# Estimating $\mu$ from "completing the square"

I'm reading "Pattern recognition a machine learning" by Bishop (link). I need some mathematical help understanding ecuation 2.71. How do we obtain $$\mu$$ from "completing the square" on the exponent of a gaussian distribution (see the paragraph following equation 2.71).

Thank & Regards

The expression inside the exponential for a normal distribution is $$-\frac{1}{2}(x-\mu)^T\Sigma^{-1}(x-\mu)$$ which he rewrites as $$-\frac{1}{2}x^T\Sigma^{-1}x+x^T\Sigma^{-1}\mu+const$$. So whenever we have any expression inside the exponential of a normal distribution, we know that the term that is linear in $$x$$ is equal to $$\Sigma^{-1}\mu$$.
Take the example above in 2.70 where he breaks apart the parts of $$x$$ into $$x_a$$ and $$x_b$$. He wants to find the condition distribution of $$x_a$$ when you have $$x_b$$ as a given. So we can rewrite the standard equation in the form of 2.70, but how can we find $$\mu_{a|b}$$ or $$\Sigma_{a|b}$$ now that $$x_b$$ is a given constant?
Well, we know from before that for a normal distribution on $$x$$, the term that is linear in $$x$$ inside the exponential is equal to $$\Sigma^{-1}\mu$$. Bishop wants to find the distribution on $$x_a$$ given $$x_b$$. So he rewrites 2.70 with $$x_b$$ as a constant and finds the term that is linear in $$x_a$$ is equal to $$x_a^T\{\Lambda_{aa}\mu_a-\Lambda_{ab}(x_b-\mu_b)\}$$ (2.74). Our previous result tells us that term is equal to $$\Sigma^{-1}_{a|b}\mu_{a|b}$$. He follows the similar process in 2.72 and 2.73 to get $$\Sigma^{-1}_{a|b}$$ and substitutes that in to find $$\mu_{a|b}$$ in 2.75.