<s>[INST] Determinants: divisibility by 6 without remainder and $n\times n$ matrix 
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*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] Regarding (2), you can reduce the matrix $A$ modulo $2$ to obtain
$$ A' = \begin{pmatrix} 1 & 0 & 0 \\ 1 & 0 & 0 \\ 1 & 1 & 1 \end{pmatrix}. $$
Clearly, $\det(A') = 0$ which shows that $\det(A) \mod 2 = \det(A') = 0$. Similarly, reducing $A$ modulo 3, we obtain
$$ A'' = \begin{pmatrix} 1 & 2 & 1 \\ 1 & 1 & 1 \\ 1 & 2 & 1 \end{pmatrix} $$
and this implies that $\det(A) \mod 3 = \det(A'') = 0$ which shows that $\det(A)$ is divible by both two and three.
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