Compute the determinant of order n + 1 (n ≥ 2) Compute the following determinant of order $n + 1 (n \geq 2)$
$$
\left[\begin{matrix}
a_0 & a_1 & a_2& \cdots & a_{n-1} & a_n \\
-x & x & 0  &\cdots  & 0 & 0\\
0 & -x & x &\cdots & 0 & 0 \\
\vdots & \vdots & \vdots & \ddots &\vdots &\vdots\\
0 & 0 & 0 & \cdots & x & 0\\
\\ 0 & 0& 0& \cdots &-x & x
\end{matrix}\right]
$$
I will be honest I got myself completely lost. I can recall facts of how you are finding determinants and how to perform cofactor expansion and that if $A$ is a triangular matrix that $$det(A) = a_{1,1}\cdot a_{2,2}\dots a_{n-1,n-1}\cdot a_{n,n}$$ and etc..
And I can't think of the way I can relate these facts what I know to solve this. Primarily because I don't know how to approach it.

I will be very thankful if you could help me solve this.


My question was answered. But I have something to add. This is my only second time asking questions in this format on platforms like stack exchange and omg I am so amazed. This is literally so cool that people just like I did can ask a question they are stuck on and get a reply in like 15 minutes??  I now this was around for a while, but I have just discovered how cool it actually is.
 A: Expansion with respect to the first row leads to
$$a_0\cdot\left[\begin{matrix}
 x & 0  &\cdots  & 0 & 0\\
-x & x &\cdots & 0 & 0 \\
\vdots & \vdots & \ddots &\vdots &\vdots\\
0 & 0 & \cdots & x & 0\\
\\ 0& 0& \cdots &-x & x
\end{matrix}\right]
-a_1\cdot\left[\begin{matrix}
-x & 0  &\cdots  & 0 & 0\\
0  & x &\cdots & 0 & 0 \\
\vdots & \vdots & \ddots &\vdots &\vdots\\
0 & 0 & \cdots & x & 0\\
\\ 0 & 0& \cdots &-x & x
\end{matrix}\right]+\dots+(-1)^na_n\cdot
\left[\begin{matrix}
-x & x & 0  &\cdots  & 0\\
0 & -x & x &\cdots & 0\\
\vdots & \vdots & \vdots & \ddots &\vdots \\
0 & 0 & 0 & \cdots & x \\
\\ 0 & 0& 0& \cdots &-x 
\end{matrix}\right]$$
which equals $$\sum\limits_{i=0}^n (-1)^ix^n(-1)^ ia_i=x^n\cdot\sum\limits_{i=0}^n a_i,$$ because these are special tridiagonal matrices, where $c_{m-1}d_{m-1}=0$ in this reccurence relation.
Determinant of each of them is equal to the product of diagonal elements.
The first $i$ diagonal terms are $-x,$ all other equal $x.$
A: There is also a simple inductive proof.
Let $A_n$ be the given matrix and replace $a_n$ by $t$ to form $A_n(t).$
Adding the last column of  $A_{n+1}$ to the previous one gives the matrix
$$\begin{bmatrix}
A_n(a_n+a_{n+1}) & \bf0\\
\bf0 & x
\end{bmatrix}$$
The rest is elementary.
A: Hint: Use cofactor expansion on the left column, which has only two nonzero entries. The cofactor of the top-left term is a diagonal matrix, and the cofactor of the other nonzero term looks a lot like an $n\times n$ matrix of the same form. Try to use this to derive the answer by induction on $n$.
