I have the following problem:
Let $X_1 , X_2 , X_3 $ random variables with normal distribution and with $\mu = (-1,1,0)^T$ , $\Sigma = \begin{pmatrix} 1 & 0 & 1\\ 0 & 3 & 1 \\ 1 & 1 & 2 \end{pmatrix} $.
If we have $Y = X_1 + 2 X_2 - 3 X_3$. Find the variance of $Y$.
I searched how to tackle this problem but I got nowhere. I get that I need to find a matrix $A$ such that $\Sigma = AA^T$ but how can you define an $A$ matrix for the distribution $Y$? I also found a similar case where they just multiplied the given $\Sigma$ by the sum of the different coefficients of the variables $X_i$ . But in the case of the distribution $Y$ it would be $0$ and that left me more confused. Any help would be very appreciated. Thanks.
Hints: Try to write down $EX_iX_j$ for all $i,j$. $$EX_1X_2-(EX_1)(EX_2)=0$$ (the first row second column element of $\Sigma$). and $$EX_1=-1,EX_2=1$$ This gives $EX_1X_2=-1$.
$$EX_1^{2}-(EX_1)^{2}=1$$ and this gives $$EX_1^{2}=2$$, and so on. Now $$EY^{2}=E(X_1+2X_2-3X_3)^{2}$$ can be obtained by just expanding the square. Varianec of $Y$ is $$EY^{2}-(EY)^{2}$$