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Let $a, b>0$, and the matrix $A_{n \times n}$ and such $$A=\begin{bmatrix} a&b&0&\cdots&0&0\\ b&a&b&\cdots&0&0\\ 0&b&a&\cdots&0&0\\ \cdots&\cdots&\cdots&\cdots&\cdots&\cdots\\ \cdots&\cdots&\cdots&b&a&b\\ \cdots&\cdots&\cdots&\cdots&b&a \end{bmatrix}$$ Find the inverse $A^{-1}$.


My idea: $$A^{-1}=\dfrac{A^{*}}{|A|}$$ and let $|A|=D_{n}$, then we have $$D_{n}=aD_{n-1}-b^2D_{n-2}$$ so $$D_{n}=\begin{cases} (n+1)\left(\dfrac{a}{2}\right)^n,a^2=4b^2\\ \dfrac{(a+\sqrt{a^2-4b^2})^{n+1}-(a-\sqrt{a^2-4b^2})^{n+1}}{2^{n-1}\sqrt{a^2-4b^2}}&a^2\neq 4b^2 \end{cases}$$ and then I fell very ugly, do you have other methods? Thank you

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  • $\begingroup$ Is it a shifting matrix? If it is there might be some cool trick to do it. $\endgroup$ – Rgkpdx Mar 17 '14 at 12:37
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Notice that

$$A=\begin{bmatrix} a&b&0&\cdots&0&0\\ b&a&b&\cdots&0&0\\ 0&b&a&\cdots&0&0\\ \cdots&\cdots&\cdots&\cdots&\cdots&\cdots\\ \cdots&\cdots&\cdots&b&a&b\\ \cdots&\cdots&\cdots&\cdots&b&a \end{bmatrix} = a \begin{bmatrix} 1&b/a&0&\cdots&0&0\\ b/a&1&b/a&\cdots&0&0\\ 0&b/a&1&\cdots&0&0\\ \cdots&\cdots&\cdots&\cdots&\cdots&\cdots\\ \cdots&\cdots&\cdots&b/a&1&b/a\\ \cdots&\cdots&\cdots&\cdots&b/a&1 \end{bmatrix} =: aX.$$

So, $A^{-1} = \frac{1}{a} X^{-1}$.

To find $X^{-1}$, apply this answer. You can also find some references here.

In case you want to search further, your matrix is tridiagonal (a special kind of band) Toeplitz matrix.

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