The inverse of a matrix (main diagonal $2$, left and right of it $-1$) I want to find inverse matrix of the following:$$A=\begin{bmatrix}2&-1&0&0&\cdots&0\\-1&2&-1&0&\cdots&0\\0&-1&2&-1&\cdots&0\\0&0&-1&2&\cdots&0\\\vdots&\vdots&\vdots&\vdots&\ddots&\vdots\\0&0&0&0&\cdots&2\end{bmatrix}$$
If possible, I want to do it using the $[A|E]\cong[E|A^{-1}]$ method (not that I know any other, except the formula with adjugated matrices which is too complicated here, also Cayley-Hamilton theorem doesn't seem useful because it seems hard to find the characteristic polynomial).
I've tried some elementary transformations but I can't even make $A$ to be $E$ matrix... Not to mention doing the same steps to get from $E$ to $A^{-1}$.
 A: If $A$ has dimensions $n \times n$, $B = A^{-1}$ has entries:
$$
B_{i,j} =
\begin{cases}
i - \frac{ij}{n+1} & \text{if } i \le j \\
j - \frac{ij}{n+1} & \text{if } j \le i
\end{cases}
$$
To verify this is the inverse, let $e_k$ be the $k$th unit vector.  Then
\begin{align*}
B A {e_k}
&= B (2e_k - e_{k+1} - e_{k-1}) \\
&= \sum_{i=1}^n (2 B_{i,k} - B_{i,k+1} - B_{i,k-1})e_i. \\
\end{align*}
If $i < k$,
\begin{align*}
2 B_{i,k} - B_{i,k+1} - B_{i,k-1}
&= 2 \left(i - \tfrac{ik}{n+1}\right)
- \left(i- \tfrac{i(k+1)}{n+1}\right)
- \left(i- \tfrac{i(k-1)}{n+1}\right)\\
&= 0.
\end{align*}
Similarly if $k < i$.  And if $i = k$, then
\begin{align*}
2 B_{k,k} - B_{k,k+1} - B_{k,k-1}
&= 2 \left(k - \tfrac{k^2}{n+1}\right)
- \left(k- \tfrac{k(k+1)}{n+1}\right)
- \left((k-1) - \tfrac{k(k-1)}{n+1}\right)\\
&= 1 + \frac{-2k^2 + (k^2 + k) + (k^2 - k)}{n+1} = 1.
\end{align*}
How to find the inverse
I used the link in Git Gud's comment.  Solving the recurrence there is pretty simple since the $b$ and $c$s and $a$s are all identical.
