If we have given that $\bf{Y} | \bf{X} \sim N(\bf{\mu}, \bf{\Sigma})$ and
$$\bf{Z} = A\bf{Y} + \bf{\varepsilon}$$
where $\bf{\varepsilon} \sim N(\bf{0}, \bf{\Omega})$, how can we show that
$$Z|X, Y \sim N(\bf{AY}, \bf{\Omega})$$.
I tried with Bayes rule but was not very helpful.
Remember that the distribution $Z|X, Y$ depends on some fixed $Y$ (some fixed outcome of the probabilistic distribution of $Y$). It means that the randomness of $Z$ is fully determined by $\varepsilon$, and $Z$ has a normal distribution with a shifted mean $AY$ and the same deviation (correlation matrix) as $\varepsilon$.
We denote by $f$ the density function.
$$f_{\mathbf{Z}|\mathbf{X,Y}}(z) = f(\mathbf{Z} = z~| \mathbf{X}=x, \mathbf{Y}=y) = f(\mathbf{AY} + \varepsilon = z~| \mathbf{X}=x, \mathbf{Y}=y) = f(\varepsilon = z - \mathbf{AY}~| \mathbf{X}=x, \mathbf{Y}=y) = f_{\varepsilon}(z-\mathbf{AY}|\mathbf{X}=x, \mathbf{Y} = y) = f_{\varepsilon}(z-\mathbf{A}y)$$ which means that $Z|X,Y$ is just shifted $\varepsilon$