<s>[INST] Find the determinant of the following; 
Find the determinant of the following matrix, and for which value of $x$ is it invertible;
  $$\begin{pmatrix}
  x & 1 & 0 & 0 & 0 & \ldots & 0 & 0 \\
  0 & x & 1 & 0 & 0 & \ldots & 0 & 0 \\
  0 & 0 & x & 1 & 0 & \ldots & 0 & 0 \\
  \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\
  \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\
  \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\
  0 & 0 & 0 & 0 & 0 & \ldots & x & 1 \\
  1 & 0 & 0 & 0 & 0 & \ldots & 0 & x 
\end{pmatrix}$$
 

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??
 [/INST] 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...
</s>