How can I find the inverse of this infinite triangular matrix? I want to find the inverse of the following matrix
\begin{bmatrix}
1&0&0&0&0&\cdots&0&0&\cdots\\
0&1&0&0&0&\cdots&0&0&\cdots\\
\binom{2}{0}&0&1&0&0&\cdots&0&0&\cdots\\
\binom{4}{1}&\binom{2}{0}&0&1&0&\cdots&0&0&\cdots\\
\binom{6}{2}&\binom{4}{1}&\binom{2}{0}&0&1&\cdots&0&0&\cdots\\
\vdots&\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\ddots\\
\binom{2(n-1)-2}{(n-1)-2}&\binom{2(n-1)-4}{(n-1)-3}&\binom{2(n-1)-6}{(n-1)-4}&\binom{2(n-1)-8}{(n-1)-5}&\binom{2(n-1)-10}{(n-1)-6}&\cdots&1&0&\cdots\\
\binom{2n-2}{n-2}&\binom{2n-4}{n-3}&\binom{2n-6}{n-4}&\binom{2n-8}{n-5}&\binom{2n-10}{n-6}&\cdots&0&1&\cdots\\
\vdots&\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\ddots&
\end{bmatrix}
in order to get a different approach to the question
How can I find the general term of this recursive sequence?
 A: Did you try to find it for the first terms such that you can justify that for 3x3 matrix or 4x4 matrix ?
If you try to do that i thought that you will find same matrix with minus forms of binomial coefficents.
If you reduce it to augmented matrix [A/I], then row reduced echelon form of A will be inverse of A such that [I/A^-1].
A: Hmm, I'm not sure where your problem is.       
Let's consider $L$ as the empty-matrix with all entries in the first principal subdiagonal set to $1$. 
Then your matrix $M$ can be written as evaluation of the power series
$$ f(x)= x^0 + 0 \cdot x + \sum_{k=2}^\infty \binom{2k-2}{k-2}x^k \tag {1.1}$$
writing
$$ M= f(L) \tag {1.2} $$
The reciprocal of $f(x)$ can be found by
$$g(x)={1\over f(x)}= 1- (1 x^2+4 x^3 + 14 x^4 + 48 x^5 + 165 x^6 + 572 x^7 + ... +c_k x^k + ...)  \tag {2.1}
$$
By this the inverse of $M$ should be definable by $g(x)$ leading to
$$ M^{-1}= g(L) = L^0 - (1 L^2 + 4 L^3 + 14 L^4 + 48 L^5 + 165 L^6 + 572 L^7 + ... +c_k L^k + ...) \tag {2.2}
$$
and I think the coefficients $c_k$ have been discussed already in your earlier question.
