What is the rank of $B$?

Let $A=(a_{ij})$ be the square matrix of size $2018$ defined by

$$a_{ij} = \begin{cases} 2 & \text{if } i+1=j\\ \frac{1}{3} & \text{if } i =j+1\\ 0 & \text{otherwise}\end{cases}$$

Let $B$ be the leading principal minor of $A$ of order $1009$ (i.e., the submatrix of $A$ formed by the first $1009$ rows and columns).

1. What is the determinant of $A$?

2. What is the rank of $B$?

My solution

For question 1:

For an $n$ dimensional matrix of the form

$$A_n =\begin{pmatrix} 0 & u & 0 & & \cdots & 0 \\ l & 0 & u & & \cdots & \\ 0 & l & 0 \\ \vdots & & & \ddots & & \vdots \\ 0 & \cdots & & & 0 & u \\ 0 & \cdots & & & l & 0 \end{pmatrix}$$

The determinant is

$${\rm det} [A_n] = \begin{cases} 0 & n=\mbox{odd} \\ (-l u)^{\frac{n}{2}} & n=\mbox{even} \end{cases}$$

For case $n=2018$, $u=2$ and $l=\frac{1}{3}$

$${\rm det}[A_{2018}] = \left( -\frac{2}{3} \right)^{1009}.$$

I'm stuck at question 2. Any hints?

Supposed your result for such an $n\times n$ determinant is correct, question 2 can be easily answered:
We have $\det B=\det A_{1009}=0$ so $\mathrm{rank} (B) <1009$, but $B$ has an $1008\times1008$ minor $A_{1008}$ with nonzero determinant, so $\mathrm{rank} (B) =1008$.