Certainly! Let's solve the system of equations using Gaussian elimination. The system is:
[tex]\[
\begin{aligned}
1. & \quad x - y + z = 4 \\
2. & \quad 5x - y + z = 6 \\
3. & \quad 3x - y + 5z = 5
\end{aligned}
\][/tex]
### Step 1: Write the augmented matrix for the system of equations.
[tex]\[
\left[\begin{array}{ccc|c}
1 & -1 & 1 & 4 \\
5 & -1 & 1 & 6 \\
3 & -1 & 5 & 5 \\
\end{array}\right]
\][/tex]
### Step 2: Use row operations to form an upper triangular matrix.
#### 2.1. Eliminate the entries below the pivot in the first column (i.e., make the first element of the first row as the pivot and eliminate below it).
- For Row 2: [tex]\(R_2 \rightarrow R_2 - 5R_1\)[/tex]
[tex]\[
\left[\begin{array}{ccc|c}
1 & -1 & 1 & 4 \\
0 & 4 & -4 & -14 \\
3 & -1 & 5 & 5 \\
\end{array}\right]
\][/tex]
- For Row 3: [tex]\(R_3 \rightarrow R_3 - 3R_1\)[/tex]
[tex]\[
\left[\begin{array}{ccc|c}
1 & -1 & 1 & 4 \\
0 & 4 & -4 & -14 \\
0 & 2 & 2 & -7 \\
\end{array}\right]
\][/tex]
#### 2.2. Eliminate the entry below the pivot in the second column (i.e., make the second element of the second row as the pivot and eliminate below it).
First, make the leading coefficient of the second row to be 1 (pivot normalization):
- Normalizing [tex]\(R_2\)[/tex]:
[tex]\[
R_2 \rightarrow \frac{R_2}{4}
\][/tex]
[tex]\[
\left[\begin{array}{ccc|c}
1 & -1 & 1 & 4 \\
0 & 1 & -1 & -3.5 \\
0 & 2 & 2 & -7 \\
\end{array}\right]
\][/tex]
- For Row 3: [tex]\(R_3 \rightarrow R_3 - 2R_2\)[/tex]
[tex]\[
\left[\begin{array}{ccc|c}
1 & -1 & 1 & 4 \\
0 & 1 & -1 & -3.5 \\
0 & 0 & 4 & 0 \\
\end{array}\right]
\][/tex]
### Step 3: Normalize the third row.
- Normalizing [tex]\(R_3\)[/tex]:
[tex]\[
R_3 \rightarrow \frac{R_3}{4}
\][/tex]
[tex]\[
\left[\begin{array}{ccc|c}
1 & -1 & 1 & 4 \\
0 & 1 & -1 & -3.5 \\
0 & 0 & 1 & 0 \\
\end{array}\right]
\][/tex]
### Step 4: Use back-substitution to find the values of [tex]\(x\)[/tex], [tex]\(y\)[/tex], and [tex]\(z\)[/tex].
#### 4.1. From the third row:
[tex]\[
z = 0
\][/tex]
#### 4.2. From the second row:
[tex]\[
y - z = -3.5 \implies y - 0 = -3.5 \implies y = -3.5
\][/tex]
#### 4.3. From the first row:
[tex]\[
x - y + z = 4 \implies x - (-3.5) + 0 = 4 \implies x + 3.5 = 4 \implies x = 0.5
\][/tex]
### Conclusion
The solution to the system of equations is:
[tex]\[
\begin{aligned}
x & = 0.5 \\
y & = -3.5 \\
z & = 0
\end{aligned}
\][/tex]
