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To solve the given augmented matrices using the "SPL OBE" (Sistem Persamaan Linear with Operasi Baris Elementer, or Gaussian Elimination method), we will apply elementary row operations to reduce each matrix to row-echelon form and find the corresponding solutions.

### a. 
The given augmented matrix is:

\[
\begin{bmatrix}
1 & 0 & -1 & | & 2 \\
2 & 1 & 1 & | & 3 \\
0 & -1 & 2 & | & 4
\end{bmatrix}
\]

**Step 1:** Subtract 2 times row 1 from row 2 to eliminate the first element of row 2.

\[
R_2 \to R_2 - 2R_1
\]

\[
\begin{bmatrix}
1 & 0 & -1 & | & 2 \\
0 & 1 & 3 & | & -1 \\
0 & -1 & 2 & | & 4
\end{bmatrix}
\]

**Step 2:** Add row 2 to row 3 to eliminate the second element of row 3.

\[
R_3 \to R_3 + R_2
\]

\[
\begin{bmatrix}
1 & 0 & -1 & | & 2 \\
0 & 1 & 3 & | & -1 \\
0 & 0 & 5 & | & 3
\end{bmatrix}
\]

**Step 3:** Divide row 3 by 5 to simplify the third row.

\[
R_3 \to \frac{1}{5}R_3
\]

\[
\begin{bmatrix}
1 & 0 & -1 & | & 2 \\
0 & 1 & 3 & | & -1 \\
0 & 0 & 1 & | & \frac{3}{5}
\end{bmatrix}
\]

**Step 4:** Use row 3 to eliminate the third element of row 1 and row 2.

\[
R_1 \to R_1 + R_3 \quad \text{and} \quad R_2 \to R_2 - 3R_3
\]

\[
\begin{bmatrix}
1 & 0 & 0 & | & \frac{13}{5} \\
0 & 1 & 0 & | & -\frac{14}{5} \\
0 & 0 & 1 & | & \frac{3}{5}
\end{bmatrix}
\]

The solution to the system is:

\[
x_1 = \frac{13}{5}, \quad x_2 = -\frac{14}{5}, \quad x_3 = \frac{3}{5}
\]

---

### b.
The given augmented matrix is:

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

This system is already simplified for the first two rows. We can eliminate the first and second elements of the third row by subtracting row 1 and adding row 2:

\[
R_3 \to R_3 - R_1 \quad \text{and} \quad R_3 \to R_3 + R_2
\]

\[
\begin{bmatrix}
1 & 0 & 0 & | & 0 \\
0 & 1 & 0 & | & 0 \\
0 & 0 & 1 & | & 4
\end{bmatrix}
\]

The solution is:

\[
x_1 = 0, \quad x_2 = 0, \quad x_3 = 4
\]

---

### c.
The given augmented matrix is:

\[
\begin{bmatrix}
1 & 2 & 3 & 4 & | & 5 \\
5 & 4 & 3 & 2 & | & 1
\end{bmatrix}
\]

**Step 1:** Subtract 5 times row 1 from row 2.

\[
R_2 \to R_2 - 5R_1
\]

\[
\begin{bmatrix}
1 & 2 & 3 & 4 & | & 5 \\
0 & -6 & -12 & -18 & | & -24
\end{bmatrix}
\]

**Step 2:** Divide row 2 by -6.

\[
R_2 \to \frac{1}{-6}R_2
\]

\[
\begin{bmatrix}
1 & 2 & 3 & 4 & | & 5 \\
0 & 1 & 2 & 3 & | & 4
\end{bmatrix}
\]

**Step 3:** Subtract 2 times row 2 from row 1 to eliminate the second element of row 1.

\[
R_1 \to R_1 - 2R_2
\]

\[
\begin{bmatrix}
1 & 0 & -1 & -2 & | & -3 \\
0 & 1 & 2 & 3 & | & 4
\end{bmatrix}
\]

The system is now simplified, and the solution is:

\[
x_1 = -3, \quad x_2 = 4
\]

---

### d.
The given augmented matrix is:

\[
\begin{bmatrix}
1 & 0 & 0 & 0 & | & 1 \\
0 & 1 & 0 & 0 & | & 2 \\
0 & 0 & 1 & 0 & | & 3 \\
0 & 0 & 0 & 1 & | & 4
\end{bmatrix}
\]

This system is already in row-echelon form, and the solution is:

\[
x_1 = 1, \quad x_2 = 2, \quad x_3 = 3, \quad x_4 = 4
\]

---

Let me know if you need further clarification or details!

### Additional Questions:
1. How can we verify if the solutions found are correct?
2. Can you explain how Gaussian elimination helps simplify systems of linear equations?
3. What are some potential pitfalls of using the Gaussian elimination method?
4. How would the process differ for larger matrices with more variables?
5. Can these solutions be checked using matrix inversion?

### Tip:
When performing Gaussian elimination, always ensure that no row with all zero elements ends up at the top of the matrix—this simplifies the process and avoids confusion.

Carilah sistem persamaan linear yang bersesuaian dengan masing-masing matriks yang diperbesar berikut!

Learn how to solve systems of linear equations using Gaussian elimination for augmented matrices. This problem involves reducing matrices to row-echelon form through elementary row operations and finding solutions for the system of equations. Suitable for undergraduate or advanced high school students.

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