Help needed on algebraic steps for Maximum Likelihood Estimation of Multivariate Normal Distribution?

The negative loglikelihood is as follows:

$$\dfrac{nd}{2} \log 2\pi + \dfrac{n}{2} \log |\Sigma| + \dfrac{1}{2}\sum_{i=1}^n(x_i-\mu)^T\Sigma^{-1}(x_i-\mu) \tag{1}$$

If I take differentiation with respect to $\mu$ on $(x_i-\mu)^T\Sigma^{-1}(x_i-\mu)$, the result becomes as follows: $$2\Sigma^{-1}\mu - 2\Sigma^{-1} = 2\Sigma^{-1}(\mu-x_i) \tag{2}$$

So

$$\dfrac{\partial l(u, \Sigma)}{\partial \mu} = \dfrac{1}{2}\sum_{i=1}^n2\Sigma^{-1}(\mu-x_i) = \Sigma^{-1}\sum_{i=1}^n(\mu-x_i) \tag{3}$$

But what I can't do with the algebraic steps for the next step to get the following result:

$$\mu_{MLE}^{*} = \dfrac{1}{n}\sum_{i=1}^n x_i \tag{4}$$

How can I go from $(3)$ to $(4)$? Hope to get algebraic steps for it.

• I have missed some part in the equation. 3. I have included $\sum_{i=1}^n$. I am horrible at doing algebraic steps... – user122358 Mar 2 '17 at 8:32
• Hope it is ok. $$\Sigma^{-1}\sum_{i=1}^n (\mu-x_i) = 0$$ $$n\mu-\sum_{i=1}^n x_i = 0$$ $$n\mu = \sum_{i=1}^n x_i$$ $$\mu = \dfrac{\sum_{i=1}^n x_i}{n}$$ $$\therefore \mu_{MLE}^{*} = \dfrac{1}{n}\sum_{i=1}^n x_i$$ – user122358 Mar 2 '17 at 8:42