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Let's break down the problem:

### Given system of linear equations:
\[
\begin{aligned}
1. &\quad x_1 + x_2 + x_3 = -3 \\
2. &\quad x_1 + 2x_2 + x_3 = 7 \\
3. &\quad 2x_1 + x_2 + x_3 = 4
\end{aligned}
\]

**a. Determine the coefficient matrix, the variable vector, and the constant vector.**

1. **Coefficient Matrix (A):**
The coefficients of the variables \(x_1\), \(x_2\), and \(x_3\) in each equation are:
\[
A = \begin{pmatrix}
1 & 1 & 1 \\
1 & 2 & 1 \\
2 & 1 & 1
\end{pmatrix}
\]

2. **Variable Vector (X):**
The variables in the system of equations are \(x_1\), \(x_2\), and \(x_3\), so the variable vector is:
\[
X = \begin{pmatrix}
x_1 \\
x_2 \\
x_3
\end{pmatrix}
\]

3. **Constant Vector (B):**
The constants on the right-hand side of the equations are:
\[
B = \begin{pmatrix}
-3 \\
7 \\
4
\end{pmatrix}
\]

Thus, the system can be written in matrix form as:
\[
A \cdot X = B
\]

---

**b. Solve the system using Gaussian elimination.**

We will apply Gaussian elimination to the augmented matrix \([A|B]\):

\[
\begin{pmatrix}
1 & 1 & 1 & | & -3 \\
1 & 2 & 1 & | & 7 \\
2 & 1 & 1 & | & 4
\end{pmatrix}
\]

### Step 1: Eliminate the first column below the pivot (1st row).
Subtract row 1 from row 2:
\[
R_2 \to R_2 - R_1 \quad \Rightarrow \quad (1, 2, 1, 7) - (1, 1, 1, -3) = (0, 1, 0, 10)
\]

Subtract 2 times row 1 from row 3:
\[
R_3 \to R_3 - 2R_1 \quad \Rightarrow \quad (2, 1, 1, 4) - 2 \times (1, 1, 1, -3) = (0, -1, -1, 10)
\]

Now, the matrix looks like:
\[
\begin{pmatrix}
1 & 1 & 1 & | & -3 \\
0 & 1 & 0 & | & 10 \\
0 & -1 & -1 & | & 10
\end{pmatrix}
\]

### Step 2: Eliminate the second column below the pivot (2nd row).
Add row 2 to row 3:
\[
R_3 \to R_3 + R_2 \quad \Rightarrow \quad (0, -1, -1, 10) + (0, 1, 0, 10) = (0, 0, -1, 20)
\]

The matrix is now:
\[
\begin{pmatrix}
1 & 1 & 1 & | & -3 \\
0 & 1 & 0 & | & 10 \\
0 & 0 & -1 & | & 20
\end{pmatrix}
\]

### Step 3: Solve for the variables.
From the third row, we get:
\[
-1 \cdot x_3 = 20 \quad \Rightarrow \quad x_3 = -20
\]

From the second row, we get:
\[
x_2 = 10
\]

From the first row, we get:
\[
x_1 + x_2 + x_3 = -3 \quad \Rightarrow \quad x_1 + 10 - 20 = -3 \quad \Rightarrow \quad x_1 = 7
\]

### Solution:
The solution to the system is:
\[
x_1 = 7, \quad x_2 = 10, \quad x_3 = -20
\]

Would you like more details or further clarification?

### Related Questions:
1. How does Gaussian elimination differ from the Gauss-Jordan method?
2. What are the conditions for a system of linear equations to have a unique solution?
3. Can every system of linear equations be solved using Gaussian elimination?
4. How does matrix inversion relate to solving linear systems?
5. What are the computational complexities of solving linear systems using different methods?

### Tip:
When using Gaussian elimination, always check for potential row swaps if a pivot is zero to avoid errors in calculation.

Diberikan system persamaan linear sebagai berikut: x1 + x2 + x3 = -3, x1 + 2x2 + x3 = 7, 2x1 + x2 + x3 = 4. Tentukan matriks koefisien SPL, vektor bilangan, dan vektor konstanta. Cari solusi dari SPL di atas menggunakan metode eliminasi Gauss!

Learn how to solve the system of linear equations using Gaussian elimination: x1 + x2 + x3 = -3, x1 + 2x2 + x3 = 7, and 2x1 + x2 + x3 = 4. This step-by-step guide covers constructing the coefficient matrix, variable vector, and constant vector, and using Gaussian elimination to find the solution.

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