We know that matrices can be inverted blockwise by using the following analytic inversion formula: \begin{equation} \begin{bmatrix} \mathbf{A} & \mathbf{C^T} \\ \mathbf{C} & \mathbf{D} \end{bmatrix}^{-1} = \begin{bmatrix} \mathbf{A}^{-1}+\mathbf{A}^{-1}\mathbf{C^T}\mathbf{S_D}^{-1}\mathbf{CA}^{-1} & -\mathbf{A}^{-1}\mathbf{C^T} \mathbf{S_D}^{-1} \\ - \mathbf{S_D} ^{-1}\mathbf{CA}^{-1} & \mathbf{S_D}^{-1} \end{bmatrix} \end{equation} with $S_D=\mathbf{D}-\mathbf{CA}^{-1}\mathbf{C^T}$ the Schur complement of the block D or alternatively, \begin{equation} \begin{bmatrix} \mathbf{A} & \mathbf{C^T} \\ \mathbf{C} & \mathbf{D} \end{bmatrix}^{-1} = \begin{bmatrix} \mathbf{S_A}^{-1} & - \mathbf{S_A}^{-1}\mathbf{C^T}\mathbf{D}^{-1} \\ - \mathbf{D}^{-1}\mathbf{C}\mathbf{S_A}^{-1} & \mathbf{D}^{-1}+\mathbf{D}^{-1}\mathbf{C}\mathbf{S_A}^{-1}\mathbf{C^TD}^{-1} \end{bmatrix} \end{equation} with $S_A=\mathbf{A}-\mathbf{C^TD}^{-1}\mathbf{C}$ the Schur complement of the block A \ what about the case when we have a $3 \times 3$ partitioned matrix as following: $\begin{bmatrix} \mathbf{P} & \mathbf{C^T} \\ \mathbf{C} & \mathbf{D} \end{bmatrix}^{-1}$ where $ \mathbf{P} = \begin{bmatrix} \mathbf{X} & \mathbf{Y} \\ \mathbf{Y^T} & \mathbf{Z} \end{bmatrix} $?

To solve this problem, I apply use Equation (2) with $S_P$ the Shur complement of the block P : \begin{equation} \begin{bmatrix} \mathbf{P} & \mathbf{C^T} \\ \mathbf{C} & \mathbf{D} \end{bmatrix}^{-1} = \begin{bmatrix} \mathbf{S_P}^{-1} & - \mathbf{S_P}^{-1}\mathbf{C^T}\mathbf{D}^{-1} \\ - \mathbf{D}^{-1}\mathbf{C}\mathbf{S_P}^{-1} & \mathbf{D}^{-1}+\mathbf{D}^{-1}\mathbf{C}\mathbf{S_P}^{-1}\mathbf{C}^T\mathbf{D}^{-1} \end{bmatrix} \end{equation} with \begin{align} S_P &= \mathbf{P}-\mathbf{C^TD}^{-1}\mathbf{C} \\ &= \begin{bmatrix} \mathbf{X} - \mathbf{C_1}^T\mathbf{D}^{-1}\mathbf{C_1} & \mathbf{Y}-\mathbf{C_1}^T\mathbf{D}^{-1}\mathbf{C_2}\\ \mathbf{Y^T}-\mathbf{C_2}^T\mathbf{D}^{-1}\mathbf{C_1} & \mathbf{Z}-\mathbf{C_2}^T\mathbf{D}^{-1}\mathbf{C_2} \end{bmatrix} \end{align} with for example $\mathbf{C} = \begin{bmatrix} \mathbf{C_1} & \mathbf{C_2} \end{bmatrix}$ The problem is that I need to inverse this $S_P$. I can apply again this partitioned matrix inversion. Is there any easier solution to inverse a $3 \times 3$ partitioned matrix ?