Inverse of the Toeplitz matrix I am working on the inverse of the sum of an identity matrix and a Toepltz matrix, and trying to find the formula for the (1,1) element of the inverse. For example, Assume $c$ is a nonzero constant, and let $$A_{t}=cI_{t}+\Omega_{t},$$ where, for $t=4$, 
$$
\Omega _{t}=\left(
\begin{array}{cccc}
1 & \rho  & \rho ^{2} & \rho ^{3} \\
\rho  & 1 & \rho  & \rho ^{2} \\
\rho ^{2} & \rho  & 1 & \rho  \\
\rho ^{3} & \rho ^{2} & \rho  & 1%
\end{array}%
\right), \quad
\text{with} \quad \left\vert \rho \right\vert <1.
$$
In general, the $(i,j)$-th element of $\Omega _{t}$ is given by $\rho ^{|i-j|}$.
Is there any way to find a general formula for the $(1,1)$-th element of $A_{t}^{-1}$ for different $t,$ say, $t=2,3,4,\ldots ?$ 
Many thanks!
 A: By Cramer's rule, $U_t=A_t^{-1}(1,1)$ is given by the ratios of two determinants. Let:
$$A_3=\left(\begin{array}{ccc}c+1&\rho&\rho^2 \\ \rho&c+1&\rho\\ \rho^2&\rho&c+1\end{array}\right),\qquad B_3=\left(\begin{array}{ccc}1&\rho&\rho^2 \\ \rho&c+1&\rho\\ \rho^2&\rho&c+1\end{array}\right)$$
and define $A_t,B_t$ in a similar fashion. By replacing the first column with the first column minus $\rho$ times the second column, then expanding along the first column, we get:
$$ \det A_{t+1} = (c+1-\rho^2) \det A_t + c\rho^2 \det B_t, $$
$$ \det B_{t+1} = (1-\rho^2) \det A_t + c\rho^2 \det B_t, $$
and by eliminating $B_t$,
$$ \det A_{t+1} = (c+1+c\rho^2-\rho^2) \det A_t -c^2\rho^2\det A_{t-1} $$
then dividing the whole line by $\det A_t$,
$$ \frac{1}{U_{t+1}} = (c+1+c\rho^2-\rho^2)-c^2\rho^2 U_{t} $$
or:

$$ U_{t+1} =\frac{1}{(c+1+c\rho^2-\rho^2)-(c^2\rho^2)U_t}$$

where obviously $U_1=\frac{1}{c+1}$. If we set $U_t=\frac{V_t}{c\rho},$
$$ V_{t+1} =\frac{1}{\frac{c+1+c\rho^2-\rho^2}{c\rho}-V_t},$$
so by setting $\beta=\frac{c+1+c\rho^2-\rho^2}{c\rho}$ and assuming $\beta \geq 2$ we have:
$$ \lim_{t\to +\infty}U_t = \frac{-\beta+\sqrt{\beta^2-4}}{2c\rho}$$
by the usual convergence properties of the negative continued fractions.
A: This is not an answer, but just to demonstrate what we're dealing with here, I've input WolframAlpha's answer for the inverses in the links below.
Click for t=2.
Click for t=3.
Click for t=4.
