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.
Best Answer
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 $$M^{-1}=\begin{bmatrix} A^{-1}& -A^{-1}BC^{-1} \\ 0 & C^{-1} \end{bmatrix}$$