Can anyone see a neat way to invert this matrix?
$$ M = \left( \begin{array}{cccccc} a & b & 0 & c & 0 & 0 \\ 2d & e & 2b & 0 & f & 0 \\ 0 & d & g & 0 & 0 & h \\ A & 0 & 0 & B & b & 0 \\ 0 & C & 0 & 2D & E & 2b \\ 0 & 0 & F & 0 & D & G \\ \end{array} \right) $$
Entries are elements of $\mathbb{R}$. It's several years since I have studied linear algebra, need to remind myself of the techniques. Due to the number of zeros here I was wondering what the most efficent technique would be to use?
Someone had an idea, but deleted it before I could finish reading it, it utilised this as it's first step:
Let \begin{align} R_1 &= \left( \begin{array}{ccc} a & b & 0 \\ 2d & e & 2b \\ 0 & d & g \end{array} \right) %%%% \quad %%%% R_2 = \left( \begin{array}{ccc} c & 0 & 0 \\ 0 & f & 0 \\ 0 & 0 & h \end{array} \right)\\ %%% V_1 &= \left( \begin{array}{ccc} A & 0 & 0 \\ 0 & C & 0 \\ 0 & 0 & F \end{array} \right) %%%% \quad %%%% V_2 = \left( \begin{array}{ccc} B & b & 0 \\ 2D & E & 2b \\ 0 & D & G \end{array} \right) \end{align}
Therefore
\begin{align} M = \left( \begin{array}{cc} R_1 & R_2 \\ V_1 & V_2 \end{array} \right) \end{align}
EDIT: Can we extend the reduced row echelon concept to this?
Leaving us with \begin{align} M^{-1}= \left( \begin{array}{cc} R_1^{-1}(I+R_2JV_1R_1) & -R_1^{-1}R_2 \\ -JV_1R_1 & J \end{array} \right) \end{align}
Where $$J = R_2(V_2-V_1R_1^{-1}R_2)^{-1}$$