Method to find the inverse of any lower triangular matrix Is there a special method to find the the inverse for a matrix which would classified as a lower or left triangular matrix for a matrix L which is n by n.  Additionally where the upper part of the matrix would also be all zeros.
where none of the diagonals are equal to zero{(1,1),  (2,2),...,(n,n)} ≠ 0 ,  or a,c,j and j in the example below.
so the determinant ≠ zero.
For example when n=4 finding the inverse of the matrix $L$ where
$$L=\begin{pmatrix}
a & 0 & 0 & 0\\
b & c & 0 & 0\\
d & e & f & 0 \\
g & h & i & j \\
\end{pmatrix}
$$
But could also work when n = 5,6,..., at least 10 (for the sake of simplicity) 
Does there exist a method to find the inverse of any sized matrix in this form?
 A: We can write $L = D(I + N)$ where $D$ is diagonal and $N$ is strictly lower triangular and nilpotent ($N^n = 0$): $N_{ij} = L_{ij}/D_{ii}$.  Then $L^{-1} = (I+N)^{-1} D^{-1}$.
$D^{-1}$ is diagonal with $(D^{-1})_{ii} = 1/D_{ii}$, and 
$(I+N)^{-1} = I + \sum_{j=1}^{n-1} (-1)^j N^j$.
A: In case anybody is looking for an actual closed form (that is, not forward substitution or a Neumann expansion), because this came up in my research at one point:
You can also work out a closed form for each element of the inverse of a triangular matrix. From Cramer's rule, we know $[A^{-1}]_{ij} = \frac{1}{det(A)}C_{ji}$, where $C_{ji}$ is the matrix of cofactors of $A$. The $ji$-cofactor is given by $(-1)^{i+j}$ times the $ji$-minor of $A$, which is defined as the determinant of the matrix obtained by eliminating the $j$th row and $i$th column of $A$. In the case of a (say, lower) triangular matrix, eliminating one row and column results in a block lower triangular matrix, where
$|C_{ji}|= \left|\begin{matrix}T_0 & \mathbf{0} & \mathbf{0} \\ A_0 & H & \mathbf{0} \\ A_1 & A_2 & T_1\end{matrix}\right|$.
Here, $T_0$ and $T_1$ are lower triangular and $H$ is lower Hessenberg. By nature of block determinants, $|C_{ji}| = |T_0||T_1||H|$. The relative size of these matrices depends on $i$ and $j$. For $i=j$, $C_{ji}$ is triangular. For $i=1$ and $j=n$, the entire matrix is Hessenberg. The determinant of triangular matrices $T_0$ and $T_1$ is just the product of the diagonals, and there is a nice form for the determinant of Hessenberg matrices given here:
https://www.math.uni-bielefeld.de/ahlswede/pub/tamm/hessen.ps
There they assume a unit upper diagonal, so for the general case it will need to be modified, but doing so is fairly straightforward. 
