I'm trying to understand how they solved this exercise.
I understood the first step of the answer obviously, but i'm a little confused about the second step. Would anyone please be able to provide some extra steps on how exactly they arrived to that equation and what properties and equations they used? Thank you.
$$\begin{align}\mathbb{V}[Z]&=\mathbb V[-X+3Y]\\[1ex]&=\mathbb{V}[X]+9\,\mathbb{V}[Y]-2\cdot3\cdot \mathbb{Cov}[X,Y]\\[1ex]&=\sigma^2(1+9)-6\,\mathbb{Cov}[X,Y]\\[2ex]34&=\sigma^2(1+9)+12\\[2ex]\sigma^2&=\frac{34-12}{10}\\[1ex]&=2.2\end{align}$$
This is because: $$\begin{align}\mathbb{V}[aX+bY]&=\mathbb{E}[(aX+bY)^2]-\mathbb{E}[aX+bY]^2\\[1ex]&={\mathbb{E}[a^2X^2]+\mathbb{E}[b^2Y^2]+\mathbb{E}[2abXY]\\-\mathbb{E}[aX]^2-\mathbb{E}[bY]^2-2\,\mathbb{E}[aX]\,\mathbb{E}[bY]}\\[1ex]&={{(\mathbb{E}[a^2X^2]-\mathbb{E}[aX]^2)}\\+{(\mathbb{E}[b^2Y^2]-\mathbb{E}[bY]^2)}\\+{(\mathbb{E}[2abXY]-2\,\mathbb{E}[aX]\,\mathbb{E}[bY])}}\\[1ex]&=a^2\,\mathbb{V}[X]+b^2\,\mathbb{V}[Y]+2ab\,\mathbb{Cov}[X,Y]\end{align}$$