How to validate the basis of the space of solutions of a system of linear equations Problem
Solve the following nonhomogeneous system of linear equations:
$$
\begin{aligned}
x_{1} + 2 x_{2} + 3 x_{3} + x_{4} &= 4,\\
2 x_{1} + 2 x_{2} + 3 x_{3} + x_{4} &= 5,\\
3 x_{1} + 3 x_{2} + 4 x_{3} + x_{4} &= 6
\end{aligned}$$
Please enter the specific solution first and then the basis in the space of solutions of the corresponding homogeneous system.
Comment of the teacher
Example. If $x_s=\begin{pmatrix} 0 \\ 1 \\ 0 \\1 \end{pmatrix}$ is a specific solution and $\left\{\begin{pmatrix} 1 \\ 2 \\ 3/2 \\ 4 \end{pmatrix}, \begin{pmatrix} 0 \\ -2 \\ 2 \\ 9 \end{pmatrix}\right\} $ is a basis of $\{x:Ax=0\}$ then please enter
[0,1,0,1],[1,2,3/2,4],[0,-2,2,9]
There is an automated system here, checking the solution.
Solution
Augmented matrix:
$$
\begin{pmatrix}
1 & 2 & 3 & 1 & 4\\
2 & 2 & 3 & 1 & 5\\
3 & 3 & 4 & 1 & 6\\
\end{pmatrix}
$$
I managed to transform the augmented matrix to the diagonal form:
$$
\begin{pmatrix}
1 & 0 & 0 & 0 &  1\\
0 & 1 & 0 & -1 & -3\\
0 & 0 & 1 & 1 &  3\\
\end{pmatrix}
$$
So, the specific solution is the last row + 0 at the bottom: $x_s=\begin{pmatrix} 1 \\ -3 \\ 3 \\ 0 \end{pmatrix}$. I checked, and it is really a solution of the original problem.
1-st equation:
$$
1 \cdot 1 + 2\cdot(-3) + 3 \cdot 3 + 1 \cdot 0 = 1 - 6 + 9 = 4
$$
2-nd equation:
$$
2 \cdot 1 + 2\cdot(-3) + 3 \cdot 3 + 1 \cdot 0 = 2 - 6 + 9 = 5
$$
3-rd equation:
$$
3 \cdot 1 + 3\cdot(-3) + 4 \cdot 3 + 1 \cdot 0 = 3 - 9 + 12 = 6
$$
But, why does the basis in the comment consist of 2 columns? I think, it must be only 1: $\begin{pmatrix} 0 \\ -1 \\ 1 \\ -1 \end{pmatrix}$ --- the 4-th column of the diagonal matrix and the last element (-1) is an element of negated identity matrix.
When I tried this solution: [1, -3, 3, 0],[0, -1, 1, -1] for the first time, the automated checker said: "incorrect". Then I wrote this question; but before posting it, I tried it again, and it said: "correct". Probably, I made a typo first time :).
How can I check, that $\begin{pmatrix} 0 \\ -1 \\ 1 \\ -1 \end{pmatrix}$ is really the basis of the space of solutions, like I did, validating the specific solution?

I'm not sure if the teacher's comment was for specifically this question, but you're right. The solution to this system lies on a line and not a plane as the teacher's comment suggests. 
To check that your answer is correct, substitute the general form back into the system. Here's an example with the first equation:
$$x_1 + 2x_2 + 3x_3 + x_4 = 4$$
$$\implies 1 + 2(-3 + x_4) + 3(3- x_4) + x_4 = 4$$
$$\implies 1 - 6 + 9 + 3x_4 - 3x_4 = 4$$
$$\implies 4 = 4$$
and so on for the equations.
