Past exam of the year 1 of the computer science course of the University of Oxford / 2019 / Discrete mathematics, continuous mathematics, and probability resit / 6 / c

Let $P=\mathbf{z}{\mathbf{z}}^T$, so $P^2=r^{2}P$ and $f^2=1+r^2$. Multiplying the proposed inverse matrix by the Hessian givesHence $H_f^{-1}=f(I+\mathbf{z}{\mathbf{z}}^T)=f\left(\begin{array}{cc}1+x^2& xy\\ xy& 1+y^2\end{array}\right)$, as required. Its existence also follows from the Hessian being a positive-definite matrix.