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Find the determinant of the following matrix, and for which value of $x$ is it invertible; $$\begin{pmatrix} x & 1 & 0 & 0 & 0 & \ldots & 0 & 0 \\ 0 & x & 1 & 0 & 0 & \ldots & 0 & 0 \\ 0 & 0 & x & 1 & 0 & \ldots & 0 & 0 \\ \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\ \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\ \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\ 0 & 0 & 0 & 0 & 0 & \ldots & x & 1 \\ 1 & 0 & 0 & 0 & 0 & \ldots & 0 & x \end{pmatrix}$$
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Now I don't really know how to procees as I get find a suitable row operations that will simplify the process so I thought I would look at cases, maybe see a pattern.

$\mathbf{2 \times 2}$

$\begin{bmatrix}x & 1\\1 & x\end{bmatrix}$
This has determinant; $x^2-1$

$\mathbf{3 \times 3}$

$\begin{bmatrix}x & 1 & 0\\0 & x & 1\\1 & 0 & x\end{bmatrix}$
This has determinant $x^3+1$

So is that the pattern? determinant is $x^n-1$ if $n$ is even,
determinant is $x^n+1$ if $n$ is odd??

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Develop the determinant by the first column...

$$\begin{vmatrix}\color{red}x&1&0&0&\ldots&0\\ \color{red}0&x&1&0&\ldots&0\\ \color{red}\ldots&\ldots&\ldots&\ldots&\ldots&\ldots\\ \color{red}1&0&0&0&\ldots&x\end{vmatrix}=$$$${}$$

$$x\begin{vmatrix}x&1&0&0&\ldots&0\\ 0&x&1&0&\ldots&0\\ \ldots&\ldots&\ldots&\ldots&\ldots&\ldots\\ 0&0&0&0&\ldots&x\end{vmatrix}+(-1)^{n-1}\begin{vmatrix}1&0&0&0&\ldots&0\\ x&1&0&0&\ldots&0\\ \ldots&\ldots&\ldots&\ldots&\ldots&\ldots\\ 0&0&0&\ldots&x&1\end{vmatrix}$$

Now check that both determinants above are triangular matrices's, so piece of cake...

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I tried using Matrix Calculator to check the pattern you have mentioned. For any random value of x, it seems that your pattern is definitely correct.

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Calculate the determinant of your matrix using the last raw expansion, i.e, $$ \det(A)=(-1)^{n+1}+(-1)^{n+n}xx^{n-1}=x^n+(-1)^{n+1}. $$

This is exactly what you guess. If $n$ odd then $\det(A)=x^n+1$ and if $n$ even then $\det(A)=x^n-1$.

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For $x=0$, the matrix is the permutation matrix for the cycle $(1,2,\dots,n)$.

Therefore, $A^n=I$, the identity matrix, and the first columns of $A^k$, for $k=0,\dots, n-1$ are the vectors of the standard basis and clearly linearly independent.

Therefore, $\lambda^n-1$ is the minimal polynomial and because of its degree $n$ the characteristic polynomial $\chi_A$ of $A$.

Your determinant is just $(-1)^n \chi_A(-x)=x^n-(-1)^n$.

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It is easy to solve this problem with laplace's formula.
At first, you chose the first row for laplace's formula and get the following
(I draw $3\times 3$ matices intead of $(n-1)\times (n-1)$, because I don't know how to draw it clearly for arbitrary $n$):
$$x\cdot\begin{bmatrix}x & 1 & 0\\0 & x & 1\\0 & 0 & x\end{bmatrix}+(-1)^{n-1}\begin{bmatrix}1 & 0 & 0\\x & 1 & 0\\0 & x & 1\end{bmatrix}=x\cdot x^{n-1}+(-1)^{n-1}=x^n+(-1)^{n-1}$$

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