<s>[INST] product of two Jacobi matrices Let
$$
A = \begin{bmatrix}
a_1 & b_1 \\
b_1 & a_2 & b_2 \\
& b_2 & \ddots & \ddots \\
& & \ddots & \ddots & b_{n-1} \\
& & & b_{n-1} & a_n
\end{bmatrix},\ B = \begin{bmatrix}
1&0&0&\cdots\\
0&a_2 & b_2 \\
0&b_2& a_3 & \ddots & \\
\vdots&\ddots  & \ddots &  b_{n-1} \\
& & b_{n-1} & a_n
\end{bmatrix},
$$
where $b_i \neq 0$ and all entries are real. The following matrix is $A^{-1} \times B$
$$
C = \begin{bmatrix}
*&*&0&0&\cdots&0\\
*&*&0&0&\cdots&0 \\
*&*&1&0&\cdots&0 \\
*&*&0&1&\cdots&0 \\
\vdots&\vdots&&\ddots&\ddots&\\
*&*&0&0&\cdots&1 \\
\end{bmatrix},
$$
As you see an identity matrix appears in $C[2 \cdots n]$ and rows 1, 2 from the third column are zero. I need to prove that it always happen. I tried this using syms in Matlab and it verified this result. I would be appreciated for any help.
update
Fonseca 2001 in his paper Explicit inverses of some tridiagonal matrices gave the following relations for a symmetric tridiagonal matrix:
$$
A^{-1} = 
\begin{bmatrix}
u_1v_1&u_1v_2&u_1v_3&\cdots&u_1v_n\\
u_1v_2&u_2v_2&u_2v_2&\cdots&u_2v_2\\
u_1v_3&u_2v_3&u_3v_3&\cdots&u_3v_3\\
\vdots\\
u_1v_n&u_2v_n&u_3v_n&\cdots&u_nv_n\\
\end{bmatrix}
$$
such that
\begin{align}
v_1&= \dfrac{1}{d_1}\\
v_k&= -\dfrac{b_{k-1}}{d_k}v_{k-1}, k=2, \cdots,n\\
d_n&=a_n\\
d_i&=a_i - \dfrac{b_ic_i}{d_{i+1}}, i=n-1, \cdots, 1\\
u_n &= \dfrac{1}{\delta_nv_n}\\
u_k&=-\dfrac{b_k}{\delta_k}u_{k+1}, k=n-1, \cdots, 1\\
\delta_1&=a_1\\
\delta_i&=a_i-\dfrac{b_{i-1}c_{i-1}}{\delta_{i-1}}, i=2, \cdots, n-1
\end{align}
I tried to replace these variables in their position to get the appropriate results, but I failed.
I would like to know is there any block version of computing inverse of a matrix? because two blocks of $A, B$ are the same, it may help.
thanks in advance.
 [/INST] You have 
$$
B=A-D,
$$
where 
$$
D=\begin{bmatrix}
a_1-1&b_1&0&0&\cdots&0\\
b_1&0&0&0&\cdots&0 \\
0&0&0&0&\cdots&0 \\
0&0&0&0&\cdots&0 \\
\vdots&\vdots&&\ddots&\ddots&\\
0&0&0&0&\cdots&0 \\
\end{bmatrix}.
$$
Then 
$$
C=A^{-1}B=I-A^{-1}D.
$$
Since $D$ has nonzero entries only at the first two columns, we have
$$
A^{-1}D= \begin{bmatrix}
*&*&0&0&\cdots&0\\
*&*&0&0&\cdots&0 \\
*&*&0&0&\cdots&0 \\
*&*&0&0&\cdots&0 \\
\vdots&\vdots&&\ddots&\ddots&\\
*&*&0&0&\cdots&0 \\
\end{bmatrix},
$$
and so 
$$
C=\begin{bmatrix}
1&0&0&0&\cdots&0\\
0&1&0&0&\cdots&0 \\
0&0&1&0&\cdots&0 \\
0&0&0&1&\cdots&0 \\
\vdots&\vdots&&\ddots&\ddots&\\
0&0&0&0&\cdots&1 \\
\end{bmatrix}-
\begin{bmatrix}
*&*&0&0&\cdots&0\\
*&*&0&0&\cdots&0 \\
*&*&0&0&\cdots&0 \\
*&*&0&0&\cdots&0 \\
\vdots&\vdots&&\ddots&\ddots&\\
*&*&0&0&\cdots&0 \\
\end{bmatrix},
$$
which is of the desired form.
</s>