Calculating Eigenvalues is only Assume that the following is used:
$$ 
A = \begin{pmatrix}
 0&  1&\\
 2&  3&\\ 
 4&  5&\\
 6&  7&\\
 8&  9& 
\end{pmatrix}
$$
Then calculating the Coveriance matrix, which, gives me:
$$ 
C = \begin{pmatrix}
 40&  40&\\
 40& 40&\\ 
\end{pmatrix}
$$
Then using the following:
$$
det = (a+b) \cdot (a+b)-4 \cdot(a \cdot b - c \cdot c),
$$
where in this case, $a = 40, b = 40, c = 40$ gives the answer:
$$
\lambda_{1} = 80, \\
\lambda_{2} = 0,
$$
These are therefore the correct Eigen values. However, using this formula, if I have the following:
$$ 
A = \begin{pmatrix}
 -4&  -2&\\
 -1&  -3&\\ 
 4&  5&\\
 6&  7&\\
 8&  9& 
\end{pmatrix},
$$
where the Covariance matrix is given: 
$$
C = \begin{pmatrix}
 99.2&  103.4&\\
 103.4& 116.8&\\ 
\end{pmatrix},
$$
gives the Eigenvalues as: 
$$
\lambda_{1} = 218.119 \\
\lambda_{2} = -15.5189
$$
When the actual values are:
$$
\lambda_{1} =211.774 \\
\lambda_{2} = 4.226
$$
Could anyone tell me where I am calculating this wrong please?
EDIT:
For
$\lambda_{1} = (a + b + det)/2 \\
 \lambda_{2} = (a + b - det)/2
$

Have you double-checked your calculations? The eigenvalues of any $2\times2$ real symmetric matrix $\pmatrix{a&c\\ c&b}$ are given by
$$\frac{a+b\pm\sqrt{(a+b)^2-4(ab-c^2)}}2.$$
Plug in the entries of $C$, I don't find any discrepancies.
