# 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.

• I have taken the liberty to modify a little your text, in particular by replacing the word "sections" by "blocks" – Jean Marie Jan 13 '18 at 18:15
• Please feel free to edit, I dont think any of my selected words match the right ones in english mathematics. – W0lfw00ds Jan 13 '18 at 20:00

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.

• Thanks for your solution. However, I couldn't find the answer to my inverse matrix $^{-1} problem.. I updated my question and fixed few things. Please see the updated question. – W0lfw00ds Jan 13 '18 at 20:00 • Aha! So you calculated what the$K\$ should be in my example! So there was a problem with my results! – W0lfw00ds Jan 13 '18 at 20:05