Inverse of matrix with very structured submatrix Does this matrix admit an easy analytic expression for its inverse?
$$\begin{bmatrix}
a_1 & 0 & 0 & 0 & 0 &0&\dots&0 \\
a_2 & 1 & -b & 0 & 0&0&\dots&0 \\
a_3 & 0 & 1 & -b & 0 & 0&\dots&0\\
a_4 & 0 & 0 & 1 & -b & 0 &\dots&0\\
\vdots \\
a_{N-1} & 0 & 0 & 0 & 0 & 0 & \dots & -b \\
a_N & 0 & 0 & 0 & 0 & 0 & \dots & 1
\end{bmatrix}$$
The special thing is that it has a very easy structure, apart from row one/column one.
 A: Yes, for $a_1\ne 0$ it does. Hints:


*

*Show that for a block-triangular matrix
$$
\begin{bmatrix}
A & 0\\B & C
\end{bmatrix}^{-1}=
\begin{bmatrix}
A^{-1} & 0\\-C^{-1}BA^{-1} & C^{-1}
\end{bmatrix}.
$$

*Show that for the $n\times n$ matrix
$$
C=I-bN,
$$
where $N$ is the matrix of all zeros except of ones on the first superdiagonal, the inverse is given by
$$
C^{-1}=\sum_{k=0}^{n-1}b^kN^k.
$$
(e.g. using the fact that $\frac{1}{1-x}=1+x^2+x^3+\ldots$ and $N^n=0$).

A: First, left-multiply by $D=\operatorname{diag}(1/a_{1},1,1,\ldots,1)$
to get
$$
B\equiv DA=\left(\begin{array}{ccccc}
1 & 0 & 0 & 0 & 0\\
a_{2} & 1 & -b\\
a_{3} &  & 1 & -b\\
\vdots &  &  & 1 & -b\\
a_{N} &  &  &  & 1
\end{array}\right).
$$
Expanding along the first row, the determinant of $B$ is (see also this page about the continuant)
$$
\det B=\det\left(\begin{array}{ccccc}
1 & -b\\
 & 1 & -b\\
 &  & \ddots & \ddots\\
 &  &  & 1 & -b\\
 &  &  &  & 1
\end{array}\right)=1.
$$
Denoting by $C\equiv(c_{ij})$ the matrix of cofactors of $B$, we
have
$$
c_{ij}=(-1)^{i+j}\det B_{-(ij)}
$$
where $B_{-(ij)}$ is the matrix $B$ with the $i$-th row and $j$-th
column removed. Lastly,
$$
A^{-1}=B^{-1}D=\frac{1}{\det B}C^{T}D=C^{T}D
$$
and so it is just a matter of computing the form of $C$. I ran out
of time, but this hint should get you started.
A: If we call your matrix $A$, then
$$
A^{-1} = 
\begin{pmatrix}
\frac{1}{a_1} & 0 & 0 & 0 & 0 & \dots \\
-\frac{1}{a_1b^2} \sum_{k=2}^Na_kb^k & 1 & b & b^2 & b^3 & \dots \\
-\frac{1}{a_1b^3} \sum_{k=3}^Na_kb^k & 0 & 1 & b & b^2 & \dots \\
-\frac{1}{a_1b^4} \sum_{k=4}^Na_kb^k & 0 & 0 & 1 & b & \dots \\
\vdots & \vdots & & \ddots & \ddots & \\
-\frac{a_N}{a_1} & 0 & 0 & 0 & \dots & 1
\end{pmatrix},
$$
which you can verify by a direct calculation.
