How to compute the determinant of this anti-triangular matrix?

Question

Compute the determinant of the following anti-triangular matrix. $$A = \begin{bmatrix} 0& 0&\dots & 0& a_{1\ n}\\ 0 & 0&\dots & a_{2\ n-1}& a_{2\ n}\\ \vdots & \vdots & \ddots & \vdots& \vdots\\ 0 & a_{n-1\ 2}& \dots & a_{n-1\ n-1}& a_{n-1\ n}\\ a_{n\ 1}& a_{n\ 2}& \dots & a_{n\ n-1}& a_{n\ n}\end{bmatrix}$$

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Since any determinant is

$$\det A = (-1)^{1+1} \alpha_{1\ 1}+(-1)^{2+1}\alpha_{1\ 2}+\dots+(-1)^{n+1}\alpha_{1\ n}$$

where $\alpha_{ij}$ are the cofactors, I have correctly computed

$$\det A = a_{1\ 1} a_{2\ 2} \cdots a_{n\ n}$$

but the solution is that times a function of $(-1)^{f(n)}$ and I really don't know how to find that.

Answer 1

Hint Like you write, inductively applying the cofactor expansion to the top row (or leftmost column) of each lower-left square submatrix gives that $$\det A = (-1)^{n - 1} a_{1,n} \cdot (-1)^{n - 2} a_{2,n-1}\cdot \cdots \cdot (-1)^1 a_{n-1,2} \cdot (-1)^0 a_{n,1} = \pm \prod_{i = 1}^n a_{i, n + 1 - i}$$ for some sign $\pm$.

Collecting powers of $-1$ in the above expansion shows that the exponent of $-1$---whose parity determines $\pm$---is $(n - 1) + (n - 2) + \cdots + 2 + 1$.

This exponent coincides with ${n \choose 2}$. Note that since ${{n + 4} \choose 2} - {n \choose 2} = 4 n + 6$ is even for all $n$, the power $(-1)^{n \choose 2}$, hence the sign $\pm$, depends only on the residue class of $n$ modulo $4$.

Answer 2

Right-multiplying by the $n \times n$ reversal matrix $\bf R$ (which is a permutation matrix and, thus, orthogonal), we reverse the order of the columns and obtain a lower triangular matrix whose determinant is easy to compute.

$$ \det ({\bf A} {\bf R}) = \det ({\bf A}) \underbrace{\det ({\bf R})}_{= \pm 1} = \prod_{i=1}^n a_{i,n+1-i}$$

Thus, what is left is determining the sign of $\det ({\bf R})$ as a function of $n$.