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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?

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  • $\begingroup$ 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. $\endgroup$ – amd Mar 7 '17 at 19:06
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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

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  • $\begingroup$ Can u please explain how you get this eliminations little but more? $\endgroup$ – Averroes2 Mar 7 '17 at 19:29
  • $\begingroup$ 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 $\endgroup$ – Omnomnomnom Mar 7 '17 at 20:26
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You can verify yourself

$$\begin{pmatrix} E_p & -A & A B \\ 0 & E_q & - B \\ 0 & 0 & E_r \end{pmatrix} \begin{pmatrix} E_p & A & 0 \\ 0 & E_q & B \\ 0 & 0 & E_r \end{pmatrix} =\begin{pmatrix} E_p & 0 & 0 \\ 0 & E_q & 0 \\ 0 & 0 & E_r \end{pmatrix}$$

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