Inverting matrix with Sherman-Morrison formula

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I am reading this book but I am struggling with a formula. Let $A$ a square matrix $N \times N$ partitioned as $\mathbf{A}=\left[\begin{array}{ll} \mathbf{P} & \mathbf{Q} \\ \mathbf{R} & \mathbf{S} \end{array}\right]$. If its inverse is partitioned in the same manner, $\mathbf{A}^{-1}=\left[\begin{array}{cc} \tilde{\mathbf{P}} & \tilde{\mathbf{Q}} \\ \widetilde{\mathbf{R}} & \widetilde{\mathbf{S}} \end{array}\right]$, then, \begin{aligned} &\widetilde{\mathbf{P}}=\left(\mathbf{P}-\mathbf{Q} \cdot \mathbf{S}^{-1} \cdot \mathbf{R}\right)^{-1} \\ &\tilde{\mathbf{Q}}=-\left(\mathbf{P}-\mathbf{Q} \cdot \mathbf{S}^{-1} \cdot \mathbf{R}\right)^{-1} \cdot\left(\mathbf{Q} \cdot \mathbf{S}^{-1}\right) \\ &\widetilde{\mathbf{R}}=-\left(\mathbf{S}^{-1} \cdot \mathbf{R}\right) \cdot\left(\mathbf{P}-\mathbf{Q} \cdot \mathbf{S}^{-1} \cdot \mathbf{R}\right)^{-1} \\ &\widetilde{\mathbf{S}}=\mathbf{S}^{-1}+\left(\mathbf{S}^{-1} \cdot \mathbf{R}\right) \cdot\left(\mathbf{P}-\mathbf{Q} \cdot \mathbf{S}^{-1} \cdot \mathbf{R}\right)^{-1} \cdot\left(\mathbf{Q} \cdot \mathbf{S}^{-1}\right) \end{aligned}

I believe it is using the Sherman-Morrison formula, presented in the previous section. It states that if we change the matrix A in the form $\mathbf{A} \rightarrow(\mathbf{A}+\mathbf{u} \otimes \mathbf{v})$ for some vector $u,v$ and $u \otimes v$ is a matrix whose $i,j$-th element is the product of the $i$th component of $u$ and the $j$-th component of $v$. Then, $(\mathbf{A}+\mathbf{u} \otimes \mathbf{v})^{-1} = \mathbf{A}^{-1}-\frac{\left(\mathbf{A}^{-1} \cdot \mathbf{u}\right) \otimes\left(\mathbf{v} \cdot \mathbf{A}^{-1}\right)}{1+\lambda}$ with $\lambda \equiv \mathbf{v} \cdot \mathbf{A}^{-1} \cdot \mathbf{u}$

Any help? I only want to figure out the formula for $\widetilde{\mathbf{P}}$. Given that one, I will try to figure out the others