If $\u = (U_1 \,\, U_2 \,\, \cdots U_n)$ is a vector of random variables, then $\Ex(\u) = \left(\Ex(U_1) \,\, \Ex(U_2) \,\, \cdots \Ex (U_n)\right)$. Meanwhile, $\Var \u$ is an $n \times n$ matrix with elements
\[\left(\Var(\u)\right)_{ij} = \Ex\left\{(U_i - \mu_i)(U_j - \mu_j)\right\}\]where $\mu_i=\Ex(U_i)$. The matrix $\Var(\u)$ is referred to as the variance-covariance matrix of $\u$.
With scalars, we know that $x^2$ is always positive (or at least, non-negative). The equivalent notion for matrices is that $\X \tp\X$ is said to be a positive definite matrix. More precisely, If $\X$ is full rank, then $\X \tp\X$ is positive definite; if $\X$ is singular (not of full rank), then $\X \tp\X$ is said to be positive semidefinite.
Letting $\A$ and $\B$ denote matrices of constants and $\x$ a random vector with mean $\bmu$ and variance $\bS$,
\[\begin{align*} \Ex(\A \tp\x) &= \A \tp\bmu \\ \Var(\A \tp\x) &= \A \tp\bS\A \\ \Ex(\x \tp\A\x) &= \bmu \tp\A\bmu + \trace(\A\bS) \\ \cov(\A \tp \x, \B \tp \x) &= \A \tp \bS \B \end{align*}\]Where the operator $\trace$ (defined for any square matrix) refers to the trace of a matrix, defined as the sum of its diagonal elements:
\[\trace(\A) = \sum_i A_{ii}\]Some basic facts about traces:
\[\begin{align*} \trace(\A\B) &= \trace(\B\A)\\ \trace(\A+\B) &= \trace(\A) + \trace(\B)\\ \trace(c\A) &= c\,\trace(\A) \end{align*}\]A further fact about traces that is not at all obvious but nonetheless useful is that if a matrix $\A$ is idempotent, then $\trace(\A)=\textrm{rank}(\A)$.
Finally, assuming that $\A \tp \x$ and $\B \tp \y$ are the same dimension, we have
\[\Var(\A \tp \x + \B \tp \y) = \A \tp \bS_A \A + \B \tp \bS_B \B + \A \tp \cov(\x, \y) \B + \B \tp \cov(\y, \x) \A.\]