Invert a symmetric partition matrix

Question

How do I invert this matrix?

$$ X = \begin{bmatrix} 1 & \mathbf{x'} \\ \mathbf{x} & I \end{bmatrix} $$

If $X$ is $n \times n$, then $I$ is the $(n-1) \times (n-1)$ identity and $x \in \mathbb R^{n-1}$ is a vector.

Thanks

Answer 1

Let's define the $n \times n$ matrix $A$ as $A = \begin{pmatrix} 1 & x^{'} \\ x & I_{n-1} \end{pmatrix}$, with $x = (x_2,x_3,\ldots,x_n)^{'}$. Then apply the follwing row operations: \begin{align} & R_i - x_iR_1 & \forall i \geq 2, \\ & \frac{1}{1-x_i^2}R_i & \forall i \geq 2, \\ & R_1 - x_iR_i & \forall i \geq 2. \end{align} Then we have found that $A$ is the product of $(n-1)$ + $(n-1)$ + $(n-1) = 3(n-1)$ elementary matrices $E_j$, and $$A^{-1} = \left( \prod_{j}^{3(n-1)} E_j \right)^{-1}.$$