x1+4x2-2x3+8x4=12
x2-7x3+2x4=-14
5x3-x4=7
x3+3x4=15 
selesaikan dengan cara obe

To solve the system of linear equations using **OBE (Operasi Baris Elementer)**, or **Gaussian Elimination**, we will use row operations to reduce the augmented matrix to its row echelon form (REF) and then to its reduced row echelon form (RREF). The given system of equations is:

\[
\begin{cases}
x_1 + 4x_2 - 2x_3 + 8x_4 = 12 \\
x_2 - 7x_3 + 2x_4 = -14 \\
5x_3 - x_4 = 7 \\
x_3 + 3x_4 = 15 
\end{cases}
\]

### Step 1: Write the Augmented Matrix

The augmented matrix for this system is:

\[
\left[ \begin{array}{cccc|c}
1 & 4 & -2 & 8 & 12 \\
0 & 1 & -7 & 2 & -14 \\
0 & 0 & 5 & -1 & 7 \\
0 & 0 & 1 & 3 & 15 
\end{array} \right]
\]

### Step 2: Apply Elementary Row Operations

We will perform a series of row operations to transform the matrix to REF and then RREF.

1. **R3 -> (1/5) * R3** to make the leading coefficient of the third row a 1:

\[
\left[ \begin{array}{cccc|c}
1 & 4 & -2 & 8 & 12 \\
0 & 1 & -7 & 2 & -14 \\
0 & 0 & 1 & -\frac{1}{5} & \frac{7}{5} \\
0 & 0 & 1 & 3 & 15 
\end{array} \right]
\]

2. **R4 -> R4 - R3** to make the leading coefficient of the fourth row 0:

\[
\left[ \begin{array}{cccc|c}
1 & 4 & -2 & 8 & 12 \\
0 & 1 & -7 & 2 & -14 \\
0 & 0 & 1 & -\frac{1}{5} & \frac{7}{5} \\
0 & 0 & 0 & \frac{16}{5} & \frac{68}{5} 
\end{array} \right]
\]

3. **R4 -> (5/16) * R4** to make the leading coefficient of the fourth row a 1:

\[
\left[ \begin{array}{cccc|c}
1 & 4 & -2 & 8 & 12 \\
0 & 1 & -7 & 2 & -14 \\
0 & 0 & 1 & -\frac{1}{5} & \frac{7}{5} \\
0 & 0 & 0 & 1 & \frac{17}{4} 
\end{array} \right]
\]

4. **R3 -> R3 + \frac{1}{5} * R4** to make the third element in the third row a 0:

\[
\left[ \begin{array}{cccc|c}
1 & 4 & -2 & 8 & 12 \\
0 & 1 & -7 & 2 & -14 \\
0 & 0 & 1 & 0 & 6 \\
0 & 0 & 0 & 1 & \frac{17}{4} 
\end{array} \right]
\]

5. **R2 -> R2 + 7 * R3** to eliminate the third element in the second row:

\[
\left[ \begin{array}{cccc|c}
1 & 4 & -2 & 8 & 12 \\
0 & 1 & 0 & 2.6 & 28 \\
0 & 0 & 1 & 0 & 6 \\
0 & 0 & 0 & 1 & \frac{17}{4} 
\end{array} \right]
\]

6. **R1 -> R1 + 2 * R3** to eliminate the third element in the first row:

\[
\left[ \begin{array}{cccc|c}
1 & 4 & 0 & 8 & 24 \\
0 & 1 & 0 & 2.6 & 28 \\
0 & 0 & 1 & 0 & 6 \\
0 & 0 & 0 & 1 & \frac{17}{4} 
\end{array} \right]
\]

7. **R1 -> R1 - 8 * R4** and **R2 -> R2 - 2.6 * R4** to make all elements in the last column of R1 and R2 zeroes:

\[
\left[ \begin{array}{cccc|c}
1 & 4 & 0 & 0 & -10 \\
0 & 1 & 0 & 0 & 17 \\
0 & 0 & 1 & 0 & 6 \\
0 & 0 & 0 & 1 & \frac{17}{4} 
\end{array} \right]
\]

### Step 3: Interpret the Results

Now, the augmented matrix is in reduced row echelon form. From this, we can read off the solutions:

\[
\begin{cases}
x_1 = -10 \\
x_2 = 17 \\
x_3 = 6 \\
x_4 = \frac{17}{4}
\end{cases}
\]

### Conclusion

The solutions for the variables \(x_1, x_2, x_3,\) and \(x_4\) are:

\[
x_1 = -10, \quad x_2 = 17, \quad x_3 = 6, \quad x_4 = \frac{17}{4}
\]

Would you like more details on any step or have any questions?

### Related Questions:
1. How is Gaussian elimination different from Gauss-Jordan elimination?
2. What are the properties of a matrix in row echelon form?
3. Can every system of linear equations be solved using Gaussian elimination?
4. How can we determine if a system has no solutions or infinitely many solutions using row reduction?
5. How does the choice of pivot element affect the speed of Gaussian elimination?

### Tip:
When performing row operations, always keep track of your steps to avoid mistakes and to make it easier to backtrack if necessary!

x1 + 4x2 - 2x3 + 8x4 = 12, x2 - 7x3 + 2x4 = -14, 5x3 - x4 = 7, x3 + 3x4 = 15 - solve using OBE

Explore how to solve the system of linear equations x1 + 4x2 - 2x3 + 8x4 = 12, x2 - 7x3 + 2x4 = -14, 5x3 - x4 = 7, and x3 + 3x4 = 15 using Gaussian Elimination and Elementary Row Operations (OBE). The step-by-step solution breaks down the process of converting the augmented matrix into its reduced row echelon form (RREF) and extracting the final solution.

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