Evaluate determinant of an $n \times n$ matrix, help

Question

I need help with this problem:

$D_{n}= \begin{vmatrix} 1 & 1 & 0 & \cdots & 0 & 0 & 0 \\ 1 & 1 & 1 & \cdots & 0 & 0 & 0 \\ 0 & 1 & 1 & \cdots & 0 & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & 1& 1 & 1 \\ 0 & 0 & 0 & \cdots & 0& 1 & 1 \end{vmatrix}$

I can't get it to squared form. It is obvious that fourth row is needed. How to find fourth row, or is there other methods?

Thanks for replies.

Answer 1

It is a special case of a tridiagonal matrix. If you denote this determinant $D_n$, developing it along the last row, you can prove the recurrence relation: $$D_n=D_{n-1}-D_{n-2}$$ The initialisation is $D_1=1$, $D_2=0$. You can easily establish that

$$\begin{cases} D_n=1&\text{if}\enspace n\equiv 0,1 \pmod6\\ D_n=0&\text{if}\enspace n\equiv 2,5 \\ D_n=-1&\text{if}\enspace n\equiv 3,4 \end{cases}$$

Answer 2

$$ \begin{vmatrix} 1&1&0&0\\ 1&1&1&0\\ 0&1&1&1\\ 0&0&1&1\\ \end{vmatrix} $$ would be an example for your problem, using gaussian elimination it is easy to get it to the unit matrix, which has 1 as determinant.