Finding the inverse of a block $2\times2$ square matrix $ \begin{bmatrix} I & A \\ A^T & 0 \end{bmatrix} $ where $A$ is a square invertible matrix. The hint I got was to rewrite the original matrix as a product of 3 matrices and use the property for inverse of product of matrices $(XYZ)^{-1} = Z^{-1} Y^{-1} X^{-1}$
$\begin{align} \begin{bmatrix} I & A \\ A^T & 0 \end{bmatrix}  &=   \begin{bmatrix} I & 0 \\ A^T & 0 \end{bmatrix}  \begin{bmatrix} I & 0 \\ 0 & -A^T A \end{bmatrix}  \begin{bmatrix} I & A \\ 0 & I \end{bmatrix} 
\end{align}\\ $
Then
$\begin{align} 
\begin{bmatrix} I & A \\ A^T & 0 \end{bmatrix}^{-1}= \begin{bmatrix} I & A \\ 0 & I \end{bmatrix} ^{-1} \begin{bmatrix} I & 0 \\ 0 & -A^T A \end{bmatrix}^{-1}  \begin{bmatrix} I & 0 \\ A^T & 0 \end{bmatrix}^{-1} 
\end{align}$
The first 2 matrices on the RHS are invertible but the third is not, so I might have gone down the wrong path with the product of 3 matrices I think.
The actual inverse for  $\begin{bmatrix} I & A \\ A^T & 0 \end{bmatrix} $ is $\begin{bmatrix} 0 & A^{-T} \\ A^{-1} & -A^{-1}A^{-T} \end{bmatrix}$ which can be found by using the formula at the end of section 3 in Lu and Shiou. I would like to know how to solve the problem without resorting to using a formula. Thanks for any tips.

Your product formula is incorrect. Computing the product you have written yields
$$
\begin{bmatrix} I & 0 \\ A^T & 0 \end{bmatrix}  \begin{bmatrix} I & 0 \\ 0 & -A^T A \end{bmatrix}  \begin{bmatrix} I & A \\ 0 & I \end{bmatrix} = 
\begin{bmatrix}
I & A\\A^T & A^TA
\end{bmatrix}.
$$
A correct formula is
\begin{align} 
\begin{bmatrix} I & A \\ A^T & 0 \end{bmatrix}  &=   
\begin{bmatrix} I & 0 \\ A^T & I \end{bmatrix}  
\begin{bmatrix} I & 0 \\ 0 & -A^T A \end{bmatrix}  \begin{bmatrix} I & A \\ 0 & I \end{bmatrix}, 
\end{align}
which coincides with the formula referenced here. Taking the inverse yields
$$
\begin{bmatrix} I & A \\ 0 & I \end{bmatrix}^{-1}
\begin{bmatrix} I & 0 \\ 0 & -A^T A \end{bmatrix}^{-1}  
\begin{bmatrix} I & 0 \\ A^T & I \end{bmatrix}^{-1} = \\
\begin{bmatrix} I & -A \\ 0 & I \end{bmatrix}
\begin{bmatrix} I & 0 \\ 0 & -A^{-1}A^{-T} \end{bmatrix}
\begin{bmatrix} I & 0 \\ -A^T & I \end{bmatrix}.
$$
Taking the product of these leads to the correct answer of
$$
\begin{bmatrix} I & A \\ A^T & 0 \end{bmatrix}^{-1} = \begin{bmatrix} 0 & A^{-T} \\ A^{-1} & -A^{-1}A^{-T} \end{bmatrix}.
$$
