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!


Since $x_1=2x_4-3x_3$ and $x_2=2x_3-3x_4\Rightarrow$
if $(x_1,x_2,x_3,x_4)\in$  nullspace($A$): $$(x_1,x_2,x_3,x_4)=(2x_4-3x_3,2x_3-3x_4,x_3,x_4)=x_3(-3,2,1,0)+x_4(2,-3,0,1)$$ So Nullspace$(A)=\langle (-3,2,1,0),(2,-3,0,1) \rangle$.

