# Recursive solution of determinants

1. Aug 9, 2008

### Chen

Hi,

I'm reading a paper where the determinant of the following matrix is solved for using some kind of recurisve method.

The matrix is given by $$M_{ij} = A \delta_{i,j} - B \delta_{i,j-1} - C \delta_{i,j+1}$$, with $$i,j = 1...N$$ and are NOT cyclic.

The author sets $$D_N = \texttt{det}$M_{(N)}$$$ and writes the equation

$$D_N = A D_{N-1} - B C D_{N-2}$$,

assumes a solution of the form $$D_N = \lambda ^N$$ and finds two solutions,

$$\lambda_{\pm} = (A \pm \sqrt{A^2 - 4 B C}) / 2$$.

He then notes the initial conditions of $$D_1 = A$$ and $$D_2 = A^2 - B C$$ and says that the answer is therefore

$$D_N = \frac{\lambda_{+}^{N+1} - \lambda_{-}^{N+1}}{\lambda_{+} - \lambda_{-}}$$.

It's the very last step I don't understand, how did he find $$D_N$$?

Thanks

2. Aug 9, 2008

### Chen

After a bit of search I see that the method to solve this kind of recurrence relations is to assume a solution of the form $$A \lambda_1^N + B \lambda_2^N$$ and find A and B from the initial conditions. However this is not exactly the form of the solution here... how come?

3. Aug 9, 2008

### HallsofIvy

Why isn't it?
$$D_N= \frac{\lambda_+}{\lambda_+-\lambda_-}\lambda_+^n+ \frac{-1}{\lambda_-(\lambda_+-\lambda_-)}\lambda_-^n$$
is exactly of the form you give.

4. Aug 9, 2008

### Chen

Yes that would be correct, a bad case of dullness. Thanks.