Calculate the determinant of the $2n \times 2n$ matrix with entries equal to zero on the main diagonal, $1$ below and $-1$ above Calculate the determinant of the $2n \times 2n$ matrix with entries equal to zero on the main diagonal, equal to $1$ below and equal to $-1$ above.
I'll denote this matrix $A_{2n}$.
So for example you have $A_{2} = \begin{bmatrix}
        0 & -1  \\
        1 & 0  \\
              \end{bmatrix} 
$ and $A_{4}= \begin{bmatrix}
        0 & -1 & -1  &-1 \\
        1 & 0  & -1  & -1\\
        1 & 1  & 0   & -1 \\
        1 & 1  & 1   & 0\\
        \end{bmatrix}
$
From trying it out by hand I think you have $\det(A_{2n})=1$ for all odd $n$ , or $=-1$ for all even $n$.
Can anyone come up with some way to prove this?
From doing these two examples my algorithm seems to be something like this:


*

*Switch the first and second rows (multiplies det by $-1$)

*If the matrix is now upper triangular, calculate the  determinant by product of main diagonal.

*If not, make a new pivot and clear that column below.

*Swap the third and fourth rows (multiplies det by $-1$).

*Repeat from 2.

 A: Start with $A_{2n}$. If you


*

*subtract $2^{nd}$ column from $1^{st}$ column.

*subtract $2^{nd}$ row from    $1^{st}$ row.

*add $1^{st}$ column to all columns except the first two columns.

*add $1^{st}$ row to all rows except the first two rows.


you will turn $A_{2n}$ into block diagonal form with
a copy of $A_2$ at the first two rows/columns.
Repeat apply the same procedure to the remaining $2n-2 \times 2n-2$ matrix, one can
transform $A_{2n}$ into a block diagonal matrix with $n$ copies of $A_2$. As a result,
$$\det A_{2n} = ( \det A_2 )^n = 1.$$
A: The determinant is always $+1$.
First note that the inverse of $A_{2n}$ is given by $B_{2n}$, where
$$B_{ij} = \begin{cases} 0 & \text{ if }i=j\\ (-1)^{j-i} & \text{ if }j>i\\ (-1)^{i-j+1} & \text{ if }j<i\end{cases}$$
This means that $\det(A) = \det(B) = \pm 1$, since both are integer matrices.
We need to rule out $-1$. This is easy though, since the eigenvalues of an even skew symmetric matrix are purely imaginary, which in turn means that the determinant is positive (it is the square of a the Pfaffian). Hence, the determinant is $+1$.
A: Add $x$ to all the entries of the matrix $A_{2n}$ and denote $C_i$ the $i$-th column of $A_{2n}$ and let $J=\underbrace{(1,\ldots,1)^T}_{2n\;\text{components}}$ hence
$$\Delta(x)=\det(C_1+xJ,\cdots ,C_{2n}+xJ)=\det(C_1+xJ,C_2-C_1,\cdots,C_{2n}-C_1)\\
=\det(C_1,C_2-C_1,\cdots,C_{2n}-C_1)+x\det(J,C_2-C_1,\cdots,C_{2n}-C_1)=\alpha+\beta x$$
hence $\Delta(x)$ is a polynomial with degree $1$.
Now it's simple to see (determinant of upper and lower triangle matrix) that $\Delta(1)=\Delta(-1)=1$ hence
$$\alpha+\beta=\alpha-\beta=1$$
so $\alpha=1$ and $\beta=0$ and finaly
$$\det A_{2n}=\Delta(0)=1$$
A: An elementary method of calculating the determinant through row operations is as follows:


*

*Add row $1$ to rows $3$ through $2n$.

*Subtract row $2$ to rows $3$ through $2n$.

*Swap rows $1$ and $2$.

*Repeat steps (1-3) for the lower right $2(n-1) \times 2(n-1)$ submatrix.  Since there are an even number of rows, this process always terminates in an upper triangular matrix whose entries alternate $\{1, -1, \ldots, 1, -1\}$.

*The determinant is then $\det A_{2n} = ((-1)^n)^2 = 1$, because there were $n$ row swaps, and there are $n$ $-1$'s on the main diagonal.


Incidentally, if we were to consider an odd number of rows, it is easy to see that the determinant is zero.
A: another way would be to evaluate the determinant to the first collumn.
From this follows that det(A(2n))=det(A(2n-1))-det(B) + det(B) - det(B) + det(B) - det(B). Because det(B) occurs an odd number of times, it's equal to det(A(2n-1)) - det(B)...
det(A(2n-1)) = 0, because A=-transpose(A) and using determinant properties:
det(-transpose(A(2n-1)))= (-1)^(2n-1) * det(transpose(A(2n-1))) = - det(A(2n-1)) = det(A(2n-1))
B is of the form : 
\begin{bmatrix}
        -1 & -1 & -1 & -1 & -1\\
        0 & -1 & -1  &-1 & -1 \\
        1 & 0  & -1  & -1 & -1\\
        1 & 1  & 0   & -1  &-1\\
        1 & 1  & 1   & 0 & -1\\
        \end{bmatrix}
with exactly 2n-1 rows and collumns.
Now if we want det(B) we can remove the (-) of the first row by placing it in front and evaluate the determinant to the first row. Out pops :
det(B) = - (det(A(2n-2)) - det(A(2n-2)) + det(A(2n-2)) - det(A(2n-2)) + det(A(2n-2)) ...).
But, again because B has an odd number of collumns, it follows that all but one term cancels and we get
det(B) = - det(A(2n-2))
Taking this together, we see that
det(A(2n)) = det(A(2n-2)
and from this follows that it's 1. (look at det (A(2)))
