I was looking at the Wikipedia article talking about the multivariate normal distribution and specifically I was looking at the section talking about the probability density function of that distribution.
In that article it is stated that if the covariance matrix $\Sigma$ of the p.d.f. of a m.n.d.
$$f_{\vec{X}}(x_1, x_2, \ldots,x_k) = \frac{1}{\sqrt{(2\pi)^k |\Sigma |}} \exp\left(-\frac{1}{2}\cdot (\vec{x} - \vec{\mu})' \cdot \Sigma^{-1} \cdot (\vec{x} - \vec{\mu}) \right)$$
is a $1 \times 1$ matrix, then the p.d.f. of the multivariate normal distribution is the same as the p.d.f. of the univariatate normal distribution, but I am not seeing why. Could you please explain it to me?
The univariate normal p.d.f. is $$ x \mapsto \frac{1}{\sqrt{2\pi \sigma^2}} \exp\left(-\frac{1}{2}\left( \frac{x-\mu} \sigma \right)^2 \right). \tag 1 $$ The expression $\left(\dfrac{x-\mu} \sigma\right)^2$ is the same as $(x-\mu) (\sigma^2)^{-1} (x-\mu)$. The $1\times1$ matrix $x-\mu$ is its own transpose, and the inverse of the $1\times1$ matrix whose entry is the number $\sigma^2$ is the $1\times1$ matrix whose entry is $(\sigma^2)^{-1}$, so $(1)$ becomes $(x-\mu)'(\sigma^2)^{-1}(x-\mu)$, which is then $(\vec x - \vec\mu)'\Sigma^{-1}(\vec x-\vec\mu)$.