Suppose that $X_1,X_2,Y$ take values in $\mathbb{R}^{p_1},\mathbb{R}^{p_2},\mathbb{R}^{q}$, respectively, and that $E[|X_1|^2+|X_2|^2+|Y|^2\lt \infty$. Show that $$\text{Cov}(AX_1+BX_2,Y)=A\text{Cov}(X_1,Y)+B\text{Cov}(X_2,Y)$$
for all appropriately shaped deterministic matrices $A$ and $B$.
Could you please verify if my solution is correct?
My solution:
$$\text{Cov}(AX_1+BX_2,Y)=E\biggl[(AX_1+BX_2-E[AX_1+BX_2])(Y-E[Y])^T\biggr]=E\biggl[(AX_1+BX_2-AE[X_1]-BE[X_2])(Y-E[Y])^T\biggr]=E\biggl[\bigl((AX_1-AE[X_1])+(BX_2-BE[X_2])\bigr)(Y-E[Y])^T\biggr]=E\biggl[(AX_1-AE[X_1])(Y-E[Y])^T+(BX_2-BE[X_2])(Y-E[Y])^T\biggr]=E\biggr[(AX_1-AE[X_1])(Y-E[Y])^T\biggr]+E\biggl[(BX_1-BE[X_2])(Y-E[Y])^T\biggr]=AE\biggl[[(X_1-E[X_1])(Y-E[Y]^T\biggr]+BE\biggl[(X_2-E[X_2])(Y-E[Y])^T\biggr]=A\text{Cov}(X_1,Y)+B\text{Cov}(X_2,Y)$$
Is everything I did correct? Would appreciate any feeback!