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I'm given matrix N by N

$$\begin{pmatrix}a & b & 0 & 0 & 0 & ... & 0 \\\ c & a & b & 0 & 0 & ... & 0 \\\ 0 & c & a & b & 0 &... & 0 \\\ ... & ... \\\ 0 & 0 & 0 & ... & c & a & b \\\ 0 & 0 & 0 & 0 & ... & c & a \end{pmatrix}$$

I need to find such $a, b$ and $c$ so that the matrix is singular

What I tried: Singularity means determinant is zero. Let determinant of the matrix be equal to $D_n$ where $n$ is the size of the matrix, in this case I derived the relation, by definition of determinants:

$D_n = a D_{n-1} - bc D_{n-2}$ - a homogenous difference equation of second order. I solved it under the conditions that $D_0 = a$ and $D_1 = a^2 - bc$. Unfortunately the answer for closed form of $D_n$ turned out to be rather complicated which lead me to believe that my approach is probably faulty. Is there a possible different approach maybe? Thank you

Makina
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  • do you know the method to write a closed formula for the Fibonacci numbers? – dcolazin Apr 28 '19 at 17:53
  • Yes, this is exactly what I did for the equation I derived above, and my solution for $D_n$ turned out to be quite a monstrosity, which I couldn't simplify. If that is truly the answer (as in: "a, b and c are such that the equation is satisfied") then sure, but I thought that maybe there is a more elegent solution with a succinct answer – Makina Apr 28 '19 at 17:56

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