@becko: and clearly, id. (1) can be easily inverted to get the formula for $\bf{M^{-1}}$

@becko: yes, for non-negative integer $a$ the matrix is always invertible, as can be seen from the determinant and from its expansion formula.

@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 $2/(4 \cdot1) \cdot 4/(8 \cdot 2) = 1/8)$.