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?
 A: Sarrus's rule works only for $3\times 3$-determinants. So you have to find another way to compute $\det A$, for example you can apply elementary transformations not changing the determinant, that is e. g. adding the multiple of one row to another:
\begin{align*}
  \det \begin{bmatrix}
1 & 2 & 3 & 4 \\
2 & 3 & 3 & 3 \\
0 & 1 & 2 & 3 \\
0 & 0 & 1 & 2
\end{bmatrix} &= 
  \det \begin{bmatrix}
1 & 2 & 3 & 4 \\
0 & -1 & -3 & -5 \\
0 & 1 & 2 & 3 \\
0 & 0 & 1 & 2
\end{bmatrix}\\
&= 
  \det \begin{bmatrix}
1 & 2 & 3 & 4 \\
0 & -1 & -3 & -5 \\
0 & 0 & -1 & -2 \\
0 & 0 & 1 & 2
\end{bmatrix}\\
&= \det \begin{bmatrix}
1 & 2 & 3 & 4 \\
0 & -1 & -3 & -5 \\
0 & 0 & -1 & -2 \\
0 & 0 & 0 & 0
\end{bmatrix}
\end{align*}
To compute the determinant of a triagonal matrix, we just have to multiply the diagonal elements, so 
$$ \det A = \det \begin{bmatrix}
1 & 2 & 3 & 4 \\
0 & -1 & -3 & -5 \\
0 & 0 & -1 & -2 \\
0 & 0 & 0 & 0
\end{bmatrix} = 1 \cdot (-1)^2 \cdot 0 = 0.
$$
A: The trick you are applying (Rule of Sarrus) only works for $ 3\times 3$ Matrices.
A: $$
 A = \begin{bmatrix}
1 & 2 & 3 & 4 \\
2 & 3 & 3 & 3 \\
0 & 1 & 2 & 3 \\
0 & 0 & 1 & 2
\end{bmatrix}
$$
$$
 P_1A = \begin{bmatrix}
1 & 2 & 3 & 4 \\
0 & -1 & -3 & -5 \\
0 & 1 & 2 & 3 \\
0 & 0 & 1 & 2
\end{bmatrix}
$$
$$
 P_2P_1A = \begin{bmatrix}
1 & 2 & 3 & 4 \\
0 & -1 & -3 & -5 \\
0 & 0 & -1 & -2 \\
0 & 0 & 1 & 2
\end{bmatrix}
$$
$$
 P_3P_2P_1A = \begin{bmatrix}
1 & 2 & 3 & 4 \\
0 & -1 & -3 & -5 \\
0 & 0 & -1 & -2 \\
0 & 0 & 0 & 0
\end{bmatrix}
$$
$$\det(P_3P_2P_1A)=\det(P_3).\det(P_2).\det(P_1).\det(A)=0$$
$$\det(P_3)\neq0,\det(P_2)\neq0,\det(P_1)\neq0$$
$$\implies \det(A)=0$$
A: 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.
A: The others have pointed out what's wrong with your solution. Let's calculate the determinant now:
\begin{align*}
\det \begin{bmatrix}
1 & 2 & 3 & 4 \\
2 & 3 & 3 & 3 \\
0 & 1 & 2 & 3 \\
0 & 0 & 1 & 2
\end{bmatrix} &\stackrel{r1 - \frac12(r2+r3+r4)}{=}
\det \begin{bmatrix}
0 & 0 & 0 & 0 \\
2 & 3 & 3 & 3 \\
0 & 1 & 2 & 3 \\
0 & 0 & 1 & 2
\end{bmatrix}
= 0.
\end{align*}
