The probability density for a multidimensional normally distributed random vector $X$ is given as $f_X(s) := (2\pi)^{-\frac{n}{2}} det(\Sigma)^{-\frac{1}{2}}e^{-\frac{1}{2} (s-\mu)^T \Sigma^{-1}(s- \mu)}$.
Let $n = 2$, $X = (X_1, X_2)$, $\mu := \begin{pmatrix}1 \\ 2\end{pmatrix}$, $\Sigma := \begin{pmatrix}6.5 & -2.5 \\ -2.5 & 6.5 \end{pmatrix} = \begin{pmatrix}2.5 & -0.5 \\ -0.5 & 2.5 \end{pmatrix}\begin{pmatrix}2.5 & -0.5 \\ -0.5 & 2.5 \end{pmatrix}$.
I just don't understand how to write down $f_X$ explicitly for the given parameters above. Could someone help me with that?
Let $(s-\mu) = (s_1- 1 ~~~ s_2-2)$
Then by direct calculation:
$det(\Sigma)^{-\frac{1}{2}} = \frac{1}{6}$
$(s-\mu)^T \Sigma^{-1}(s- \mu) = \frac{13}{72}(s_1-1)^2 + \frac{13}{72}(s_2-2)^2 + \frac{10}{72}(s_1-1)(s_2-2)$
Finally $\displaystyle f(s) = \frac{1}{\sqrt{2 \pi}} \frac{1}{6} e^{-\frac{1}{2}\left(\frac{13}{72}(s_1-1)^2 + \frac{13}{72}(s_2-2)^2 + \frac{10}{72}(s_1-1)(s_2-2)\right)}$
Another option:
A simplified version for the bivariate normal is available. For example, see this, page 3.
Note that $\mu_X = 1$, $\mu_Y = 2$, $\sigma_X = \sqrt{\frac{13}{2}}$, $\sigma_Y = \sqrt{\frac{13}{2}}$ and $\rho = -\frac{5}{13}$
Substitute the values and you are done. You can check the direct calculations by the second method.