<s>[INST] Finding Determinants Recursively 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?
 [/INST] It should
$$\begin{pmatrix} D_2\\ D_1\end{pmatrix} = \begin{pmatrix} 7\\ 3\end{pmatrix}$$
and not
$$\begin{pmatrix} D_2\\ D_1\end{pmatrix} = \begin{pmatrix} 3\\ 7\end{pmatrix}$$
</s>