How to switch rows in matrix L when decomposing matrix A into PA = LU? 
Find the permutation matrix $P$, the lower triangular matrix $L$ and the upper triangular matrix $U$ such that $$ PA=LU $$
Given $$ A=  \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6  \\ -2 & -3 & -4 & -5 & -6 & -7 \\ 3 & 7 & 11 & 16 & 21 & 27 \\ -4 & -5 & -5 & -5 & -5 & -5 \end{pmatrix}$$

I came this far
$$ U = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6  \\ 0 & 1 & 2 & 3 & 4 & 5 \\ 0 & 0 & 0 & 1 & 2 & 4 \\ 0 & 0 & 1 & 2 & 3 & 4 \end{pmatrix}$$
and
$$ L= \begin{pmatrix} 1 & 0 & 0 & 0   \\ -2 & 1 & 0 & 0  \\ 3 & 1 & 1 & 0  \\ -4 & 3 & 0 & 1 \end{pmatrix}$$
The last step I have to make is to change the fourth row with the third one but I do not know exactly how to change the entries in the lower triangular matrix L. Can anyone explain what exactly I have to switch in L?

Using row reduction, we arrived at
$$U = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6  \\ 0 & 1 & 2 & 3 & 4 & 5 \\ 0 & 0 & 0 & 1 & 2 & 4 \\ 0 & 0 & 1 & 2 & 3 & 4 \end{pmatrix}$$
As an alternate approach to the excellent write up by @Ian (+1), you could have reversed the row reduction steps, including the swap, as $$A = E_1^{-1}E_2^{-1}E_3^{-1}E_4^{-1}U$$
This results in
$$L = \begin{pmatrix} 1 & 0 & 0 & 0 \\ -2 & 1 & 0 & 0 \\ 3 & 1 &0 & 1 \\ -4 & 3 & 1 & 0 \end{pmatrix}$$
We see that $L$ is not lower triangular and we just need to swap row three and four, resulting in
$$L = \begin{pmatrix} 1 & 0 & 0 & 0 \\ -2 & 1 & 0 & 0  \\ -4 & 3 & 1 & 0 \\3 & 1 &0 & 1\end{pmatrix}$$
That swap requires a permutation matrix
$$P = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 &0 & 1 \\ 0 & 0 & 1 &0\end{pmatrix}$$
Now we can verify
$$PA = LU$$
You can also verify that $$A = PLU = P^T LU = P^{-1} LU$$
See, for example, How can LU factorization be used in non-square matrix?
