Finding a nullspace of a matrix. I am given the following matrix $A$ and I need to find a nullspace of this matrix.

$$A =
 \begin{pmatrix}
  2&1&4&-1 \\
  1&1&1&1 \\
  1&0&3&-2  \\
  -3&-2&-5&0
 \end{pmatrix}$$

I have found a row reduced form of this matrix, which is:
$$A' =
 \begin{pmatrix}
  1&0&3&-2 \\
  0&1&-2&3 \\
  0&0&0&0  \\
  0&0&0&0
 \end{pmatrix}$$
And then I used the formula $A'x=0$, which gave me:
$$A'x =
 \begin{pmatrix}
  1&0&3&-2 \\
  0&1&-2&3 \\
  0&0&0&0  \\
  0&0&0&0
 \end{pmatrix}
\begin{pmatrix}
  x_1 \\
  x_2 \\
  x_3  \\
  x_4
 \end{pmatrix}=
\begin{pmatrix}
  0 \\
  0 \\
  0  \\
  0
 \end{pmatrix}$$
Hence I obtained the following system of linear equations:
$$\begin{cases} x_1+3x_3-2x_4=0 \\ x_2-2x_3+3x_4=0 \end{cases}$$
So I just said that $x_3=\alpha$, $x_4=\beta$ and the nullspace is:
$$nullspace(A)=\{2\beta-3\alpha,2\alpha-3\beta,\alpha,\beta) \ | \ \alpha,\beta \in \mathbb{R}\}$$
Is my thinking correct? Thank you guys!

Everything is right. But after finding the equations \begin{cases} x_1+3x_3-2x_4=0 \\ x_2-2x_3+3x_4=0 \end{cases} This $x_1$ and $x_2$ are pivot variables and $x_3$ and $x_4$ are free variables. The number of non zero rows is the rank of the matrix, in our case $2$ and hence nullity of the matrix is $2$. since $dim(W) = Rank + Nullity$. Now since $nullity =2$, the usual basis is $\{(1,0),(0,1)\}$. Hence Substitute $x_3=1$ and $x_4=0$ in two equations, we will get $x_1=-3$ and $x_2 =2$ and we get a point $(-3,2,1,0)$ and again substitute $x_3=0$ and $x_4=1$ in two equations, we will get $x_1 = 2$ and $x_2 =-3$ and we will get another point $(2,-3,0,1)$. These two points $(-3,2,1,0)$ and $(2,-3,0,1)$ serves as a basis for the nullspace of the matrix.   
