Finding pattern in matrix inverse I want to see if any pattern that could be formulated exist across the rows (or equivalently , columns) of the matrix inverse (in the middle , let's name it $A^{-1}$) so that an analytical  formula that relates $o_n$ and $b_n$ (perhaps along with other b's) exists (for any positive integer n )
$$
 \begin{pmatrix}
o_1 \\
o_2 \\
: \\
: \\
: \\
o_n 
\end{pmatrix} = 
 \begin{pmatrix}
1 & 0 & 0  & 0 & ... & 0 \\
w & 1 & 0   & 0 & \ddots & 0  \\ 
i & w & 1  & 0 & \ddots & 0 \\ 
i & i & w & 1 & \ddots & 0 \\ 
: & \ddots & \ddots  & \ddots & \ddots  & \ddots \\ 
i & i & i & ... & w & 1 
\end{pmatrix}^{-1}
\begin{pmatrix}
b_1 \\
b_2 \\
: \\
: \\
: \\
b_n 
\end{pmatrix} 
$$
The matrix inverse $A^{-1}$ is a lower triangular matrix with diagonal full of 1 , subdiagonal full of w and others are i .
when $n = 6$ , the matrix inverse becomes
$$ \left( \begin{array}{cccccc} 1 & 0 & 0 & 0 & 0 & 0 \\ -w & 1 & 0 & 0 & 0 & 0 \\ w^2-i & -w & 1 & 0 & 0 & 0 \\ -w^3+2 i w-i & w^2-i & -w & 1 & 0 & 0 \\ w^4-3 i w^2+2 i w+i^2-i & -w^3+2 i w-i & w^2-i & -w & 1 & 0 \\ -w^5+4 i w^3-3 i w^2-3 i^2 w+2 i w+2 i^2-i & w^4-3 i w^2+2 i w+i^2-i & -w^3+2 i w-i & w^2-i & -w & 1 \end{array} \right) $$
I recommend matrix calculator https://www.dcode.fr/matrix-inverse to quickly find matrix inverse symbolically .
Since in this case it doesn't matter whether we look at rows or columns , by looking at the first column , I could only write down
$$  \left\{ \begin{array}{ccc}
a_1 & = & 1 \\
a_2 & = & a_1 (-w) \\
a_3 & = & a_2 (-w) - i  \\
a_4 & = & a_3 (-w) + iw - i\\
a_5 & = & a_4 (-w) + iw - iw^2 - i + i^2\\
a_6 & = & a_5 (-w) + iw - iw^2 +iw^3 - 2i^2w - i + 2i^2\\
\end{array} \right.
$$
Edit : After reading @metamorphy 's answer
Below is my note ,
since the "1"-diagonal of $S^k$ will lie on the $k^{th}$ subdiagonal , we may visualize $A^{-1}$ and write down
$$
o_n =  ( \alpha_+\lambda_{+}^0 + \alpha_- \lambda_{-}^{0} ) b_n + 
( \alpha_+\lambda_{+}^{1} + \alpha_- \lambda_{-}^{1} ) b_{n-1} + ... + 
( \alpha_+\lambda_{+}^{n-1} + \alpha_- \lambda_{-}^{n-1} ) b_{1} 
$$
$$
 = \sum_{k=0}^{n-1} ( \alpha_+\lambda_{+}^{k} + \alpha_- \lambda_{-}^{k} ) b_{n-k} 
$$
with
$$ \alpha_+ = \frac{\lambda_{+} - 1 }{\lambda_{+} - \lambda_{-}} ,\qquad \alpha_- = \frac{\lambda_{-} - 1 }{\lambda_{-} - \lambda_{+}} 
$$

Let $S$ be the $n\times n$ matrix with $S_{ij}=\begin{cases}1,&i=j+1\\0,&\text{otherwise}\end{cases}$.
The matrix being inverted is $A=I+wS+iS^2(I-S)^{-1}$, hence $A^{-1}=f(S)$ with $$f(z)=\frac{1-z}{1-(1-w)z-(w-i)z^2}.$$
Next, we find the partial fraction decomposition of $f(z)$, say $$f(z)=\frac{\alpha_+}{1-\lambda_+ z}+\frac{\alpha_-}{1-\lambda_- z},\qquad\lambda_\pm=\frac12\left(1-w\pm\sqrt{(1+w)^2-4i}\right)$$ (this is possible if $(1+w)^2\neq4i$; use continuity or symbolic argument otherwise). Then $$A^{-1}=\sum_{k=0}^{n-1}(\alpha_+\lambda_+^k+\alpha_-\lambda_-^k)S^k$$ because $S^k=0$ for $k\geqslant n$; this gives a closed form of the elements on $k$-th diagonals of $A^{-1}$.
