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I am reading the book, Applied Linear Algebra and Matrix Analysis.
When I was doing the exercise of Section2.5 Problem 29, I was puzzled at solving it.
Here is the problem description:

Give a 2*2 block matrix $M = \begin{bmatrix}A&B\\0&C \end{bmatrix}$, where the blocks A and C are invertible matrices, find a formula for $M^{-1}$ in terms of $A$, $B$, and $C$.

And the answer is:

Assume $M^{-1}$ has the same form as $M$ and solve for the blocks in $M$ using $MM^{−1}$ = I.

But I still confused about how to make it right.
So I do some searches on the net and I find a paper give explicit inverse formulae for 2 × 2 block matrices with three different partitions.
BUT, as the paper mentioned, all the blocks must be nonsingular, which means they all have a matrix inverse. It is not subject to the conditions, which only say the blocks $A$ and $C$ are invertible matrices.
SO, I want to know if there is a better way which is subject to conditions.
I will appreciate it if anyone help me.

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    $\begingroup$ well... have you tried to let $M^{-1}=\begin{pmatrix} E& F \\ 0 & G \end{pmatrix}$, compute $MM^{-1}=I$ and solve for $E,F,G$? $\endgroup$ – Surb Mar 22 at 7:45
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    $\begingroup$ @Surb You are right, I made a stupid mistake that I forgot $M^{-1}$ is an upper triangular matrix. I calculate as you said and I get $M^{-1} = \begin{bmatrix}A^{-1} & -A^{-1}BC^{-1}\\ 0 & C^{-1} \end{bmatrix}$. Next time I won't foget it :( $\endgroup$ – Bowen Peng Mar 22 at 7:56
  • $\begingroup$ This is why we do exercises, to practice and remember the trick for next time :). I think your upper right block is wrong. $\endgroup$ – Surb Mar 22 at 7:58
  • $\begingroup$ @Surb Yeah, I correct it now. Maybe I need to do more practices. :( $\endgroup$ – Bowen Peng Mar 22 at 7:59
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A informal argument I let you add the details on dimensions etc.:

Let $$M^{-1}=\begin{bmatrix} E& F \\ 0 & G \end{bmatrix}$$ Then, $$\begin{bmatrix}I & 0 \\ 0 & I \end{bmatrix}=I=MM^{-1} = \begin{bmatrix}A&B\\0&C \end{bmatrix}\begin{bmatrix} E& F \\ 0 & G \end{bmatrix}=\begin{bmatrix} AE & AF+BG \\ 0 &CG\end{bmatrix}$$ Thus, we must have $AE=I$, $CG=I$ and $AF+BG=0$. It follows that $E=A^{-1}$ and $G=C^{-1}$. Now, $AF+BG=0$ becomes $AF+BC^{-1}=0$ so that $F=-A^{-1}BC^{-1}$. We conclude that $$M^{-1}=\begin{bmatrix} A^{-1}& -A^{-1}BC^{-1} \\ 0 & C^{-1} \end{bmatrix}$$

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Hint. As suggested by Surb, let $M^{-1}=\begin{bmatrix} E& F \\ 0 & G \end{bmatrix}$ then, by definition of inverse matrix, $MM^{-1}=I$, i.e. $$\begin{bmatrix}A & B\\0 & C\end{bmatrix} \begin{bmatrix}E & F\\0 & G\end{bmatrix} = \begin{bmatrix}I & 0\\0 & I\end{bmatrix}$$ and it follows that $$AE=I,\quad CG=I,\quad AF+BG=0.$$ Can you take it from here and find $E$, $F$, and $G$ in terms of $A$, $B$, and $C$?

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You do exactly as the suggestion says: $$ \begin{bmatrix}X&Y\\0&Z\end{bmatrix}\begin{bmatrix}A&B\\0&C\end{bmatrix} =\begin{bmatrix}XA&XB+YC\\0&ZC\end{bmatrix} $$ So you need $X=A^{-1} $, $Z=C^{-1} $, and $A^{-1}B+YC=0$. This last equality gives $Y=-A^{-1}BC^{-1} $. So $$ M^{-1}=\begin{bmatrix}A^{-1}&-A^{-1}BC^{-1}\\0&C^{-1}\end{bmatrix}. $$

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    $\begingroup$ Yep. Thanks! $ $ $\endgroup$ – Martin Argerami Mar 22 at 8:01

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