# How to inverse a block diagonal matrix?

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.

Block diagonal matrices can be inverted block by block. See also [*].

$$\begin{bmatrix} 40 & 0 & 0 & 0 \\ 0 & 80 & 100 & 0 \\ 0 & 40 & 120 & 0 \\ 0 & 0 & 0 & 60\end{bmatrix}^{-1} = \begin{bmatrix} ^{-1} & \begin{matrix} 0 \quad & 0\quad \end{matrix} & 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} & \begin{bmatrix} 80 & 100 \\ 40 & 120 \end{bmatrix}^{-1} & \begin{matrix} 0 \\ 0 \end{matrix} \\ 0 & \begin{matrix} 0\quad & 0\quad \end{matrix} & ^{-1} \end{bmatrix}.$$

Can you take it from here?

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.