Past exam of the mathematical methods I course of the University of Oxford / 2016 resit / 9 / iii

Because $A$ is symmetric, translating $\mathbf{r}=\mathbf{x}+\mathbf{a}$ givesThe linear term vanishes when $\boxed{2A\mathbf{a}+\mathbf{b}=0}$. Solving gives $\boxed{\mathbf{a}=(-1,0,0{)}^{\mathsf{T}}}$. Since $\mathbf{b}=-2A\mathbf{a}$, the remaining constant isThis is completing the square for a quadratic form.