Need help calculating this determinant using induction This is the determinant of a matrix of ($n \times n$) that needs to be calculated:
\begin{pmatrix}
3 &2 &0 &0 &\cdots &0 &0 &0 &0\\
1 &3 &2 &0 &\cdots &0 &0 &0 &0\\
0 &1 &3 &2 &\cdots &0 &0 &0 &0\\
0 &0 &1 &3 &\cdots &0 &0 &0 &0\\
\vdots &\vdots &\vdots&\ddots &\ddots &\ddots&\vdots &\vdots&\vdots\\
0 &0 &0 &0 &\cdots &3 &2 &0 &0\\
0 &0 &0 &0 &\cdots &1 &3 &2 &0\\
0 &0 &0 &0 &\cdots &0 &1 &3 &2\\
0 &0 &0 &0 &\cdots &0 &0 &1 &3\\
\end{pmatrix}
The matrix follows the pattern as showed.
I have to calculate it using induction (we haven't learnt recursion so far).
Thanks

You have a tridiagonal matrix.
It's determinant can be written as a recurrence relation:
$$\det A \stackrel{\textrm{def}}{=} f_n = a_nf_{n-1} + c_{n-1}b_{n-1}f_{n-2}.$$
Define $f_0 = 1, f_{-1} = 0$.
However, your $a_n, b_n, c_n$ values are constant, so
$$ \det A = 3f_{n-1}+ 2 f_{n-2}.$$
Thus, you have
$$\begin{align*}
n = 1: & f_1 = 3f_0 + 2f_{-1} = 3 \\
n = 2: & f_2 = 3f_1 + 2f_0 = 3\cdot 3 + 2 = 11 \\
n = 3: & f_3 = 3f_2 + 2f_1 = 33 + 6 = 39 \\
\vdots & 
\end{align*}
$$
