Past exam of the mathematics course of the University of Cambridge / 2025 / ib / Paper 3 / 19H / d / Solution

The matrix is antisymmetric, so the value is zero. For a common candidate $p=(a,b,c,d{)}^T$, the column-player inequalities are $Ap\le 0$:The first three imply $a+d\le b\le c\le a-2d$, hence $d=0$ and $a=b=c$. Normalization givesIndeed $Ap=(0,0,0,-1{)}^T/3\le 0$, and antisymmetry gives the corresponding row inequalities.