Variance and covariance matrix

Question

Let $(X,Y)$ ~ $N\left(\begin{bmatrix}2\\3\end{bmatrix}\right.$,$\left.\begin{bmatrix}7 & 2\\2 & 3\end{bmatrix}\right)$ . Find distribution of $T = 3X + 4Y$.

I know if $X$ ~ $N(\mu, \Sigma)$ and $Z = AX$ then $Z$ ~ $N(A\mu, A\Sigma A^T)$

I think that the $A$ is something like $\begin{bmatrix}3 & 4\end{bmatrix}$, then the $\mu = \begin{bmatrix}3 & 4\end{bmatrix}$ $\begin{bmatrix}2\\3\end{bmatrix} = \begin{bmatrix}18\end{bmatrix}$

but I got a problem to find $\Sigma$ for $T$.

Answer 1

Your thinking about finding $\mu_T$ is okay.

Finding "covariance matrix" $\Sigma$ is a matter of calculating:

$\left[\begin{array}{cc} 3 & 4\end{array}\right]$$\left[\begin{array}{cc} 7 & 2\\ 2 & 3 \end{array}\right]$$\left[\begin{array}{c} 3\\ 4 \end{array}\right]$

resulting in a $1\times 1$-matrix. Its unique entry is the variance of $T$.