For every $\lambda\in\mathbb R^2$, the scalar $\lambda^TX$ is normal with mean $\lambda^T\mu$ and variance $\lambda^TV\lambda$. The <moment-generating function of a normal distribution> therefore gives
$$
\boxed{M_X(\lambda)=\exp\left(\lambda^T\mu+\frac12\lambda^TV\lambda\right)}.
$$
Thus $\mu$ and $V$ determine the moment-generating function on all of $\mathbb R^2$. By the <uniqueness theorem for moment-generating functions>, they uniquely determine the <probability distribution> of $X$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / ia / Paper 2 / 11F / a / ii / Solution

If two random vectors have moment-generating functions that are finite and equal on a neighborhood of the origin, then they have the same <probability distribution>.

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 Incoming links: Uniqueness theorem for moment-generating functions