= Solution
Use the divisor-$n$ <sample covariance matrix>, since this is the normalization that enters the <multivariate normal> <likelihood function>:
$$
\bar x=\frac1n\sum_{j=1}^n x_j,\qquad
S=\frac1n\sum_{j=1}^n(x_j-\bar x)(x_j-\bar x)^T.
$$
For a <covariance matrix> that is a <positive-definite matrix> $V$, the product of the <multivariate normal densities> gives
$$
-2\ell(\mu,V)=np\log(2\pi)+n\log|V|+
\sum_{j=1}^n(x_j-\mu)^TV^{-1}(x_j-\mu).
$$
To separate the <sample mean> from the <sample covariance matrix>, expand $x_j-\mu=(x_j-\bar x)+(\bar x-\mu)$. The cross terms sum to zero because $\sum_j(x_j-\bar x)=0$. Also $u^TV^{-1}u=\operatorname{tr}(V^{-1}uu^T)$ by the <trace> identity. Therefore
$$
\boxed{-2\ell(\mu,V)=np\log(2\pi)+n\log|V|+
n\operatorname{tr}(V^{-1}S)+n(\bar x-\mu)^TV^{-1}(\bar x-\mu).}
$$
The printed expression suppresses the parameter-independent constant $np\log(2\pi)$; dropping this constant has no effect on <maximum likelihood estimation> or the <likelihood-ratio test>.
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