Q:

Finding all solutions of a system of linear equations in $\mathbb{R}$

So I have to find all solutions of the following equations:
$$
a+2b+c-3d=3;
$$
$$
-2a+b-2c+d=-1;
$$
$$
2a+2c-2d=2;
$$
$$
a+3b+c-4d=4
$$
so I found out using the Gauss elimination, that the equation has no solutions,
but after solving it the normal way, I found that one possible solution could be:
$d=0, b=1, a=1-c$ and $c$ can be anything. But I am not sure if that's really a solution or if it doesn't count, because than there would be an unlimited amount of solutions.

A:

Your question is:
$$\mathbf Ax=\small\begin{bmatrix}1&2&1&-3\\-2&1&-2&1\\2&0&2&-2\\1&3&1&-4\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\\x_4\end{bmatrix}=\begin{bmatrix}3\\-1\\2\\4\end{bmatrix}$$
The reduced row echelon form of $\mathbf A$ is:
$$\text{rref(A)}=\tiny\begin{bmatrix}1&0&1&-1\\0&1&0&-1\\0&0&0&0\\0&0&0&0\end{bmatrix}$$
So you have two linearly dependent rows (only two pivot columns). Therefore the $\text{rank(A)}=2$, which implies that the cardinality of the column space is $2.$ We know that by the rank-nullity theorem that $\text{rank(A)+ ker(A) = number of columns(A)}$. Therefore the nullity is $2.$  The matrix is rank deficient, and you have two free variables.
Solving the system:

Find the reduced row echelon form of the augmented matrix (pivots boxed):

$$\small\begin{bmatrix}1&2&1&-3&\vert&3\\-2&1&-2&1&\vert&-1\\2&0&2&-2&\vert&2\\1&3&1&-4&\vert&4\end{bmatrix}\underset{\text{Rref}}{\rightarrow}\begin{bmatrix}\boxed{1}&0&\color{blue}{1}&\color{blue}{-1}&\vert&1\\0&\boxed{1}&\color{blue}{0}&\color{blue}{-1}&\vert&1\\0&0&\color{blue}{0}&\color{blue}{0}&\vert&0\\0&0&\color{blue}{0}&\color{blue}{0}&\vert&0\end{bmatrix}$$
The free variables correspond to the column vectors without a pivot row (in blue). They will be multiplied by $x_3$ and $x_4$ after solving the system, and the choice of $x_3$ and $x_4$ will make no difference, because they will be scalars in front of a vector in the null space of $\mathbf A.$ So we can even give them different names to identify them as free variables - for example, $x_3=s$ and $x_4=t.$ With this, we want to solve $x_1$ and $x_2,$ keeping in mind:
$$x_1\begin{bmatrix}1\\0\\0\\0\end{bmatrix}+x_2\begin{bmatrix}0\\1\\0\\0\end{bmatrix}+s\begin{bmatrix}1\\0\\0\\0\end{bmatrix}+t\begin{bmatrix}-1\\-1\\0\\0\end{bmatrix}=\begin{bmatrix}1\\1\\0\\0\end{bmatrix}$$
The reduced echelon spells out:
\begin{align}1\,x_1 + 0\,x_2 +1s\,-1\,t&=1;\quad x_1=1+0-1s+1t\\
0\,x_1 + 1\,x_2 +0\,s-1t&=1; \quad x_2 =0 + 1 +0\,s+1t\\
&x_3=s\\
&x_4=t
\end{align}
We end up with
$$\underset{\text{vectors }\in\text{ Col(A)}}{\underbrace{\begin{bmatrix}1\\0\\0\\0\end{bmatrix}+\begin{bmatrix}0\\1\\0\\0\end{bmatrix}}}+\underset{\text{null space of A}}{\underbrace{s\begin{bmatrix}-1\\0\\1\\0\end{bmatrix}+t\begin{bmatrix}1\\1\\0\\1\end{bmatrix}}}$$
$s$ and $t$ are free variables. Scalars in $\mathbb R.$

