Calculating determinant matrix with size of n

Question

we got the following matrix in order of $n$x$n$:

$$\begin{pmatrix} 1 & 0 & . & . & . & 0 & 1\\ 1 & 1 & 0 & . & . & . & 0\\ 0 & 1 & 1 & 0 & . & . & .\\ . & . & . & . & . & . & .\\ . & . & . & . & . & . & .\\ 0 & . & . & 0 & 1 & 1 & 0\\ 0 & . & . & . & 0 & 1 & 1\\ \end{pmatrix}$$

I need to calculate the determinant of the matrix. I started to do a row reduction as follows:

R2 = R2 - R1,
R3 = R3 - R2,
...,
Rn = Rn - Rn-1

and from what I have seen I have reached the following matrix:

$$\begin{pmatrix} 1 & 0 & . & . & . & 0 & 1\\ 0 & 1 & 0 & . & . & . & -1\\ 0 & 0 & 1 & 0 & . & . & 1\\ . & . & . & . & . & . & -1\\ . & . & . & . & . & . & 1\\ 0 & . & . & 0 & 0 & 1 & -1\\ 0 & . & . & . & 0 & 0 & 2\\ \end{pmatrix}$$

So I believe that when $n$ is odd then the determinant of $|A|$ is $|A|=2$ and when its double then its $|A|=0$

but I am not sure I solved it the right way.

Answer 1

Your matrix is the sum between an identity matrix and a circulant matrix, so the characteristic polynomial is given by: $$ p(\lambda)=(1-\lambda)^n-(-1)^n \tag{1}$$ and the determinant is given by $(-1)^n p(0)$, so it is $2$ if $n$ is odd and zero otherwise.