matrices - Computing determinant of this matrix

I have a very specific kind of matrix and I have to find the formula to find the determinant of these matrix.

a(i,j)=a if(i==j) and a(i,j)=0 if(floor(i/2)=floor(j/2) and i!=j) and n is odd

$$ A=(a_{i,j})_{n \times n}=\left(
\begin{array}{ccccc}
a&0&b&b \cdots &b&b\\
0& a&b&b \cdots& b&b\\
b& b& a&0 \cdots& b&b\\
b& b& 0&a \cdots& b&b\\
\vdots& \vdots& \vdots& \ddots&\cdots\\
b&b&b & \cdots&b&a

\end{array}
\right)? $$

Answer

The problem is closely related to this one posted and answered yesterday, with $p\leftarrow2$, $np\leftarrow n-1$, $c\leftarrow a/b$, with the entire matrix divided by $b$.

Since according to a comment under the question $n$ is odd, we need to deal with the last component separately. All eigenvectors in the linked question except for the one filled with $1$s sum to $0$. Thus we can append a $0$ to them to obtain eigenvectors of the present matrix. That leaves a two-dimensional subspace to be dealt with, spanned by the vector $x$ that has a $1$ in the last component and the vector $y$ that has $1$s everywhere else. Applying $A$ to these vectors yields $Ax=ax+by$ and $Ay=(n-1)bx+((n-3)b+a)y$. Thus the product of the remaining two eigenvalues is

$$
\left|\matrix{a&b\\(n-1)b&(n-3)b+a}\right|=((n-3)b+a)a-(n-1)b^2\;.
$$

Multiplying this by the $n-2$ eigenvalues from the linked question, with the above substitutions and the factor $b^{n-2}$ that was divided out, yields the determinant of the present matrix:

\begin{align}
\det A
&=
\left(((n-3)b+a)a-(n-1)b^2\right)b^{n-2}\left(\frac ab\right)^{(n-1)/2}\left(\frac ab-2\right)^{(n-3)/2}
\\
&=
\left(((n-3)b+a)a-(n-1)b^2\right)a^{(n-1)/2}\left(a-2b\right)^{(n-3)/2}\;.

\end{align}