Past exam of the mathematics course of the University of Cambridge / 2025 / ia / Paper 1 / 2C (Vectors and Matrices)

Expansion gives $\det A=(a-3)(a+2)$. Thus uniqueness fails precisely for $a=-2$ and $a=3$.

Solvability requires $b$ to be orthogonal to the left nullspace. For $a=-2$ one may take $n=(-3,-2,2{)}^T$; for $a=3$ one may take $n=(1,-1,1{)}^T$.

For $b=(2,b,0{)}^T$, compatibility gives $b=-3$ when $a=-2$ and $b=2$ when $a=3$. The respective solution families areandwhere $t\in \mathbb{R}$.