I'd like to get the determinant $|P_n|$ with :
$$P_n=
\begin{bmatrix}
p+q & q & 0 & \dots & 0 & 0 \\
p & p+q & \ddots & \ddots & \vdots &\vdots \\
0 & p & \ddots & \ddots & 0 &\vdots\\
\vdots & 0 & \ddots & \ddots & q & 0\\
\vdots & \vdots & \ddots & \ddots & p+q & q\\
0 & 0 & \dots & 0 & p & p+q &
\end{bmatrix}
$$
With $p$ and $q$ real numbers that differ from each other and $n \geq 2 $.
As you can see $P_n$ is a classical tridiagonal matrix, therefore we can get easily the recurrence formula :
$|P_{n+2}|-(p+q)|P_{n+1}|+pq|P_n|=0$
From this formula we can get the equation : $r^2-(p+q)r+pq=0$ which give $\Delta = (p-q)^2$
So I get $r_1 = p$ and $r_2 = q$, therefore $|P_n| = c_1p^n + c_2q^n$.
As $|P_2| = (p+q)^2-pq = p^2 + pq + q^2$ I can get $c_1$ and $c_2$ but they do not work with $|P_3|= p^3+pq^2+qp^2+q^3$
I'm kind of stuck from this now, is there a mistake in my method ?
Best Answer
You would need $|P_1|$ and $|P_2|$ to solve for $c_1$, $c_2$. On solving, you get $c_1 = \frac{p} {p-q}$ and $c_2 = \frac{q} {q-p}$ and these will work for $|P_3|$.