For $n \ge 2$, $A$ and $A^{-1}$ are $n \times n$ positive definite matrices. Now, for a scalar $\alpha > 0$ and $(n-1)$-dimensional column vector $\beta$ and $(n-1) \times (n-1)$ $\Delta$ matrix.
Given that
$$ A^{-1} = \begin{bmatrix} \alpha & \beta^T \\\ \beta & \Delta \end{bmatrix} $$
$\tilde{A}$ is the $(n-1) \times (n-1)$ submatrix, and I would like to prove the following holds,
$$ \tilde{A}^{-1} = \Delta - \dfrac{\beta\beta^T}{\alpha} $$
I have tried to decompose the $A^{-1}$ to something like
$$ \begin{align*} \begin{bmatrix} \alpha & \beta^T \\\ \beta & \Delta \end{bmatrix}^{-1} & = \begin{bmatrix} I & O \\\ \beta/\alpha & I \end{bmatrix}^{-1} \begin{bmatrix} \alpha & \beta^T \\\ O & \Delta - \beta\beta^T/\alpha \end{bmatrix}^{-1} \\\ \end{align*} $$
Yet, I am not sure where to keep going at this point, what should I focus on for this moment?
Write $A=\begin{bmatrix}a & b^T\\ b & \tilde A\end{bmatrix}$. Now use the block multiplication to have this two equations:
$$b\beta^T + \tilde A \Delta = I$$ and $$\alpha b + \tilde A \beta = 0.$$
Now multiply the first equality with $\alpha$ and the second equality with $\beta^T$ and take the difference of two equations to have:
$$\tilde A \left(\alpha \Delta - \beta \beta^T\right) = \alpha I.$$