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Aug 2, 2017 at 15:09 comment added valle Yes, thank you. I accepted your answer.
Aug 2, 2017 at 15:09 vote accept valle
Aug 2, 2017 at 15:01 comment added G Cab @becko; is everyhing (enough) clear now?
Jul 26, 2017 at 21:52 history edited G Cab CC BY-SA 3.0
edited typo
Jul 26, 2017 at 20:13 comment added G Cab @becko: I added some notes to better clarify the meaning of matrix $\bf E$, and to show that inversion of (1) just involves simple manipulations around the same set of basic matrices.
Jul 26, 2017 at 20:08 history edited G Cab CC BY-SA 3.0
added notes on inversion
Jul 26, 2017 at 13:22 comment added valle Yes, but you still have to invert $\mathbf{B}_h$, $\mathbf{St}_{2h}$ and $\mathbf{I}_h + \overline{\mathbf{E}_h}$, which I'm not sure how to do. In your notation, what does $(\mathbf{I}_h + \overline{\mathbf{E}_h})^\mathbf{a}$ mean?
Jul 26, 2017 at 13:09 comment added G Cab @becko: and clearly, id. (1) can be easily inverted to get the formula for $\bf{M^{-1}}$
Jul 26, 2017 at 13:06 history edited G Cab CC BY-SA 3.0
added tags to formulas
Jul 26, 2017 at 12:58 comment added G Cab @becko: yes, for non-negative integer $a$ the matrix is always invertible, as can be seen from the determinant and from its expansion formula.
Jul 26, 2017 at 11:59 comment added valle +1 Oh you are correct. I had a typo in the program I wrote to test this. Sorry :) ... This shows that the determinant is always positive, so the matrix is always invertible.
Jul 26, 2017 at 11:50 comment added G Cab @becko before posting, I tested both formulas for various values and are both ok. In particular for $a=1$ and $h=1$, the det. should in fact be $1/8$ and the last formula gives $2/(4 \cdot1) \cdot 4/(8 \cdot 2) = 1/8)$.
Jul 25, 2017 at 23:37 review First posts
Jul 26, 2017 at 0:52
Jul 25, 2017 at 23:35 history answered G Cab CC BY-SA 3.0