The correct answer is: A. Only the trivial solution x1 = x2 = x3 = x4 = 0 exists.
To solve this system of equations, we can use Gaussian elimination. First, we can eliminate $x_1$ by adding $-3$ times the first equation to the second equation:
$$-3x_1 – 6x_2 – 3x_3 – 12x_4 = -6$$
This gives us the following system of equations:
$$x_1 + 2x_2 + x_3 + 4x_4 = 2$$
$$0x_2 + 3x_2 + 0x_3 + 9x_4 = 0$$
We can then eliminate $x_2$ by adding $-\frac{3}{3}x_2$ to the second equation:
$$0x_2 + 0x_2 + 0x_3 + 6x_4 = 0$$
This gives us the following system of equations:
$$x_1 + 2x_2 + x_3 + 4x_4 = 2$$
$$0x_2 + 0x_2 + 0x_3 + 0x_4 = 0$$
Since the second equation is now a trivial equation, we know that $x_2 = x_3 = x_4 = 0$. Substituting this into the first equation, we get $x_1 = 2$. Therefore, the only solution to the system of equations is $x_1 = x_2 = x_3 = x_4 = 0$. This is the trivial solution.
The other options are incorrect because:
- Option B is incorrect because there is a solution.
- Option C is incorrect because there is only one solution.
- Option D is incorrect because there is not more than one solution.