From the MIT OCW Linear Algebra (18.06) final exam, question 9:
For square matrices with 3's on the diagonal, 2s on the diagonal above, and 1s on the diagonal below:
$$A_1=\begin{pmatrix} 3 \end{pmatrix}, A_2=\begin{pmatrix} 3 & 2 \\ 1 & 3\\ \end{pmatrix}, A_3=\begin{pmatrix} 3&2&0\\ 1&3&2\\ 0&1&3\\ \end{pmatrix},... $$
The determinant of $A_n$, $D_n$, can be defined recursively as: $$D_n=aD_{n-1}+bD_{n-2}$$
Find $D_5$ using eigenvalues.
I found $D_n=3D_{n-1}-2D_{n-2}$ and used $$\begin{pmatrix} D_n\\ D_{n-1}\\ \end{pmatrix} = \begin{pmatrix} a&b\\ 1&0\\ \end{pmatrix} \begin{pmatrix} D_{n-1}\\ D_{n-2}\\ \end{pmatrix} $$ and $\begin{pmatrix} D_2\\ D_1\\ \end{pmatrix} =\begin{pmatrix} 7\\ 3\\ \end{pmatrix}$
to solve:
$$\begin{pmatrix} D_n\\ D_{n-1}\\ \end{pmatrix} =(-1)1^n\begin{pmatrix}1\\1\end{pmatrix} + 2^n\begin{pmatrix}2\\1\end{pmatrix} $$
Plugging in $n=5$ gives $D-5=63$, but the answer key says $D_5=207$.
What am I doing wrong?
It should $$\begin{pmatrix} D_2\\ D_1\end{pmatrix} = \begin{pmatrix} 7\\ 3\end{pmatrix}$$ and not $$\begin{pmatrix} D_2\\ D_1\end{pmatrix} = \begin{pmatrix} 3\\ 7\end{pmatrix}$$