Solve the system of equestions

1.       x1+x2+x3+x4= 0,   2x1+x2-x4= 0,  x1+3x2+2x3+4x4= 0.

We are given the following system of equations:

\[
1. \quad x_1 + x_2 + x_3 + x_4 = 0
\]
\[
2. \quad 2x_1 + x_2 - x_4 = 0
\]
\[
3. \quad x_1 + 3x_2 + 2x_3 + 4x_4 = 0
\]

### Step 1: Write the system in matrix form

We can represent the system as:

\[
\begin{bmatrix}
1 & 1 & 1 & 1 \\
2 & 1 & 0 & -1 \\
1 & 3 & 2 & 4
\end{bmatrix}
\begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
x_4
\end{bmatrix}
=
\begin{bmatrix}
0 \\
0 \\
0
\end{bmatrix}
\]

### Step 2: Solve using Gaussian elimination or row reduction

We will perform row operations on the augmented matrix:

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

**(R2 - 2R1 → R2):**
\[
\begin{bmatrix}
1 & 1 & 1 & 1 & 0 \\
0 & -1 & -2 & -3 & 0 \\
1 & 3 & 2 & 4 & 0
\end{bmatrix}
\]

**(R3 - R1 → R3):**
\[
\begin{bmatrix}
1 & 1 & 1 & 1 & 0 \\
0 & -1 & -2 & -3 & 0 \\
0 & 2 & 1 & 3 & 0
\end{bmatrix}
\]

**(R2 → -R2):**
\[
\begin{bmatrix}
1 & 1 & 1 & 1 & 0 \\
0 & 1 & 2 & 3 & 0 \\
0 & 2 & 1 & 3 & 0
\end{bmatrix}
\]

**(R3 - 2R2 → R3):**
\[
\begin{bmatrix}
1 & 1 & 1 & 1 & 0 \\
0 & 1 & 2 & 3 & 0 \\
0 & 0 & -3 & -3 & 0
\end{bmatrix}
\]

### Step 3: Back substitution

From the last row, we have:
\[
-3x_3 - 3x_4 = 0 \quad \Rightarrow \quad x_3 + x_4 = 0 \quad \Rightarrow \quad x_3 = -x_4
\]

Substitute \(x_3 = -x_4\) into the second row:
\[
x_2 + 2x_3 + 3x_4 = 0 \quad \Rightarrow \quad x_2 + 2(-x_4) + 3x_4 = 0 \quad \Rightarrow \quad x_2 = -x_4
\]

Substitute \(x_2 = -x_4\) and \(x_3 = -x_4\) into the first row:
\[
x_1 + x_2 + x_3 + x_4 = 0 \quad \Rightarrow \quad x_1 - x_4 - x_4 + x_4 = 0 \quad \Rightarrow \quad x_1 = x_4
\]

### Step 4: General solution

The solution to the system is:
\[
x_1 = x_4, \quad x_2 = -x_4, \quad x_3 = -x_4, \quad x_4 = x_4
\]

Thus, the general solution can be written as:
\[
(x_1, x_2, x_3, x_4) = (x_4, -x_4, -x_4, x_4) = x_4(1, -1, -1, 1)
\]
where \(x_4\) is a free parameter.

### Conclusion:
The solution is a family of solutions of the form:
\[
(x_1, x_2, x_3, x_4) = t(1, -1, -1, 1), \quad t \in \mathbb{R}
\]

Let me know if you need further details or have any questions!

Here are some follow-up questions related to this:

1. What happens if we add a fourth equation to the system?
2. How can we visualize the solution in 4-dimensional space?
3. How does Gaussian elimination help in solving larger systems of equations?
4. How can we use matrices to express dependent or independent systems?
5. What does the free parameter \(t\) represent geometrically?

**Tip:** For underdetermined systems (fewer equations than variables), solutions often involve free parameters, indicating a family of solutions rather than a unique one.

Solve the system of equations: x1 + x2 + x3 + x4 = 0, 2x1 + x2 - x4 = 0, x1 + 3x2 + 2x3 + 4x4 = 0.

Learn how to solve a system of linear equations with four variables using Gaussian elimination. This step-by-step solution covers essential linear algebra concepts, such as row reduction and handling free variables. Suitable for high school students (Grades 10-12) and beyond.

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