Read off $\sigma_1^2=4$, $\sigma_2^2=1$, $\rho=0.8$ from the bivariate normal density

If the joint pdf of $(X,Y)$ is $f(x,y)=\frac1{2.4\pi}\exp\bigl\{-\frac1{0.72}\bigl[\frac{x^2}4-1.60\frac{xy}2+y^2\bigr]\bigr\}$ (the booklet prints the middle term as "$1.60\,\frac{X^y}2$"), then the variance covariance matrix $\Sigma$ is: