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I need help finding an explicit formula for the determinant of a $10\times10$ matrix below

$$A=\begin{bmatrix} a & 1 & & & & \vdots\\ 1 & a & 1 & & & \vdots\\ & 1 & a & 1 & & \vdots\\ & & 1 & a & 1 & \vdots\\ & & & 1 & \ddots & 1\\ \cdots & \cdots & \cdots & \cdots & 2 &a \end{bmatrix}$$


My attempt:

I observed that

$$A=\underbrace{\begin{bmatrix} a & 1 & & & & \vdots\\ 1 & a & 1 & & & \vdots\\ & 1 & a & 1 & & \vdots\\ & & 1 & a & 1 & \vdots\\ & & & 1 & \ddots & 1\\ \cdots & \cdots & \cdots & \cdots & 1 &a \end{bmatrix}}_{B}+\underbrace{\begin{bmatrix} 0 & 0 & & & & \vdots\\ 0 & 0 & 0 & & & \vdots\\ & 0 & 0 & 0 & & \vdots\\ & & 0 & 0 & 0 & \vdots\\ & & & 0 & \ddots & 0\\ \cdots & \cdots & \cdots & \cdots & 1 & 0 \end{bmatrix}}_{C}$$

The formula for $\det B$ is already known in this question and $\det C=0$.

I can use the formula from this answer:

$$\det A=\det B\left(1+\sum_{k=1}^{n-1}C_{i_{k+1}}{}^{[i_{k+1}}\cdots C_{i_n}{}^{i_n]}\right)$$

Now my problem is that I am not sure how to unpack the weird symbols in the sum. I am not entirely familiar with tensor language used in the linked answer. So what is the explicit formula for the summation?

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    $\begingroup$ I prefer just use cofactor extension through last row, then I believe you can get fomula with $\det B_9$ and $\det B_8$. $\endgroup$ Commented Mar 21 at 2:45

1 Answer 1

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Thanks to the hint provided by @Myungheon Lee, using Laplace expansion,

$$\det A=2\det B_8-a\det B_9.$$

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    $\begingroup$ What are $B_8$ and $B_9$? $\endgroup$ Commented Mar 21 at 7:13

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