Question

Let \(\mathbf A_n\) be a \(n\times n\) matrix with the \((i,j)\)th entry defined by
\[a_{ij}=\begin{cases}0&\text{for }i=j\\[1ex]\dfrac{1}{j-i}&\text{otherwise }\end{cases}\]For example,
\[\mathbf A_3=\begin{bmatrix}0&1&\dfrac12\\[1ex]-1&0&1\\[1ex]-\dfrac12&-1&0\end{bmatrix}\quad\text{and}\quad\mathbf A_4=\begin{bmatrix}0&1&\dfrac12&\dfrac13\\[1ex]-1&0&1&\dfrac12\\[1ex]-\dfrac12&-1&0&1\\[1ex]-\dfrac13&-\dfrac12&-1&0\end{bmatrix}\]Show that \(\det\mathbf A_n=0\) if \(n\) is odd. Is there a closed form for the determinant if \(n\) is even?

Answer 1

oh this is interesting

Answer 2

1 cofactor down for A3 is A2 and A1 is a cofactor of that