Problem with LU decomposition I have this matrix:
$$
A =\begin{bmatrix}1 & -2 & 3\\ 2 & -4 & 5 \\ 1 & 1 & 2\end{bmatrix}
$$
After I decomposit it, I get:
$$
L = \begin{bmatrix}1 & 0 & 0\\1 & 1 & 0\\ 2 & 0 & 1 \end{bmatrix}\\
U = \begin{bmatrix} 1 & -2 & 3\\ 0 & 3 & -1\\ 0 & 0 & -1\end{bmatrix}\\
P = \begin{bmatrix}1 & 0 & 0\\0 & 0 & 1\\0 & 1 & 0\\\end{bmatrix}
$$
Now I need to solve $Ax=b$ using: $b_1 = \begin{bmatrix}2\\0\\1 \end{bmatrix}$ and $b_2 = \begin{bmatrix}-1\\1\\-1 \end{bmatrix}$
But I'm stuck and don't know how to permute the vectors and every time I try I get a different result...
As I checked in Octave, the solution of $Ax=b_1$ is: $x_1 = \begin{bmatrix}-8\\1\\4 \end{bmatrix}$, but I can't get that solution.
Can you please explain me how to get this done? Thank you!

All of your work is correct for $A = PLU$.
Now, we want to solve $Ax = b$ for $b_1$ and $b_2$, so we use forward and back substitution as: 
$$Ly = Pb, Ux = y$$
Forward substitution yields:
$$Ly = \begin{bmatrix}1 & 0 & 0\\1 & 1 & 0\\ 2 & 0 & 1 \end{bmatrix}\begin{bmatrix}y_1\\y_2\\y_3 \end{bmatrix}=Pb = \begin{bmatrix}1 & 0 & 0\\0 & 0 & 1\\0 & 1 & 0\\\end{bmatrix}\begin{bmatrix}2\\0\\1 \end{bmatrix} = \begin{bmatrix}2\\1\\0 \end{bmatrix} \implies y = \begin{bmatrix}2\\-1\\-4 \end{bmatrix}$$
Next, we use back substitution to solve $Ux = y$, yielding:
$$Ux = \begin{bmatrix} 1 & -2 & 3\\ 0 & 3 & -1\\ 0 & 0 & -1\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3 \end{bmatrix}=y=\begin{bmatrix}2\\-1\\-4 \end{bmatrix} \implies \begin{bmatrix}x_1\\x_2\\x_3 \end{bmatrix} = \begin{bmatrix}-8\\1\\4 \end{bmatrix}$$
Rinse and repeat for $b_2$, yielding:
$$x_2 = \begin{bmatrix}6\\-1\\-3 \end{bmatrix}$$
