What am I doing wrong in calculating this determinant? I have matrix:
$$
 A = \begin{bmatrix}
1 & 2 & 3 & 4 \\
2 & 3 & 3 & 3 \\
0 & 1 & 2 & 3 \\
0 & 0 & 1 & 2
\end{bmatrix}
$$
And I want to calculate $\det{A}$, so I have written:
$$
\begin{array}{|cccc|ccc}
1 & 2 & 3 & 4 & 1 & 2 & 3 \\
2 & 3 & 3 & 3 & 2 & 3 & 3 \\
0 & 1 & 2 & 3 & 0 & 1 & 2 \\
0 & 0 & 1 & 2 & 0 & 0 & 1
\end{array}
$$
From this I get that:
$$
\det{A} = (1 \cdot 3 \cdot 2 \cdot 2 + 2 \cdot 3 \cdot 3 \cdot 0 + 3 \cdot 3 \cdot 0 \cdot 0 + 4 \cdot 2 \cdot 1 \cdot 1) -  (3 \cdot 3 \cdot 0 \cdot 2 + 2 \cdot 2 \cdot 3 \cdot 1 + 1 \cdot 3 \cdot 2 \cdot 0 + 4 \cdot 3 \cdot 1 \cdot 0) = (12 + 0 + 0 + 8) - (0 + 12 + 0 + 0) = 8
$$
But WolframAlpha is saying that it is equal 0. So my question is where am I wrong?

The method that you're using works just fine for $3\times 3$ matrices, but fails to work with $n\times n$ matrices for other $n$. You're going to have to do it another way.
For example, expanding the deteriminant along the first column, we find that $$\begin{align}\det A &=1\cdot\det\left[\begin{array}{ccc}3 & 3 & 3\\1 & 2 & 3\\0 & 1 & 2\end{array}\right]-2\cdot\det\left[\begin{array}{ccc}2 & 3 & 4\\1 & 2 & 3\\0 & 1 & 2\end{array}\right]+0\cdot\det\left[\begin{array}{ccc}3 & 3 & 3\\2 & 3 & 4\\0 & 1 & 2\end{array}\right]-0\cdot\det\left[\begin{array}{ccc}3 & 3 & 3\\2 & 3 & 4\\1 & 2 & 3\end{array}\right]\\ &= \det\left[\begin{array}{ccc}3 & 3 & 3\\1 & 2 & 3\\0 & 1 & 2\end{array}\right]-2\det\left[\begin{array}{ccc}2 & 3 & 4\\1 & 2 & 3\\0 & 1 & 2\end{array}\right].\end{align}$$
At that point, you can use your method of calculating determinants of $3\times 3$ matrices to get the rest of the way.
