Assume $A$ is a symmetric nonsingular positive semidefinite nonnegative $3 \times 3$-matrix. Let $\gamma_0$ be the maximal $\gamma \geq 0$ for which $A-\gamma E_{33}$ is positive semidefinite. Show that $\gamma_0=\frac{detA}{detA_{33}}$, where $$A=\begin{pmatrix} a_{11}&a_{12}&a_{13}\\ a_{12}&a_{22}&a_{23}\\ a_{13}&a_{23}&a_{33} \end{pmatrix},$$ $$A_{33}=\begin{pmatrix} a_{11}&a_{12}\\ a_{12}&a_{22}\\ \end{pmatrix},$$ $$E_{33}=\begin{pmatrix} 0&0&0\\ 0&0&0\\ 0&0&1 \end{pmatrix}.$$
I need some help with that. Is it possible to show with characteristic polynomial? I couldn't come to the end.
A matrix is positive semidefinite if and only if all its minors are nonnegative. In particular $A-\gamma E_{33}$ is positive semidefinite if and only if its determinant is nonnegative, where $$\det(A-\gamma E_{33})=a_{13}\det A_{13}-a_{23}\det A_{23}+(a_{33}-\gamma)\det A_{33}=\det A-\gamma\det A_{33}.$$ So it follows that the maximal value of $\gamma$ for which $A-\gamma E_{33}$ is positive semidefinite is $\frac{\det A}{\det A_{33}}$.