# How to find the inverse of this Block Matrix?

Find the inverse of this matrix:

$\begin{pmatrix} E_p & A & 0 \\ 0&E_q & B \\ 0 &0 &E_r \end{pmatrix} \in GL_{p+q+r}(K)$

$GL$= General linear group

$K$= field

$E_r, E_p, E_q$ = identity matrix

$A, B$ = matrices

I tried to look at the special case $p=q=r=1$ and get

$\begin{pmatrix} 1 & 0 & 0 \\ -A&1 & 0 \\ AB &-B &1 \end{pmatrix} \in GL_{p+q+r}(K)$

Is this right? How can I generalise this?

• As long as you’re careful to keep things in the right order when multiplying, you can pretty much manipulate block matrices as if the blocks were scalars. – amd Mar 7 '17 at 19:06

Using block row operations, we have $$\left[ \begin{array}{ccc|ccc} E_p & A & 0& E_p & 0 & 0\\ 0 & E_q & B& 0 & E_q & 0\\ 0 & 0 & E_r& 0 & 0 & E_r \end{array} \right]\to\\ \left[ \begin{array}{ccc|ccc} E_p & A & 0& E_p & 0 & 0\\ 0 & E_q & 0& 0 & E_q & -B\\ 0 & 0 & E_r& 0 & 0 & E_r \end{array} \right]\to\\ \left[ \begin{array}{ccc|ccc} E_p & 0 & 0& E_p & -A & AB\\ 0 & E_q & 0& 0 & E_q & -B\\ 0 & 0 & E_r& 0 & 0 & E_r \end{array} \right]\to$$ the matrix on the right is the inverse
• For example: to get from the first line to the second, I multiply on the left by $$\pmatrix{E_p\\&E_q&-B\\ && E_r}$$ then use block-matrix multiplication – Omnomnomnom Mar 7 '17 at 20:26
$$\begin{pmatrix} E_p & -A & A B \\ 0 & E_q & - B \\ 0 & 0 & E_r \end{pmatrix} \begin{pmatrix} E_p & A & 0 \\ 0 & E_q & B \\ 0 & 0 & E_r \end{pmatrix} =\begin{pmatrix} E_p & 0 & 0 \\ 0 & E_q & 0 \\ 0 & 0 & E_r \end{pmatrix}$$