Looking at the $2 \times 2$ and $3 \times 3$ forms of this matrix, we see that:
$\det \begin{bmatrix} t & -a_0 \\ -1 & t-a_1 \end{bmatrix} = t(t-a_1) - a_0 = t^2 - a_1t - a_0$
and, by expansion along the first row:
$\det \begin{bmatrix} t & 0 & -a_0 \\ -1 & t & -a_1 \\ 0 & -1 & t-a_2 \end{bmatrix} = t \times\det \begin{bmatrix} t & -a_1 \\ -1 & t-a_2 \end{bmatrix} + (-a_0) \det\begin{bmatrix} -1 & t \\ 0 & -1 \end{bmatrix}$
$= t[t(t-a_2) - a_1] - a_0 = t^3 - a_2t^2 - a_1t - a_0 $
So it looks like:
$\det \begin{bmatrix} t & 0 & 0 & \cdots & 0 & -a_0 \\ -1 & t & 0 & \cdots & 0 & -a_1 \\ 0 & -1 & t & \cdots & 0 & -a_2 \\ 0 & 0 & -1 & \cdots & 0 & -a_3 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & t & -a_{n-1} \\ 0 & 0 & 0 & \cdots & -1 & t-a_n \end{bmatrix} = t^{n+1} - a_nt^n - a_{n-1}t^{n-1} - ... - a_2t^2 - a_1t - a_0$
Which we can prove by induction.
Assume that:
$\det \begin{bmatrix} t & 0 & 0 & \cdots & 0 & -a_0 \\ -1 & t & 0 & \cdots & 0 & -a_1 \\ 0 & -1 & t & \cdots & 0 & -a_2 \\ 0 & 0 & -1 & \cdots & 0 & -a_3 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & t & -a_{n-2} \\ 0 & 0 & 0 & \cdots & -1 & t-a_{n-1} \end{bmatrix} = t^{n} - a_{n-1}t^{n-1} - a_{n-2}t^{n-2} - ... - a_2t^2 - a_1t - a_0$
Then, by expansion along the first row:
$\det \begin{bmatrix} t & 0 & 0 & \cdots & 0 & -a_0 \\ -1 & t & 0 & \cdots & 0 & -a_1 \\ 0 & -1 & t & \cdots & 0 & -a_2 \\ 0 & 0 & -1 & \cdots & 0 & -a_3 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & t & -a_{n-1} \\ 0 & 0 & 0 & \cdots & -1 & t-a_n \end{bmatrix} = t \det \begin{bmatrix} t & 0 & 0 & \cdots & 0 & -a_1 \\ -1 & t & 0 & \cdots & 0 & -a_2 \\ 0 & -1 & t & \cdots & 0 & -a_3 \\ 0 & 0 & -1 & \cdots & 0 & -a_4 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & t & -a_{n-1} \\ 0 & 0 & 0 & \cdots & -1 & t-a_n \end{bmatrix} $
$+ (-1)^{n+1} \times (-a_0)(-1)^n $
$ = t[t^{n} - a_{n}t^{n-1} - a_{n-1}t^{n-2} - ... - a_3t^2 - a_2t - a_1] + (-1)^{2n+1} a_0$
$ = t^{n+1} - a_nt^n - a_{n-1}t^{n-1} - ... - a_2t^2 - a_1t - a_0$
Proof complete.
Your argument is correct.
With a little effort (but not much), you should be able to modify your argument to make it more geometric.
Here is the the two dimensional version:
If we have a parallelogram spanned by the vectors $\vec{a}$, $\vec{b}$,
then the area of
the parallelogram is the same as the area of the rectangle spanned by $\vec{a}, \vec{b}',$
where $\vec{b}'$ is the component of $\vec{b}$ that is orthogonal to $\vec{a}$,
i.e. $$\vec{b}' = \vec{b} - \dfrac{(\vec{a} \cdot \vec{b})}{(\vec{a} \cdot \vec{a})} \vec{a}.$$
Now the process of going from the matrix $\Bigl( \, \vec{a} \quad \vec{b} \, \Bigr) $ to the matrix $\Bigl( \, \vec{a} \quad \vec{b}' \, \Bigr)$ involves
right-multiplying by the matrix $\begin{pmatrix} 1 & - (\vec{a} \cdot \vec{b})/
(\vec{a}\cdot\vec{a}) \\ 0 & 1\end{pmatrix}$, whose determinant is $1$.
Thus we see that $\Big( \, \vec{a} \quad \vec{b} \, \Bigr)$
and $\Bigl( \, \vec{a} \quad \vec{b}' \, \Bigr)$ have the same
determinant, and also describe parallelograms with the same area,
the latter being a rectangle.
This reduces you to checking the relationship between area and determinant in the case of a rectangle. Rotating this rectangle, you can make its edges parallel to the coordinate axes. Again, a rotation matrix has determinant one, so you are reduced
to checking the relationship between determinants and areas in the case
of a rectangle whose sides are parallel to the coordinate axes. this
is pretty obvious, and so we are done.
In the three dimensional case, you can argue similarly: you first of all reduce to the case where one face is a rectangle, and you then reduce to the case when the third side is perpendicular to the rectangular face, so that the whole thing
is a cuboid. These steps involve right multiplying by matrices
which are upper triangular with $1$'s down the diagonal, which thus have det $= 1$.
Now applying a bunch of rotations (again, each has det $= 1$), you can make
your cuboid have sides parallel to the coordinate axes, at which point the
formula is again pretty obvious.
Note that if we allow ourselves one more step --- namely, multiplying by a diagonal matrix (which, geometrically, is a rescaling of each of the coordinate axes) --- then we can start with any (non-degenerate) parallelepiped and convert it to
the standard cuboid with unit length sides sitting at the origin.
In linear algebra terms, this can be restated as the fact that any matrix
with non-zero determinant can be written as the product of a diagonal matrix,
some rotation matrices, and an upper triangular matrix with $1$'s down the diagonal.
Combining the diagonal and upper triangular matrix with $1$'s down the diagonal,
we obtain an upper triangular matrix with non-zero entries down the diagonal.
In other words, any matrix with non-zero determinant can be written as a product
of some rotation matrices with an upper triangular matrix.
This is usually called the $QR$ decomposition in linear algebra textbooks;
in more theoretical treatments it is called the Iwasawa decomposition.
So what I have just given is a geometric description of the $QR$ decomposition.
The difference between what I've described and your argument is that you use
elementary matrices, while I use just upper triangular matrices with $1$'s down
the diagonal. These arise from the geometric process of projecting one vector
to make it perpendicular to another, which is where the geometric perspective is
coming from.
Best Answer
In this particular case the middle column is equal to the average of the first and 3rd, which automatically means, that there is a column of zero after a couple of equivalent transformations, which makes it zero.