Determinant 4x4 This looks very simple but I guess I made a mistake somwhere because the result should be 17 
$$\begin{vmatrix} 1 & 2 & 4 & -2 \\ 0 & 1 & 2 & 0 \\ 1 & -2 & 0 & 1 \\ 2 & -1 & -1 & 1 \end{vmatrix}$$ 
First I added  negative 1st row to 3rd
$\begin{vmatrix} 1 & 2 & 4 & -2 \\ 0 & 1 & 2 & 0 \\ 0 & -4 & -4 & 3 \\ 2 & -1 & -1 & 1 \end{vmatrix}$
Then I multiplicate 2nd row times $4$, so the whole determinant will be multiplied with $0.25$.
Then I added 3rd row to the second
$\begin{vmatrix} 1 & 2 & 4 & -2 \\ 0 & 0 & 4 & 0 \\ 0 & -4 & -4 & 3 \\ 2 & -1 & -1 & 1 \end{vmatrix}$
Now I cross 3rd column and 2nd row which would be $4((-1)^5)$ plus the rest which would be $-2(-5)$ = $10$ not $17$
Thanks for help
 A: You start with
$$\begin{vmatrix} 1 & 2 & 4 & -2 \\ 0 & 1 & 2 & 0 \\ 1 & -2 & 0 & 1 \\ 2 & -1 & -1 & 1 \end{vmatrix}$$ 
and then, to the third row, you subtract the first, giving
$$\begin{vmatrix} 1 & 2 & 4 & -2 \\ 0 & 1 & 2 & 0 \\ 0 & -4 & -4 & 3 \\ 2 & -1 & -1 & 1 \end{vmatrix}$$ 
then duplicate 2nd row to get
$$\begin{vmatrix} 1 & 2 & 4 & -2 \\ 0 & 2 & 4 & 0 \\ 0 & -4 & -4 & 3 \\ 2 & -1 & -1 & 1 \end{vmatrix}$$ 
and then, to the 2nd row, add the 3rd, to get
$$\begin{vmatrix} 1 & 2 & 4 & -2 \\ 0 & -2 & 0 & 3 \\ 0 & -4 & -4 & 3 \\ 2 & -1 & -1 & 1 \end{vmatrix}$$ 
notice how the last element of the 2nd row now is a $3$.
Using Laplace formula over the 1st column we get
$$1\times \begin{vmatrix}-2 & 0 & 3 \\ -4 & -4 & 3 \\  -1 & -1 & 1 \end{vmatrix} - 2\times \begin{vmatrix}  2 & 4 & -2 \\ -2 & 0 & 3 \\ -4 & -4 & 3 \end{vmatrix}$$ 
Because
$$\begin{vmatrix}-2 & 0 & 3 \\ -4 & -4 & 3 \\  -1 & -1 & 1 \end{vmatrix} = 2, \begin{vmatrix}  2 & 4 & -2 \\ -2 & 0 & 3 \\ -4 & -4 & 3 \end{vmatrix} = -16$$ 
we get
$$0.5\times\left(1\times 2 - 2\times (-16) \right) = 17$$
