How does this determinant calculation work?

Question

Given that $a_0, a_1,...,a_{n-1} \in \mathbb{C}$ I am trying to understand how the following calculation for the determinant of the following matrix follows: $$ \text{det} \begin{bmatrix} x & 0 & 0 & ... & 0 & a_0 \\ -1 & x & 0 & ... & 0 & a_1 \\ 0 & -1 & x & ... & 0 & a_2 \\ . \\ . \\ . \\ 0 & 0 & 0 & ... & -1 & x + a_{n-1} \\ \end{bmatrix} \\ = (x) \text{ det} \begin{bmatrix} x & 0 & 0 & ... & 0 & a_1 \\ -1 & x & 0 & ... & 0 & a_2 \\ 0 & -1 & x & ... & 0 & a_3 \\ . \\ . \\ . \\ 0 & 0 & 0 & ... & -1 & x + a_{n-1} \\ \end{bmatrix} + \text{det} \begin{bmatrix} 0 & 0 & ... & 0 & a_0 \\ -1 & x & ... & 0 & a_2 \\ . \\ . \\ . \\ 0 & 0 & ... & -1 & x + a_{n-1} \\ \end{bmatrix} \\ = x(x^{n-1} + a_{n-1}x^{n-2}+...+a_1) + (-1)^{n-1}\text{det} \begin{bmatrix} -1 & x & ... & 0 & a_2 \\ . \\ . \\ . \\ 0 & 0 & ... & -1 & x + a_{n-1} \\ 0 & 0 & ... & 0 & a_0 \\ \end{bmatrix} $$

I do not understand: (1) how the determinant can be broken up into the sum of the determinants of the 2 smaller matrices and (2) how are the determinants of the 2 smaller matrices what they are?

Answer 1

The determinant was expanded along the first column, so you get $$ \begin{bmatrix} \color{red}{x} & 0 & 0 & ... & 0 & a_0 \\ \color{blue}{-1} & x & 0 & ... & 0 & a_1 \\ 0 & -1 & x & ... & 0 & a_2 \\ . \\ . \\ . \\ 0 & 0 & 0 & ... & -1 & x + a_{n-1} \\ \end{bmatrix} \\ = (\color{red}{x}) \text{ det} \begin{bmatrix} x & 0 & 0 & ... & 0 & a_1 \\ -1 & x & 0 & ... & 0 & a_2 \\ 0 & -1 & x & ... & 0 & a_3 \\ . \\ . \\ . \\ 0 & 0 & 0 & ... & -1 & x + a_{n-1} \\ \end{bmatrix} - (\color{blue}{-1})^{1+2} \text{det} \begin{bmatrix} 0 & 0 & ... & 0 & a_0 \\ -1 & x & ... & 0 & a_2 \\ . \\ . \\ . \\ 0 & 0 & ... & -1 & x + a_{n-1} \\ \end{bmatrix} $$ After this just keep expanding along the first column.

Answer 2

If $A=(a_{i,j}) \in M_n(\mathbb{C})$, $$\det(A) = \sum_{k=1}^{n} (-1)^ka_{1,k} \det(\Delta_{1,k}) $$ Where $\Delta_{1,k}$ is A minus the column and the line of $a_{1,k}$

Here the sum only has 2 terms because others are equals to $0$.