<s>[INST] Determinants: divisibility by 6 without remainder and $n\times n$ matrix 
*

*Compute the determinant of \begin{pmatrix}
 1 & 2 & 3 & ...& n\\
 -1 & 0 & 3 & ...& n\\
  -1 & -2 & 0 & ...& n\\
...& ...& ...& ...& \\
-1 & -2 & -3 & ...& n 
 \end{pmatrix}
after some elementary raw operation, one can reach: 
\begin{pmatrix} 1 & 2 & 3 & ...& n\\
 -2 & -2 & 0 & ...& 0\\
0& -2& -3& ...&0& \\
...& ...& ...& ...& \\
0 & 0 & 0 & 1-n & n 
 \end{pmatrix}
but I don't sure how to proceed.

*Why the det. of the following matrix is divisible by 6 without remainder?
\begin{pmatrix} 2^0 & 2^1 & 2^2 \\ 4^0 & 4^1 & 4^2\\ 5^0 & 5^1 & 5^2 \end{pmatrix}
So I know that I have to show that its det. is divisible by $2$ and $3$, or equivalently that the sum of its digits divisible by $3$ and last digit is even. But I don't sure how to start the process. 
Thank you.
 [/INST] *

*Add the first row to the other rows:
$$
\pmatrix{
 1 & 2 & 3 & 4 & 5 \\
-1 & 0 & 3 & 4 & 5 \\
-1 & -2 & 0 & 4 & 5 \\
-1 & -2 & -3 & 0 & 5 \\
-1 & -2 & -3 & -4 & 5 \\
}
\to
 \pmatrix{
 1 & 2 & 3 & 4 & 5 \\
 0 & 2 & * & * & * \\
 0 & 0 & 3 & * & * \\
 0 & 0 & 0 & 4 & * \\
 0 & 0 & 0 & 0 & 2\times 5 \\
}
$$
So the determinant is $2(n!)$.

*The matrix is a Vandermonde matrix and its determinant is $(4-2)(5-2)(5-4)=6$.
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