# Give a 2*2 block matrix $M = \begin{bmatrix}A&B\\0&C \end{bmatrix}$ and find a formula for $M^{-1}$ in terms of $A$, $B$, and $C$

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 $$C$$.

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.

• well... have you tried to let $M^{-1}=\begin{pmatrix} E& F \\ 0 & G \end{pmatrix}$, compute $MM^{-1}=I$ and solve for $E,F,G$? – Surb Mar 22 at 7:45
• @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 :( – Bowen Peng Mar 22 at 7:56
• This is why we do exercises, to practice and remember the trick for next time :). I think your upper right block is wrong. – Surb Mar 22 at 7:58
• @Surb Yeah, I correct it now. Maybe I need to do more practices. :( – Bowen Peng Mar 22 at 7:59

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 $$M^{-1}=\begin{bmatrix} A^{-1}& -A^{-1}BC^{-1} \\ 0 & C^{-1} \end{bmatrix}$$
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 $$C$$?
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 $$M^{-1}=\begin{bmatrix}A^{-1}&-A^{-1}BC^{-1}\\0&C^{-1}\end{bmatrix}.$$
• Yep. Thanks!  – Martin Argerami Mar 22 at 8:01