$$ L: \mathbb{R}^4 → \mathbb{R}^4 $$ defined by $$ L \begin{pmatrix} \begin{bmatrix} x\\y\\z\\w \end{bmatrix} \end{pmatrix} = \begin{bmatrix} 1&2&1&3\\ 2&1&-1&2\\ 1&0&0&-1\\ 4&1&-1&0 \end{bmatrix} \begin{bmatrix} x\\y\\z\\w \end{bmatrix} $$
I'm interested in looking for its kernel Ker(L), find a basis for Ker(L) and the dimension of Ker(L). Same as for its range, range(L) and its dimension range(L).
I'm not really confident with my answer so I'm asking so that I could verify if my solution is correct.
What I did is equate the matrix above to a zero matrix.
$$
\begin{bmatrix}
1&2&1&3\\
2&1&-1&2\\
1&0&0&-1\\
4&1&-1&0
\end{bmatrix}
\begin{bmatrix}
x\\y\\z\\w
\end{bmatrix}
=
\begin{bmatrix}
0\\0\\0\\0
\end{bmatrix}
\Rightarrow
\left[
\begin{array}{cccc|c}
1&2&1&3&0\\
2&1&-1&2&0\\
1&0&0&-1&0\\
4&1&-1&0&0
\end{array}
\right]
$$
Then I looked for its RREF and the result is
$$
\left[
\begin{array}{cccc|c}
1&0&0&-1&0\\
0&1&0&\frac{8}{3}&0\\
0&0&1&\frac{-4}{3}&0\\
0&0&0&0&0
\end{array}
\right]
$$
and decided the following:
$$ \begin{bmatrix}
x = r\\
y = \frac{8}{3}r\\
z = \frac{-4}{3}r\\
w = r \end{bmatrix}
\Rightarrow r
\begin{bmatrix}
1\\
\frac{8}{3}\\
\frac{-4}{3}\\
1 \end{bmatrix}
\Rightarrow \ker(L)
\begin{bmatrix}
1\\
\frac{8}{3}\\
\frac{-4}{3}\\
1 \end{bmatrix}
$$
And from there I'm not sure if I did right, and I'm confused on where next I should go in finding the rest of the problem.
I apologize if its a stupid question.
For the method you are using the key point is that row operations preserve the solution for the linear system of equations. For the solution you are almost right, $Ker(L)$ has dimension one but for its basis you have a wrong sign for $\frac83$ and $-\frac43$, that is
$$r \begin{bmatrix} 1\\ \color{red}{-\frac{8}{3}}\\ \color{red}{\frac{4}{3}}\\ 1 \end{bmatrix}$$
From the REEF we also know that the image/range has dimension $3$ and we can take the first three columns (or any other triple) of the original matrix as its basis.