Let: $$A＝\begin{pmatrix} 1& 1 &1\\2 & 1&1\\1&0&2\end{pmatrix} $$
and we have : $$\text{det}(A)＝\begin{vmatrix} 1& 1 &1\\2 & 1&1\\1&0&2\end{vmatrix}＝\begin{vmatrix} 1 & 1\\0 & 2\end{vmatrix}-\begin{vmatrix} 2 & 1 \\1& 2\end{vmatrix}＋\begin{vmatrix} 2 & 1\\1 & 0\end{vmatrix}＝-1 $$
but using wolfram alpha : $\text{det}(A)=-2$
Thank you for all
You appear to have just made a mistake in calculating the last step (with the $2\times 2$ determinants). We have $$ \begin{align*} \begin{vmatrix} 1 & 1 \\ 0 & 2 \end{vmatrix} - \begin{vmatrix} 2 & 1 \\ 1 & 2 \end{vmatrix} + \begin{vmatrix} 2 & 1 \\ 1 & 0 \end{vmatrix} &= \left(1\times 2 - 1 \times 0\right) - \left(2\times 2 - 1\times 1\right) + \left(2\times 0 - 1 \times 1 \right)\\ &= \left(2 - 0\right) - \left(4 - 1\right) + \left(0 - 1\right) \\ &= 2 - 3 - 1 \\ &= -2. \end{align*} $$