Inverse of a triangular-by-block $3 \times 3$ matrix Let's assume that matrix blocks $A$, $B$ and $C$ are known.
I need to find blocks $X$, $Y$ and $Z$. I also need to know which of these blocks  are nonsingular.
$
\begin{bmatrix} I & 0 & 0 \\ C & I & 0 \\ A & B & I \end{bmatrix}^{-1} = 
\begin{bmatrix} I & 0 & 0 \\ Z & I & 0 \\ X & Y & I \end{bmatrix}
$

My try so far: 
$
K^{-1} = \begin{bmatrix} I & 0 & 0 \\ C & I & 0 \\ A & B & I \end{bmatrix}^{-1}  
H = \begin{bmatrix} I & 0 & 0 \\ Z & I & 0 \\ X & Y & I \end{bmatrix}
$
$ K^{-1} = H$
$ K^{-1}K = HK$
$ I = HK$  

Above you can see that I was able to change the problem to form $ I = HK $.
I should now draw the matrixes again into linear format like following: 
$
\begin{bmatrix} I & 0 & 0 \\ 0 & I & 0 \\ 0 & 0 & I \end{bmatrix} =
\begin{bmatrix} I & 0 & 0 \\ Z & I & 0 \\ X & Y & I \end{bmatrix}
\begin{bmatrix} I & 0 & 0 \\ C & I & 0 \\ A & B & I \end{bmatrix}
$
Everything seems fine above, except that I'm not sure I can present the last part $K$ with $ \begin{bmatrix} I & 0 & 0 \\ C & I & 0 \\ A & B & I \end{bmatrix} $
, as it was originally inverse matrix $ K^{-1} = \begin{bmatrix} I & 0 & 0 \\ C & I & 0 \\ A & B & I \end{bmatrix}^{-1} $. I don't think you can just remove the $^{-1}$ and keep the contents the same. However, I saw some content related to this problem that did it like this. I'm guessing if it's all wrong.
 A: Here is an answer that uses the following formula :
$$\tag{1}(\frak{I}+M)^{-1}=I-M+M^2-M^3+...$$
that is especialy interesting when there is a finite number of terms in the expansion.
Let us use formula (1) with $ \frak {I}:=\begin{bmatrix} I & 0 & 0 \\ 0 & I & 0 \\ 0 & 0 & I \end{bmatrix}$ and $\frak {M}:=\begin{bmatrix} 0 & 0 & 0 \\ C & 0 & 0 \\ A & B & 0 \end{bmatrix}$ :
$$(\frak{I}+M)^{-1}=\begin{bmatrix} I & 0 & 0 \\ 0 & I & 0 \\ 0 & 0 & I \end{bmatrix}-\begin{bmatrix} 0 & 0 & 0 \\ C & 0 & 0 \\ A & B & 0 \end{bmatrix}+\begin{bmatrix} 0 & 0 & 0 \\ C & 0 & 0 \\ A & B & 0 \end{bmatrix}^2$$
(other terms are zero matrices), giving:
$$(\frak{I}+M)^{-1}=\begin{bmatrix} I & 0 & 0 \\ 0 & I & 0 \\ 0 & 0 & I \end{bmatrix}-\begin{bmatrix} 0 & 0 & 0 \\ C & 0 & 0 \\ A & B & 0 \end{bmatrix}+\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ BC & 0 & 0 \end{bmatrix}=$$
$$\begin{bmatrix} \ I & \ 0 & \ \ \ 0 \\ -C & \ I & \ \ \ 0 \\ (BC-A) & -B & \ \ \ I \end{bmatrix}$$
which is the desired inverse matrix.
Remark: this method is easily extensible to any size.
