Simple determinant calculation

Question

\begin{vmatrix} 1 & -2 & 2 \\ 1 & -2 & -2\\ 2 & -2 & 1 \end{vmatrix}

It is fairly easy I just want to reassure the steps are correct:

\begin{vmatrix} 1 & -2 & 2 \\ 1 & -2 & -2\\ 2 & -2 & 1 \end{vmatrix} Taking out $-2^{n}$ when $n$ is the number of rows/columns in this case $n=1$ (the second column)

$$-2 \begin{vmatrix} 1 & 1 & 2 \\ 1 & 1 & -2\\ 2 & 1 & 1 \end{vmatrix} $$

Reducing column 2 from column 1 as no effect on the determinant

$$-2 \begin{vmatrix} 0 & 1 & 2 \\ 0 & 1 & -2\\ 1 & 1 & 1 \end{vmatrix} $$

Using Laplace expansion on the first column

$$(-2)*1[1*(-2)-2*1]=8.$$

Answer 1

\begin{vmatrix} 1 & -2 & 2 \\ 1 & -2 & -2\\ 2 & -2 & 1 \end{vmatrix} $$R'_1=R_1-R_2 \space R'_2=R_2-R_3 $$ $$\begin{vmatrix} 0 & 0 & 4 \\ -1 & 0 & -3\\ 2 & -2 & 1 \end{vmatrix} =8$$