If $W$ and $V$ are standard independent normal variables find the joint distribution of $3W+2V$ and $2W-3V$.
Since $W$ and $V$ are standard independent normal variables then would it just be the probability density function $f(3w+2v, 2w-3v)=\phi(3w+2v)\phi(2w-3v)=\dfrac{1}{\sqrt{2\pi}}e^{-.5(3w+2v)^2-.5(2w-3v)^2}$ where the mean is $0$ and standard deviation is $1$?
Let $X=\begin{pmatrix}W\\V\end{pmatrix}$, $A=\begin{pmatrix}3&2\\2&-3\end{pmatrix}$, and $Y=\begin{pmatrix}3W+2V\\2W-3V\end{pmatrix}=AX$. So if $X\sim N(0,I_2)$, then $Y\sim N(A0,AI_2A')=N(0,AA')$. Then, you can use the standard formula for multivariate normals. In this case, $$ AA'=A^2=\begin{pmatrix}13&0\\0&13\end{pmatrix} $$ so the correlation is $0$ and $3W+2V$ and $2W-3V$ turn out to be independent. Either has a marginal density that is normal with mean $0$ and variance $13=3^2+2^2$. As for the joint density, $$ f(s,t)=\frac{1}{26\pi}\exp\left(-\frac{1}{2}\left(\frac{s^2}{13}+\frac{t^2}{13}\right)\right). $$