Find the solution of the following system of equations using Gaussian elimination method and show the steps. 
x-y+z = 4 
5x-y+z = 6 
3x-y+5z = 5


To solve the given system of equations using Gaussian elimination, we will perform row operations to convert the augmented matrix into an upper triangular form and then use back-substitution to find the solution. Here is the system of equations:

\[
\begin{cases}
x - y + z = 4 \\
5x - y + z = 6 \\
3x - y + 5z = 5
\end{cases}
\]

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

\[
\begin{bmatrix}
1 & -1 & 1 & | & 4 \\
5 & -1 & 1 & | & 6 \\
3 & -1 & 5 & | & 5 \\
\end{bmatrix}
\]

### Step 2: Make the first element in the first column (pivot) equal to 1
The first row already has a 1 as the leading entry in the first column, so no row operation is needed.

### Step 3: Make all entries below the pivot in the first column zero
To make the entries below the first pivot (1 in the first row) zero, we perform the following operations:

- \( R_2 = R_2 - 5R_1 \)
- \( R_3 = R_3 - 3R_1 \)

Applying these row operations:

\[
R_2 = [5, -1, 1, |, 6] - 5 \cdot [1, -1, 1, |, 4] = [0, 4, -4, |, -14]
\]
\[
R_3 = [3, -1, 5, |, 5] - 3 \cdot [1, -1, 1, |, 4] = [0, 2, 2, |, -7]
\]

So the matrix becomes:

\[
\begin{bmatrix}
1 & -1 & 1 & | & 4 \\
0 & 4 & -4 & | & -14 \\
0 & 2 & 2 & | & -7 \\
\end{bmatrix}
\]

### Step 4: Make the second pivot (4 in the second row, second column) equal to 1
To make this entry 1, we divide the entire second row by 4:

\[
R_2 = \frac{1}{4} R_2 = [0, 1, -1, |, -\frac{7}{2}]
\]

The matrix is now:

\[
\begin{bmatrix}
1 & -1 & 1 & | & 4 \\
0 & 1 & -1 & | & -\frac{7}{2} \\
0 & 2 & 2 & | & -7 \\
\end{bmatrix}
\]

### Step 5: Make all entries below the pivot in the second column zero
To eliminate the entry in the third row, second column (2), we use the operation:

- \( R_3 = R_3 - 2R_2 \)

Calculating this:

\[
R_3 = [0, 2, 2, |, -7] - 2 \cdot [0, 1, -1, |, -\frac{7}{2}] = [0, 0, 4, |, 0]
\]

The matrix is now:

\[
\begin{bmatrix}
1 & -1 & 1 & | & 4 \\
0 & 1 & -1 & | & -\frac{7}{2} \\
0 & 0 & 4 & | & 0 \\
\end{bmatrix}
\]

### Step 6: Make the third pivot (4 in the third row, third column) equal to 1
To do this, divide the entire third row by 4:

\[
R_3 = \frac{1}{4} R_3 = [0, 0, 1, |, 0]
\]

The matrix is now in row echelon form:

\[
\begin{bmatrix}
1 & -1 & 1 & | & 4 \\
0 & 1 & -1 & | & -\frac{7}{2} \\
0 & 0 & 1 & | & 0 \\
\end{bmatrix}
\]

### Step 7: Back-substitute to find the values of \( x \), \( y \), and \( z \)

From the third row, we have:
\[
z = 0
\]

Substitute \( z = 0 \) into the second row:
\[
y - z = -\frac{7}{2} \Rightarrow y = -\frac{7}{2}
\]

Finally, substitute \( y = -\frac{7}{2} \) and \( z = 0 \) into the first row:
\[
x - y + z = 4 \Rightarrow x - \left(-\frac{7}{2}\right) = 4 \Rightarrow x + \frac{7}{2} = 4 \Rightarrow x = 4 - \frac{7}{2} = \frac{8 - 7}{2} = \frac{1}{2}
\]

### Solution
The solution to the system is:
\[
x = \frac{1}{2}, \quad y = -\frac{7}{2}, \quad z = 0
\]

Would you like more details on any part of this process?

---

Here are five related questions for further exploration:

1. How would the solution change if one of the equations in the system was altered?
2. What happens if two rows are linearly dependent? Can we still find a unique solution?
3. How does Gaussian elimination differ from Gauss-Jordan elimination?
4. How can Gaussian elimination be applied to systems with no solutions?
5. How can we use matrices and determinants to verify the uniqueness of solutions?

**Tip:** To quickly identify row operations in Gaussian elimination, look at pivot positions and aim to simplify entries below and above them sequentially.

Find the solution of the following system of equations using Gaussian elimination method and show the steps. x-y+z = 4, 5x-y+z = 6, 3x-y+5z = 5

Discover the steps to solve the system of linear equations x - y + z = 4, 5x - y + z = 6, 3x - y + 5z = 5 using Gaussian elimination. This solution walks through transforming the augmented matrix to row echelon form and using back-substitution, covering essential algebraic concepts in Grades 10-12.

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