In the derivation of the joint pdf of $f_\textbf{X}(\pmb{x})$, where $\textbf{X}=\pmb\mu+A\pmb Z$ and $\textbf{X}\sim~N_n(\pmb\mu,\Sigma)$, there is a step I do not understand.
In particular, it is the fact that if $\Sigma^{-1}=(A^{-1})^TA^{-1}$, then $|A^{-1}|=|\Sigma|^{-1/2}$. This shows up during the derivation of the joint pdf as follows:
$$f_{\textbf{X}}(\textbf{x})=|A^{-1}|(2\pi)^{-n/2}exp\bigg(-\frac{1}{2} (\pmb x-\pmb \mu)^T(A^{-1})^TA^{-1}(\pmb x-\pmb \mu)\bigg) $$
Now since $\Sigma=AA^T$ is invertible we have $\Sigma^{-1}=(AA^T)^{-1}=(A^T)^{-1}A^{-1}=(A^{-1})^TA^{-1}$ and thus,
$$f_{\textbf{X}}(\textbf{x})=(2\pi)^{-n/2}|\Sigma|^{-1/2}exp\bigg(-\frac{1}{2} (\pmb x-\pmb \mu)^T\Sigma^{-1}(\pmb x-\pmb \mu)\bigg) $$
I can't understand how $|\Sigma|^{-1/2}=|A^{-1}|$?