So the question asks:
For $n \geq 2$, compute the determinant of the following matrix: $$ B = \begin{bmatrix} -X & 1 & 0 & \cdots & 0 & 0 \\ 0 & -X & 1 & \ddots & \vdots & \vdots \\ \vdots & \ddots & \ddots & \ddots & 0 & \vdots \\ \vdots & & \ddots & \ddots & 1 & 0 \\ 0 & \cdots & \cdots & 0 & -X & 1 \\ a_0 & a_1 & \cdots & \cdots & a_{n-2} & (a_{n-1} - X) \end{bmatrix} $$
Looking at the $2 \times 2$ and $3 \times 3$ forms of this matrix:
$\det \begin{bmatrix} -X & 0 \\ 0 & (a_1-X) \end{bmatrix} = -X(a_1-X) - 0 = X^2 - a_1X $
by expansion along the first row:
$\det \begin{bmatrix} -X & 1 & 0 \\ 0 & -X & 0 \\ 0 & 0 & (a_2-X) \end{bmatrix} = (-X) \times\det \begin{bmatrix} -X & 0 \\ 0 & a_2-X \end{bmatrix} - 1 \det\begin{bmatrix} 0 & 0 \\ 0 & a_2-X \end{bmatrix}$
$= (-X)[(-X)(a_2-X) -0] - 0 = X^3 - a_2X^2 $
So it looks like:
$\det \begin{bmatrix} -X & 1 & 0 & \cdots & 0 & 0 \\ 0 & -X & 1 & \ddots & \vdots & \vdots \\ \vdots & \ddots & \ddots & \ddots & 0 & \vdots \\ \vdots & & \ddots & \ddots & 1 & 0 \\ 0 & \cdots & \cdots & 0 & -X & 1 \\ a_0 & a_1 & \cdots & \cdots & a_{n-2} & (a_{n-1} - X) \end{bmatrix} = X^{n} - a_{n-1}X^{n-1} - a_{n-2}X^{n-2} ... - a_1X$
Does this look right? Is "prove by induction" valid to use here?