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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!

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

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  • $\begingroup$ Thank you very much, that helps! $\endgroup$ – pprime May 15 at 7:44

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