Inverse of an upper bidiagonal Toeplitz matrix I have a matrix with the following structure
$$\left[\begin{array}{cccccc|c}
    -1      & 1-b & 0 & \dots & 0 & 0 & b \\
    0       & -1 & 1-b & \dots & 0 & 0 & b \\
    \cdots \\
    0 & 0 & 0 & \dots &-1 & 1-b & b \\
    0       & 0 & 0 & \dots & 0 & -1 & 1 \\ \hline
    0 & 0 & 0 & \dots & 0 & 0 & -1
\end{array}\right]$$
I have to find the inverse of this matrix. I begin by using the block matrix inversion formula
$$\begin{bmatrix}\mathbf {A} &\mathbf {B} \\\mathbf {C} &\mathbf {D} \end{bmatrix}^{-1}=\begin{bmatrix}\left(\mathbf {A} -\mathbf {BD} ^{-1}\mathbf {C} \right)^{-1}&-\left(\mathbf {A} -\mathbf {BD} ^{-1}\mathbf {C} \right)^{-1}\mathbf {BD} ^{-1}\\-\mathbf {D} ^{-1}\mathbf {C} \left(\mathbf {A} -\mathbf {BD} ^{-1}\mathbf {C} \right)^{-1}&\quad \mathbf {D} ^{-1}+\mathbf {D} ^{-1}\mathbf {C} \left(\mathbf {A} -\mathbf {BD} ^{-1}\mathbf {C} \right)^{-1}\mathbf {BD} ^{-1}\end{bmatrix}$$
where
$$\mathbf{A} = \begin{bmatrix}
    -1       & 1-b & 0 & \dots & 0 & 0  \\
    0       & -1 & 1-b & \dots & 0 & 0  \\
    \cdots \\
    0 & 0 & 0 & \dots &-1 & 1-b  \\
    0       & 0 & 0 & \dots & 0 & -1 
\end{bmatrix}$$
$$\mathbf{B} = \begin{bmatrix} b \\ b \\ \vdots \\ b \\ 1 \end{bmatrix}$$
$$\mathbf{C} = \begin{bmatrix} 0 & 0 & 0 & \dots & 0 & 0 \end{bmatrix}$$
$$\mathbf{D} = \begin{bmatrix} -1 \end{bmatrix}$$
Given that $\mathbf{C}$ and $\mathbf{D}$ I can simplify the formula as
$$\begin{bmatrix}\mathbf {A} &\mathbf {B} \\\mathbf {C} &\mathbf {D} \end{bmatrix}^{-1}=\begin{bmatrix}\mathbf{A}^{-1} & \mathbf {A}^{-1} \mathbf {B}\\ \mathbf{0} & -1\end{bmatrix}$$
The only part left is to solve for $\mathbf{A}$ which I believe can be called an upper bidiagonal Toeplitz matrix. Unfortunately, I have not been able to find a formula to compute the inverse for the same.

Elements $m_{ij}$ of the inverse matrix $M=A^{-1}$ are:
\begin{align} 
m_{ij}
&=
\begin{cases}
-(1-b)^{j-i}, &j\ge i,
\\
0, &j<i.
\end{cases}
\end{align}
