

\[
\begin{pmatrix}
1 & -2 & 4 & | & 0 \\
0 & 1 & -2 & | & 5 \\
3 & -5 & 13 & | & 32 \\
\end{pmatrix}
\]

#### **Operation 2:** Subtract 3 times Row 1 from Row 3 to eliminate the first entry in the third row:

\[
R_3 \rightarrow R_3 - 3R_1
\]

\[
\begin{pmatrix}
1 & -2 & 4 & | & 0 \\
0 & 1 & -2 & | & 5 \\
0 & 1 & 1 & | & 32 \\
\end{pmatrix}
\]

#### **Operation 3:** Subtract Row 2 from Row 3 to eliminate the second entry in the third row:

\[
R_3 \rightarrow R_3 - R_2
\]

\[
\begin{pmatrix}
1 & -2 & 4 & | & 0 \\
0 & 1 & -2 & | & 5 \\
0 & 0 & 3 & | & 17 \\
\end{pmatrix}
\]

Now, the matrix is in **row-echelon form**.

### Step 3: Write the equivalent system of equations

The equivalent system of equations corresponding to the row-echelon form is:

\[
\begin{aligned}
1) & \quad x - 2y + 4z = 0 \\
2) & \quad y - 2z = 5 \\
3) & \quad 3z = 17 \\
\end{aligned}
\]

### Step 4: Solve the system using back substitution

#### **Solve for \(z\):**
From the third equation:

\[
3z = 17
\]

\[
z = \frac{17}{3}
\]

#### **Solve for \(y\):**
Substitute \(z = \frac{17}{3}\) into the second equation:

\[
y - 2\left(\frac{17}{3}\right) = 5
\]

\[
y - \frac{34}{3} = 5
\]

\[
y = 5 + \frac{34}{3} = \frac{15}{3} + \frac{34}{3} = \frac{49}{3}
\]

#### **Solve for \(x\):**
Substitute \(y = \frac{49}{3}\) and \(z = \frac{17}{3}\) into the first equation:

\[
x - 2\left(\frac{49}{3}\right) + 4\left(\frac{17}{3}\right) = 0
\]

\[
x - \frac{98}{3} + \frac{68}{3} = 0
\]

\[
x - \frac{30}{3} = 0
\]

\[
x = \frac{30}{3} = 10
\]

### Final Solution
The solution to the system is:

\[
\boxed{\left( x, y, z \right) = \left( 10, \frac{49}{3}, \frac{17}{3} \right)}
\]

Would you like any details or have any questions?

### Related Questions
1. What is the Gaussian elimination method, and how is it applied to solve systems of equations?
2. How can row operations affect the solution of a system of linear equations?
3. What is the difference between row-echelon form and reduced row-echelon form?
4. How can you verify the solution obtained from Gaussian elimination?
5. What are some special cases to consider when using Gaussian elimination (e.g., no solution, infinite solutions)?
6. Can Gaussian elimination be used for non-square systems of equations? If so, how?
7. How is Gaussian elimination related to matrix inversion?
8. How do pivoting techniques improve the stability of Gaussian elimination?

### Tip
Always double-check each row operation for accuracy, as small mistakes can easily propagate through the solution process.

Learn how to solve a system of linear equations using the Gaussian elimination method. This detailed step-by-step solution covers the process of transforming the equations into row-echelon form and solving for the variables x, y, and z. Suitable for advanced high school students or college-level learners.

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