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Let $ab=1$. Find

$$\begin{vmatrix} c & a & a^2 & ... & a^{n-1} \\ b & c & a & \dots & a^{n-2} \\ b^2 & b & c& \dots &a^{n-3} \\ \vdots & \vdots & \vdots & \ddots & \vdots\\ b^{n-1} & b^{n-2} & b^{n-3} & \dots & c \end{vmatrix}$$

I tried to decompose by line however it does not lead to a solution

Rodrigo de Azevedo
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Vertum
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    Have you tried solving it for low dimensions at first? For example $n=3$? –  Aug 26 '21 at 13:17
  • Oh sure. I even know the answer, but I don’t know how to solve it in general. If $n=2$ we have $(c-1) (c+1)$. If $n=3$ we have $(c-1)^2 (c+2)$. If $n=4$ analogous $(c-1)^3 (c+3)$. So if $n=N$ $(c-1)^{N-1} (c+N-1)$ – Vertum Aug 26 '21 at 13:31
  • Do you know that the determinant is multilinear wrt. the rows/columns? –  Aug 26 '21 at 13:32
  • Yes, but I don’t know how it can help. – Vertum Aug 26 '21 at 13:38
  • You can take the answer or you can multiply the second column with $b$, the third with $b^2$ and so on (the approaches are equivalent). How does your matrix look now? Maybe you are able to get something like in this question: https://math.stackexchange.com/questions/2756296/how-do-i-proof-that-the-determinant-of-this-matrix-has-the-form-z-1n-1-z?noredirect=1&lq=1 –  Aug 26 '21 at 13:39

3 Answers3

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Hint. Consistently subtract from each previous row the next one multiplied by $a$, and then add to each subsequent column the previous one multiplied by $a$. The result will be a triangular matrix.

kabenyuk
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Let $M$ be the matrix at hand. Its entries have the form

$$M_{ij} = (c-1)\delta_{ij} + b^{i-j} = (c-1)\delta_{ij} + b^i a^j$$ where $\delta_{ij} = \begin{cases}1, & i = j\\ 0,& i \ne j\end{cases}$ is the Kronecker delta. We can express $M$ as a rank-1 update of the diagonal matrix $(c-1)I_n$:

$$M = (c-1)I_n + uv^T \quad\text{ where }\quad \begin{cases} u &= (b,b^2,\ldots,b^n)^T\\ v &= (a,a^2,\ldots,a^n)^T \end{cases} $$

When $c \ne 1$, by matrix determinant lemma, we have:

$$\det(M) = (c-1)^n \det\left(I_n + \frac{1}{c-1}uv^T\right) = (c-1)^n\left( 1 + \frac{1}{c-1}v^Tu\right)$$ Since $v^T u = \sum\limits_{k=1}^n a^k b^k = \sum\limits_{k=1}^n 1 = n$, we obtain $$\det(M) = (c-1)^n\left( 1 + \frac{n}{c-1}\right) = (c-1)^{n-1}(c + n - 1)\tag{*1}$$

When $c = 1$, $M = uv^T$ implies the columns of $M$ are linearly dependent. This forces $\det(M) = 0$ and formula $(*1)$ works at $c = 1$.

achille hui
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Let's add a third answer for fun. Since $ab=1$, we can write

$$ \det \begin{pmatrix} c&a&a^2&\cdots&a^{n-1} \\ b& c & a &\dots& a^{n-2}\\ b^2 & b & c &\cdots & a^{n-3}\\ \vdots &\vdots & \vdots & &\vdots\\ b^{n-1} & b^{n-2} &b^{n-3}& \cdots &c \end{pmatrix} =\frac{1}{b}\cdot\frac{1}{b^2}\cdots\frac{1}{b^{n-1}}\det \begin{pmatrix} c&1&1&\cdots&1\\ b &bc&b&\cdots &b\\ b^2 &b^2& b^2c&\cdots &b^2\\ \vdots&\vdots&\vdots& &\vdots\\ b^{n-1} & b^{n-1}& b^{n-1}&\cdots& b^{n-1}c \end{pmatrix}\\ =\det \begin{pmatrix} c & 1 & 1& \cdots & 1\\ 1&c&1&\cdots&1\\ 1&1&c&\cdots&1\\ \vdots &\vdots&\vdots & & \vdots \\ 1&1&1&\cdots&c \end{pmatrix}=(c-1)^{n-1}(c+n-1). $$ where you can use any of these answers: How do I proof that the determinant of this matrix has the form $(z-1)^{n-1} (z+(n-1))$?