Calculating determinant Dn

Question

Do i have to calculate this determinant Dn with recurrence formulas because i have difficulties with first and last row having 1 and 2, so i can't have Dn-1!

\begin{pmatrix} 1 & 2 & 0 & . & . & . & 0 & 0 & 0\\ 5 & 8 & 3 & . & . & . & 0 & 0 & 0\\ 0 & 5 & 8 & . & . & . & 0 & 0 & 0\\ . & . & . & . & . & . & . & . & .\\ 0 & 0 & 0 & . & . & . & 8 & 3 & 0\\ 0 & 0 & 0 & . & . & . & 5 & 8 & 3\\ 0 & 0 & 0 & . & . & . & 0 & 2 & 1\\ \end{pmatrix}

Answer 1

Let $E_n$ be the determinant of the $n \times n$ matrix with 5s, 8s and 3s along the three central diagonals. Multiplying out this determinant using the first and last rows gives $$E_{n+2}=8E_{n+1}-15E_n$$ and we obtain the formula $$E_n=\frac {1}{2} (5^{n+1}-3^{n+1}).$$

Let $D_n$ be the determinant of the given matrix, where $n\ge 3$. Multiplying out this determinant using the first and last rows gives $$E_{n-2}-16E_{n-3}+60E_{n-4}=\frac {1}{2} (5^{n-2}-7\times 3^{n-2}).$$

Answer 2

Result to answer query

$$\begin{pmatrix} 1 & 2 & 0\\ 5 & 8 & 3 \\ 0 & 2 & 1\\ \end{pmatrix}$$ has determinant $1\times (8\times 1-3 \times 2)-2\times (5\times 1-3 \times 0)=-8.$