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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 $|\Sigma|^{-1/2}=|A^{-1}|$?

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  • $\begingroup$ Please insert the definition of $|B|$ for a given matrix $B$... $\endgroup$ – dan_fulea May 15 at 16:27
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    $\begingroup$ That's what you get taking determinant of both sides. $\endgroup$ – StubbornAtom May 15 at 16:28
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    $\begingroup$ Hint: a matrix and its transpose have the same determinant $\endgroup$ – David Reed May 15 at 16:33
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    $\begingroup$ I guess $A$ is square matrix, then $|\Sigma|^{-1/2}=|A^{-1}|$ follows from $|A|=|A^T|$ $\endgroup$ – kludg May 15 at 16:35
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    $\begingroup$ It's also a theorem that $|A^{-1}| = |A|^{-1} $. This is actually easy enough to prove in a comment: $1 = |I| = |A^{-1}A| = |A^{-1}||A| \to |A^{-1}| = |A|^{-1} $ $\endgroup$ – David Reed May 15 at 16:49

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