The factor-solve method For what values of $a_1,a_2,...,a_n$ is the $n\times n$ matrix nonsingular.
$A=\begin{bmatrix} 
a_1 & 1 & 0 &.&.&.& 0 & 0 \\
a_2 & 0 & 1 &.&.&.& 0 & 0 \\
. & . & . &.&.&.& . & . \\
. & . & . &.&.&.& . & . \\
. & . & . &.&.&.& . & . \\
a_{n-2} & 0 & 0 &.&.&.& 1 & 0 \\
a_{n-1} & 0 & 0 &.&.&.& 0 & 1 \\
a_{n} & 0 & 0 &.&.&.& 0 & 0 \\
\end{bmatrix}$
 A: Pf 1:
$$\det(A) = (-1)^{n-1}a_{n} \det(I_{n \times 1}) = (-1)^{n-1}a_{n} \ne 0 \iff a_n \ne 0$$
QED

Pf 2:
If you move the first column all the way to the right, you would have an augmented matrix for the linear system
$$Bx=a \tag{*}$$
where
$$a = [a_1, \cdots, a_n]^T \in \mathbb R^{n, 1}$$
$$x = [x_1, \cdots, x_{n-1}]^T \in \mathbb R^{n \times 1,1}$$
and $B \in \mathbb R^{n, n \times 1}$ is $I_{n \times 1}$ with an extra row of zeroes below it ie $$[a | B] = A$$
Now observe that the last row of $(*)$ is:
$$0x_1 + 0x_2 + \cdots 0x_{n-1} = a_{n}$$
This is consistent iff $$a_n = 0$$
Now observe that


*

*$$[\text{x satisfies} \ (*) \iff \ \text{x satisfies} \ (I_{n \times 1}x = [a_1, \cdots, a_{n-1}]^T]) \iff [a_n = 0]$$

*$$[(*) \ \text{has a unique solution} \ \iff \ (I_{n \times 1}x = [a_1, \cdots, a_{n-1}]^T \ \text{has a unique solution})]\iff [a_n = 0]$$

*$$[B \ \text{has full rank} \iff I_{n \times 1} \text{has full rank}] \iff [a_n = 0]$$

*$$B \ \text{has full rank} \iff A \ \text{has full rank}$$
Thus
$$[A \ \text{has full rank} \iff I_{n \times 1} \text{has full rank}] \iff [a_n = 0]$$
$$\to [A \ \text{has full rank}] \iff [a_n = 0] \tag{1}$$
$$\to [\det(A) \ne 0] \iff [a_n = 0] \tag{2}$$
where
$(1)$ is true because $I_{n \times 1}$ is invertible and hence $\text{has full rank}$
$(2)$ is true because $A$ is square. Note that $det(B)$ is meaningless.
QED
