eigenvectors of a matrix Good Day,
I have a matrix of
\begin{bmatrix}
28 & 10\\ 
10 & 19
\end{bmatrix}
I have found the eigenvalues...
first eigenvalue (v1) : 24 + 5sqrt5 = rd off to 35.18
second eigenvalue (v2) : 24 - 5sqrt5 = rd off to -12.81

Following the page above,
if eigenvalue = 35.18, -6.18x+10y = 0 and 10x-16.18y=0
if eigenvalue = 12.82, 16.18x+10y = 0 and 10x+6.18y=0
How should I proceed from here to get the Eigenvector?
With egreg's help, I am able to derive to [(sqrt{5}+1)/2   1] and [(sqrt5-1)/2   1].
There's another part of the question as shown below:

This is my working currently:

Is my current working correct (before getting matrix of abcd)?
I have expanded to the following...
$\frac{Yn}{Xn} = \frac{[[(\frac{\sqrt{5}+1}{2})(24+5\sqrt{5})^{n}+(\frac{\sqrt{5}-1}{2})(24+5\sqrt{5})^{n}]+90[(\frac{\sqrt{5}+1}{2})(\frac{1-\sqrt{5}}{2})(24+5\sqrt{5})^{n}+(\frac{\sqrt{5}-1}{2})(\frac{\sqrt{5}+1}{2})(24-5\sqrt{5})^{n}]]}{[(24+5\sqrt{5})^{n}+(24+5\sqrt{5})^{n}]+90[(\frac{1-5\sqrt{5}}{2})(24+5\sqrt{5})^{n}+(\frac{\sqrt{5}+1}{2})(24-5\sqrt{5})^{n}]}$
How can I simplify it?
Thanks.

You shouldn't round, if you can avoid it. Your matrix is
\begin{bmatrix}
29 & 10 \\
10 & 19
\end{bmatrix}
whose characteristic polynomial is
$$
X^2 - 48X + 451
$$
The roots are given by the formula
$$
\frac{48\pm\sqrt{48^2-4\cdot 451}}{2}
$$
so they are $24+5\sqrt{5}$ and $25-5\sqrt{5}$, so you computed correctly. An eigenvector relative to $25+5\sqrt{5}$ is a non zero solution of
$$
\begin{bmatrix}
29-(24+5\sqrt{5}) & 10 \\
10 & 19-(24+5\sqrt{5})
\end{bmatrix}
\begin{bmatrix}
x_1 \\
x_2
\end{bmatrix}=
\begin{bmatrix}
0\\
0
\end{bmatrix}
$$
However, since we know that only one equation is sufficient, because by definition of eigenvalue the matrix has rank less than $2$, we can simply solve
$$
(5-5\sqrt{5})x_1+10x_2=0
$$
or
$$
(1-\sqrt{5})x_1+2x_2=0
$$
We can obviously give any (non zero) value to $x_2$, so we set $x_2=1$ and so
$$
x_1=\frac{2}{\sqrt{5}-1}=\frac{2}{\sqrt{5}-1}\frac{\sqrt{5}+1}{\sqrt{5}+1}
=\frac{2(\sqrt{5}+1)}{5-1}=\frac{\sqrt{5}+1}{2}
$$
and an eigenvector is
\begin{bmatrix}
\frac{\sqrt{5}+1}{2}\\[2ex]
1
\end{bmatrix}
Similarly for the other eigenvector.
