Two-dimensional distribution of random variable I have random variable $X \sim N\left( \begin{bmatrix} 1 \\ -1 \end{bmatrix}, \begin{bmatrix} 9 & 1 \\ 1 & 4 \end{bmatrix} \right)$. And I need to get distribution of random variable $Y$. Where 
$$Y_1 = 2X_1 - X_2 + 4$$
and 
$$Y_2 = 3X_1 + X_2 - 3$$
I used to do this kind of calculations with one-dimensional distribution. But I've never seen a matrix one before. Couldn't find anything on the internet too. Could you help me with that?
 A: $$
\begin{bmatrix} X_1 \\ X_2 \end{bmatrix} \sim N\left( \begin{bmatrix} 1 \\ -1 \end{bmatrix}, \begin{bmatrix} 9 & 1 \\ 1 & 4 \end{bmatrix} \right)
$$
$$
\begin{bmatrix} Y_1 \\ Y_2 \end{bmatrix} = \begin{bmatrix} 2 & -1 \\ 3 & 1 \end{bmatrix} \begin{bmatrix} X_1 \\ X_2 \end{bmatrix} + \begin{bmatrix} 4 \\ -3 \end{bmatrix} \tag 1
$$
So
$$
\operatorname{E}\begin{bmatrix} Y_1 \\ Y_2 \end{bmatrix} = \begin{bmatrix} 2 & -1 \\ 3 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ -1 \end{bmatrix} + \begin{bmatrix} 4 \\ -3 \end{bmatrix} = \begin{bmatrix} 7 \\ -1 \end{bmatrix}
$$
and
$$
\operatorname{var}\begin{bmatrix} Y_1 \\ Y_2 \end{bmatrix} = \begin{bmatrix} 2 & 3\\ -1 & 1 \end{bmatrix} \begin{bmatrix} 9 & 1 \\ 1 & 4 \end{bmatrix} \begin{bmatrix} 2 & -1 \\ 3 & 1 \end{bmatrix}.
$$
The matrix in the middle is the variance of $X.$ It is multiplied on the left by one matrix and on the right by the transpose of that matrix. That "one matrix" is the one by which $X$ was multiplied in line $(1)$ above.
The resulting random vector $Y$ also has a bivariate normal distribution.
