TITLE: How to inverse a block diagonal matrix?
QUESTION [6 upvotes]: Given a matrix $$x = \begin{bmatrix} 
40 & 0 & 0 & 0\\
0 & 80 & 100 & 0 \\
0 & 40 & 120 & 0 \\
0 & 0 & 0 & 60\end{bmatrix}$$
How to find the inverse of that matrix?
What I know: $\det(x) = ac-bd$, 
inverse of a 2x2 matrix: $$x^{-1} = \frac{1}{\det(x)}\cdot \begin{bmatrix} d &-b\\ -c &a\end{bmatrix}.$$
There is a lot of content online; however none of them has a specific numerical example.

REPLY [10 votes]: Block diagonal matrices can be inverted block by block. See also [*].
In your example:
$$\begin{bmatrix} 40 & 0 & 0 & 0 \\ 0 & 80 & 100 & 0 \\ 0 & 40 & 120 & 0 \\ 0 & 0 & 0 & 60\end{bmatrix}^{-1}
= \begin{bmatrix} [40]^{-1} & \begin{matrix} 0 \quad & 0\quad \end{matrix} & 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} & \begin{bmatrix} 80 & 100 \\ 40 & 120 \end{bmatrix}^{-1} & \begin{matrix} 0  \\ 0 \end{matrix} \\
0 & \begin{matrix} 0\quad & 0\quad \end{matrix} & [60]^{-1}
\end{bmatrix}.
$$
Can you take it from here?

REPLY [2 votes]: Guide:
If you have a matrix of the form of 
$$diag(D_1, D_2, D_3),$$
where each block is invertible, then its inverse is $$diag(D_1^{-1},D_2^{-1}, D_3^{-1}).$$
You should verify this.
In your question $D_2$ is $2$ by $2$ and the other two blocks are scalar.