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