Consider the degree-$n$ polynomial
$$
f(x_1,\ldots,x_n)=\prod_{i=1}^n\left(\sum_{j=1}^na_{ij}x_j-b_i\right).
$$
To form the square-free monomial $x_1\cdots x_n$, one must choose each variable exactly once from the $n$ factors. Such choices are indexed by permutations, so
$$
[x_1\cdots x_n]f
=\sum_{\sigma\in S_n}\prod_{i=1}^na_{i,\sigma(i)}
=\operatorname{perm}A.
$$
This coefficient is nonzero by hypothesis. Apply the <Combinatorial Nullstellensatz> with every $d_i=1$ and the given two-element sets $S_i$. It supplies $x\in\prod_iS_i$ with $f(x)\ne0$. Every factor is then nonzero, so
$$
(Ax)_i\ne b_i
$$
for all $i$. This is <coordinate avoidance from a nonzero permanent>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 161 / 3 / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-161/3/ii/solution