Help solving this recurrence relation I wanted to resolve the determinant of the next (nxn) matrix via recurrence relations:

$$
\begin{vmatrix}
a &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp;.... 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0\\ 
1 &amp; a &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp;.... 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\ 
0 &amp; 1 &amp; a &amp; 1 &amp; 0 &amp; 0 &amp;.... 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0\\ 
0 &amp; 0 &amp; 1 &amp; a &amp; 1 &amp; 0 &amp;.... 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0\\ 
.. &amp; .. &amp; .. &amp; .. &amp; .. &amp; .. &amp;..... &amp; .. &amp; .. &amp; ..\\ 
0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; .... 0 &amp; 1 &amp; a &amp; 1 &amp; 0\\ 
0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; .... 0 &amp; 0 &amp; 1 &amp; a &amp; 1\\ 
0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; .... 0 &amp; 0 &amp; 0 &amp; 1 &amp; a\\ 
\notag
\end{vmatrix}
$$

After analyzing the matrix I found the recurrence relation:
$$
D_{n}-a*D_{n-1}  +D_{n-2}=0
$$

So the polynomial that describes this recurrence is:
$$
P(\lambda) = \lambda^2 - a * \lambda + 1
$$
The roots will be:
$$
\lambda_1 = \frac{a}{2} + \frac{\sqrt{a^2-4}}{2}\\
\lambda_2 = \frac{a}{2} - \frac{\sqrt{a^2-4}}{2}
$$
To resolve the recurrence I need 2 constants (C1 &amp; C2) that satisfy:
$$
D_n = C_1*(\frac{a}{2} + \frac{\sqrt{a^2-4}}{2})^n + C_2*(\frac{a}{2} - \frac{\sqrt{a^2-4}}{2})^n 
$$
With the initial conditions
$$
D_1 = a\\
D_2 = a^2 - 1
$$
The problem is I don't know how to resolve the equation system generated by substituting the initial conditions on the function.

Any type of help is appreciated.
You plug your initial conditions in, so you get
$$D_1=a = C_1*(\frac{a}{2} + \frac{\sqrt{a^2-4}}{2}) + C_2*(\frac{a}{2} - \frac{\sqrt{a^2-4}}{2})\\D_2=a^2-1 = C_1*(\frac{a}{2} + \frac{\sqrt{a^2-4}}{2})^2 + C_2*(\frac{a}{2} - \frac{\sqrt{a^2-4}}{2})^2$$
These are two equations in the two unknowns $C_1,C_2$.  The usual substitution technique will work, though it will be messy.  You can use the recurrence to evaluate $D_0=1=C_1+C_2$.  Using that with the first gives $$a=(C_1+C_2)\frac a2+(C_1-C_2)\frac{\sqrt{a^2-4}}{2}\\C_1-C_2=\frac a{\sqrt{a^2-4}}\\C_1=\frac 12(1+\frac a{\sqrt{a^2-4}})\\C_2=\frac 12(1-\frac a{\sqrt{a^2-4}})$$
To solve for the constants we just evaluate the expression for $D_n$ at $n=1$ and $n=2$, as well as use the information that $D_1=a$ and $D_2=a^2-1$ to get
$$C_1\left(\frac{a}{2}+\frac{\sqrt{a^2-4}}{2}\right)+C_2\left(\frac{a}{2}-\frac{\sqrt{a^2-4}}{2}\right)=a$$
$$C_1\left(\frac{a}{2}+\frac{\sqrt{a^2-4}}{2}\right)^2+C_2\left(\frac{a}{2}-\frac{\sqrt{a^2-4}}{2}\right)^2=a^2-1$$
Now, since $a$ is just some constant, we can treat this as a linear system and write it as
$$A\vec{c}=\left[\begin{array}{cc}
\left(\frac{a}{2}+\frac{\sqrt{a^2-4}}{2}\right) &amp; \left(\frac{a}{2}-\frac{\sqrt{a^2-4}}{2}\right) \\
\left(\frac{a}{2}+\frac{\sqrt{a^2-4}}{2}\right)^2 &amp; \left(\frac{a}{2}-\frac{\sqrt{a^2-4}}{2}\right)^2
\end{array}\right]
\left[\begin{array}{c}
C_1 \\
C_2
\end{array}\right]=
\left[\begin{array}{c}
a \\
a^2-1
\end{array}\right]$$
Which has solution
$$
\left[\begin{array}{c}
C_1 \\
C_2
\end{array}\right]
=
\frac{\text{adj }A}{\det A}\left[\begin{array}{c}
a \\
a^2-1
\end{array}\right]
$$
Where
$$\det A=\left(\frac{a}{2}+\frac{\sqrt{a^2-4}}{2}\right)\left(\frac{a}{2}-\frac{\sqrt{a^2-4}}{2}\right)^2-\left(\frac{a}{2}-\frac{\sqrt{a^2-4}}{2}\right)\left(\frac{a}{2}+\frac{\sqrt{a^2-4}}{2}\right)^2$$
$$=\frac{2}{a-\sqrt{a^2-4}-2}$$
and
$$\text{adj }A=\left[\begin{array}{cc}
\left(\frac{a}{2}-\frac{\sqrt{a^2-4}}{2}\right)^2 &amp; -\left(\frac{a}{2}-\frac{\sqrt{a^2-4}}{2}\right) \\
-\left(\frac{a}{2}+\frac{\sqrt{a^2-4}}{2}\right)^2 &amp; \left(\frac{a}{2}+\frac{\sqrt{a^2-4}}{2}\right)
\end{array}\right]$$