# (Tridiagonal) Inverse of a matrix

Given this $$n \times n$$ matrix: $$A= \left(\begin{matrix}a_1&a_1&...&a_1\\a_1&a_2&...&a_2\\\vdots& &\ddots &\vdots\\a_1&a_2&...&a_n\end{matrix}\right)$$

How can I show that the inverse of this type of matrices are tridiagonal?

Help would be nice! Thank you!

I would prove explicit formulae. For $$n>1$$, $$(A^{-1})_{k,k+1}=(A^{-1})_{k+1,k}=\frac{1}{a_k-a_{k+1}}\qquad(1\leqslant k with $$a_0=0,a_{n+1}=\infty$$ in the latter one, so that $$(A^{-1})_{1,1}=\frac{a_2}{a_1(a_2-a_1)},\qquad(A^{-1})_{n,n}=\frac{1}{a_n-a_{n-1}}.$$