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.
Best Answer
You can notice that $A=-I + (1-b)N$ where $N$ is the standard nilpotent matrix, so
$$A^{-1} = - \left( I + (1-b)N + (1-b)^2 N^2 + \cdots + (1-b)^{n-1} N^{n-1} \right)$$
and the powers of $N$ are easily computed.