<s>[INST] How to prove that when $n$ is even, and $A = (a_{ij})_{i,j}$, $1\leq i,j\leq n$ to show that $\mathrm{det} A = 1$? How to prove that when $n$ is even, and $A = (a_{ij})_{i,j}$, $1\leq i,j\leq n$ with $$a_{ij}=\begin{cases} 1& i<j,\\ 0 & i=j,\\ -1& i>j,\end{cases}$$
to show that $\mathrm{det} A = 1$ ?
 [/INST] Subtract the second from the first row, the third from the second row, ..., the $n$th row from the $n-1$st row and you will get the following matrix (omitted entries are all $0$):
$$\begin{pmatrix}
1 & 1 & 0 & 0 & 0&\dots & 0 & 0&0\\
0 & 1 & 1 & 0 & 0 &\dots & 0 & 0&0  \\
0 & 0 & 1 & 1 & 0 & \dots & 0 & 0&0\\
 & & & \ddots & \ddots \\ &&&&\ddots&\ddots
\\ & & & &&\ddots&\ddots
\\&&&&&&\ddots&\ddots
\\
0 & 0 &0&0&0&\dots&0&1&1 \\
-1 & -1 & -1&-1&-1&\dots & -1 &-1 &0
\end{pmatrix}$$
now add the first, third, fifth, ..., $n-1$st row to the last row and you get
$$\begin{pmatrix}
1 & 1 & 0 & 0 & 0&\dots & 0 & 0&0\\
0 & 1 & 1 & 0 & 0 &\dots & 0 & 0&0  \\
0 & 0 & 1 & 1 & 0 & \dots & 0 & 0&0\\
 & & & \ddots & \ddots \\ &&&&\ddots&\ddots
\\ & & & &&\ddots&\ddots
\\&&&&&&\ddots&\ddots
\\
0 & 0 &0&0&0&\dots&0&1&1 \\
0 & 0& 0&0&0&\dots & 0 &0 &1
\end{pmatrix}$$ 
Nice! Now we have an upper diagonal matrix so the determinant is just the product of the diagonal entries which are all $1$ so we are done.
</s>