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 $\mathrm{det} A = 1$ ?
 A: 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.
A: Let $L$ be the lower triangular matrix with all entries in the lower triangular part equal to $1$. Then $A=L^T-L$ and
$S_n:=L^{-1}A(L^{-1})^T=L^{-1}(L^T-L)(L^{-1})^T=L^{-1}-L^{-T}$. Note that
$$
L^{-1}=\pmatrix{1\\ -1&1\\ &-1&\ddots\\ &&\ddots&\ddots\\ &&&-1&1}
\ \Rightarrow
\ S_n:=\pmatrix{0&1\\ -1&0&1\\ &-1&\ddots&\ddots\\ &&\ddots&\ddots&1\\ &&&-1&0}.
$$
$S_n$ is a tridiagonal Toeplitz matrix. By Laplace expansion along the first row, we see that $\det(S_n)=\det(S_{n-2})$. Therefore, $\det(S_n)=\det(S_{n-2})=\cdots=\det(S_2)=\det\pmatrix{0&1\\ -1&0}=1$. As $\det(L)=1$, we have $\det(A)=\det(S_n)=1$.
Alternatively, using Schur complement, if we write $A_n=A=\pmatrix{A_2&e_2e_{n-2}^T\\ -e_{n-2}e_2^T&A_{n-2}}$ where $e_k\in\mathbb R^k$ denotes the column vector containing $k$ ones, then
\begin{aligned}
\det(A_n)
&=\det(A_2)\det(A_{n-2}+e_{n-2}e_2^TA_2^{-1}e_2e_{n-2}^T)\\
&=\det(A_2)\det(A_{n-2}-e_{n-2}\underbrace{e_2^TA_2e_2}_{=0}e_{n-2}^T)\\
&=\det(A_2)\det(A_{n-2})=\det(A_{n-2}).
\end{aligned}
Therefore $\det(A_n)=\det(A_{n-2})=\cdots=\det(A_2)=1$.
The determinant can also be evaluated using Sylvester determinant theorem by considering $B=A+ee^T$, but I shall omit the proof here.
