Let $X$ ~ $N\left(\begin{bmatrix}1\\-1\end{bmatrix}\right.$,$\left.\begin{bmatrix}9 & 1\\1 & 4\end{bmatrix}\right)$. Find distribution of $Y$ when $Y_1 = 2X_1 - X_2 + 4$, $Y_2 = 3X_1 + X_2 -3$.
I know if $X$ ~ $N(μ,Σ)$ and $Z=AX$ then $Z$ ~ $N(Aμ,AΣA^T)$, but I have no idea what to do with the $4$ and $-3$.
My idea to find $A\mu$ is to do it like $\begin{bmatrix}2 & -1\\3 & 1\end{bmatrix}$$\begin{bmatrix}1\\-1\end{bmatrix}$ + $\begin{bmatrix}4\\-3\end{bmatrix}$, but then I'm not sure how to find $A\Sigma A^T$. Can I just do it like $\begin{bmatrix}2 & -1\\3 & 1\end{bmatrix}$$\begin{bmatrix}9 & 1\\1 & 4\end{bmatrix}$$\begin{bmatrix}2 & 3\\-1 & 1\end{bmatrix}$ $+$ $\begin{bmatrix}4\\-3\end{bmatrix}$ ?
Let $A = \begin{bmatrix}2 & -1\\3 & 1\end{bmatrix}$. Then $Y = AX + \begin{bmatrix}4\\-3\end{bmatrix}$.
So $EY = A\mu + \begin{bmatrix}4\\-3\end{bmatrix}$ , as you have, but that is not $A\mu$.
Addition of a constant (vector) does not change the covariance, so $EY = \text{Cov}(AX) = A\text{Cov}(X)A^T$