$$X=(X_1,X_2)$$ $$E(X_1)=1,\ E(X_2)=2$$
$$ C^2_X = \begin{bmatrix} 1&-1\\ -1&2\\ \end{bmatrix} $$
X is normal random vector, which means it is a bivariate normal random variable. In other words, $X_1$ and $X_2$ are jointly normal.
If we were to use a function $g:ℝ^2 \rightarrow ℝ^3: g(X) = (X_1+X_2, X_1-X_2,2X_2)=XB,$
$$where \ \ B= \begin{bmatrix} 1&1&0\\ 1&-1&2\\ \end{bmatrix} $$
then, $$E(g(X))= \begin{bmatrix} 3&-1&4\\ \end{bmatrix} $$
and,
$$ C^2_{g(X)} = \begin{bmatrix} 1&-1&2\\ -1&5&-6\\ -1&-6&8\\ \end{bmatrix} $$
now, $$det(C^2_{g(X)})=|C^2_{g(X)}|= 0$$
The question asks to find distribution of Z. My friend said something about transformation of random variable from $ℝ^2$ to $ℝ^3$ doesn't have a density because it doesn't span $ℝ^3$. How does one explain why the density function of g(X) doesn't exist? Would one be able to explain this using the fact the determinant is $0$? What is the distribution of g(X)? Thanks in advance!