Calculate the following determinant:

Question

I want to calculate the determinant of the following matrix $$\begin{pmatrix} \alpha +\beta&\alpha\beta&0&\cdots&0&0\\ 1&\alpha + \beta &\alpha \beta&\cdots&0& 0\\ 0&1&\alpha + \beta & \cdots &0&0\\ .&.&.&.&.&.\\ .&.&.&.&.&.\\ .&.&.&.&.&.\\ .&.&.&.&.&.\\ 0&0&0&0&1&\alpha + \beta\end{pmatrix}$$

Here is my attempt

Is this correct?

I used Laplace

Answer 1

Yes we have that

$$D_n=(\alpha+\beta)D_{n-1}-\alpha\beta D_{n-1}$$

and then we can proceed by recurrence with

• $D_1=\alpha+\beta$
• $D_2=(\alpha+\beta)^2-\alpha\beta$

Answer 2

The matrices satisfy the recurrence relation \begin{eqnarray*} \Delta_n=( \alpha+\beta) \Delta_{n-1}-\alpha \beta \Delta_{n-2}. \end{eqnarray*} It easy to show by induction that \begin{eqnarray*} \Delta_n=\sum_{i=0}^{n} \alpha^{i}\beta^{n-i}= \frac{\alpha^{n}- \beta^{n}}{\alpha- \beta}. \end{eqnarray*}