Conditional multivariate Gaussian distribution I am trying to understand how 'completing the square' method is used to determine mean and co-variance of multivariate Gaussian distribution but I am not able to make much sense of the following step
$$−((x−μ)^TΣ^-(x − μ))/2 = −(x^TΣ^−x + x^TΣ^-μ + const)/2$$
where x and μ are vectors and Σ is co-variance matrix.
What does this step indicate and how is it derived? 
P.S. I am not well versed with linear algebra which might be a reason for not being able to comprehend this fully, if you can point me to certain properties from linear algebra that are being used here, that will be very helpful
 A: First, here's how you would complete the square with scalars (no matrices!)
$$ax^2 - 2bx = a\left(x-\frac{b}{a} \right)^2 - \frac{b^2}{a}$$
Please verify this for yourself. 
Now, think intuitively about how you could rewrite this expression for matrices. Generally, a quadratic term $az^2$ for scalar values corresponds to a term $z^T A z$ when in matrix form. This is sometimes called a "quadratic form". Another thing is that we can't divide matrices, but we can multiply by the inverse, so a fraction like $\frac{b}{a}$ is usually converted to $A^{-1} b$. 
Now we can rewrite the scalar complete-the-square equation in matrix form.
$$x^TA x - 2b^Tx = \left(x-A^{-1}b \right)^TA\left(x-A^{-1}b \right) - b^TA^{-1}b$$
Please expand this equation and verify that it is actually correct, and that the ad-hoc reasoning we did above actually works.
Now replace $b = \Sigma^{-1} \mu$ and $A = \Sigma$, and $\text{const} = b^TA^{-1} b$. You should arrive at something which resembles equation in the question.
Some linear algebra tips which might help: The matrix $A$ or $\Sigma$ is generally treated as symmetric without loss of generality, so $A = A^T$. Also I believe you wrote $\Sigma^-$ in your post but meant $\Sigma^{-1}$
