Questions on asymmetric nature of mass and stiffness matrices EDIT: I now know the answer to my second question after re-reading relevant parts of my textbook and reading Eli's answer.
I was under the impression that mass and stiffness matrices are always symmetric (as I have been taught in my course and in textbook as well) but while solving questions from another textbook, I came to know that the matrices can be asymmetric as well. This has confused me now. 
My first question is why then have we been taught that they are always symmetric all this while? Now, I had a look at this question on [Physics.SE] and the first answer to this makes it clear that it is only the symmetric component of a matrix $\left(\frac{M+M^T}{2}\right)$ that contributes to the kinetic or potential energy through the quadratic form because $x^TAx=0\forall x$ for a skew symmetric matrix. 

So is it that, when being taught, we are told that the mass and stiffness matrices have to always be symmetric because they implicitly have in mind that even if they are asymmetric, it is only the symmetric component that is of importance and which contributes to kinetic and potential energy?

This leads me to my second question, which can well be asked independently as well but I thought this ties in well here, so I am asking it here (I will make a separate post for this if users think that is better). How do we know that the matrices $M$ and $K$ appearing in $T=\frac{1}{2}\dot u^TM\dot u$ and $V=\frac{1}{2}u^TKu$ are indeed the matrices which will also appear in the final matrix form of equations of motion for a multi DOF linear system, i.e., $M\ddot x+Kx=0$?
I also came to know that for deriving the differential equation governing motion of multi DOF system, Lagrange's equations will always lead to symmetric mass and stiffness matrices while the method using free body diagrams and Newton's law may give asymmetric matrices as well. So, is the matrix that we get from Lagrange's equation the symmetric component (I have mentioned before in my question what symmetric component is) of the matrix obtained from the free body diagram method? 
 A: your second question:
If you have two generalized coordinate $(q_1\,,q_2)$ the general case for the kinetic energy is:
$$T=\frac{1}{2}\,a\,\dot{q}_1^2+\frac{1}{2}\,b\,\dot{q}_2^2\pm\,c\,\dot{q}_1\,\dot{q}_2\tag 1$$
The mass matrix M is :
$$M=\frac{\partial}{\partial \vec{\dot{q}}}\left(\frac{\partial\,T}{\partial \vec{\dot{q}}}\right)\tag 2$$
with $\vec{\dot{q}}=\left(\dot{q}_1\,,\dot{q}_2\right)^T$
thus:
$$M= \left[ \begin {array}{cc} a&\pm\,c\\\pm\,c&b
\end {array} \right]
$$
The mass matrix M is always symmetric 
if  a,b,c are only depend on the system parameter and the  masses then M is constant.
If one of the parameter a or b or c depend on the generalized coordinates $q_i$ , then is the mass matrix not constant $M=M(q_i)$, but still symmetric!
Stiffness matrix K
for a linear spring forces the potential energy is:
$$U=\frac{1}{2}\,a\,{q}_1^2+\frac{1}{2}\,b\,\dot{q}_2^2\pm\,c\,{q}_1\,{q}_2\tag 3$$
Stiffness matrix K is:
$$K=\frac{\partial}{\partial \vec{{q}}}\left(\frac{\partial\,U}{\partial \vec{{q}}}\right)\tag 4$$ 
with $\vec{{q}}=\left({q}_1\,,{q}_2\right)^T$
thus:
$$K= \left[ \begin {array}{cc} a&\pm\,c\\\pm\,c&b
\end {array} \right]
$$
If the parameters a,b,c are constant,the stiffness matrix K is symmetric and constant.
Edit:
Example: Double Pendulum

with the position vectors :
$$\vec{R}_1=\left[ \begin {array}{c} \rho\,\cos \left( q_{{1}} \right) 
\\\rho\,\sin \left( q_{{1}} \right) \end {array}
 \right]
$$
$$\vec{R}_2=\left[ \begin {array}{c} \rho\,\cos \left( q_{{1}} \right) +\rho\,
\cos \left( q_{{2}} \right) \\ \rho\,\sin \left( q_{
{1}} \right) +\rho\,\sin \left( q_{{2}} \right) \end {array} \right] 
$$
you have two generalized coordinates $\vec{q}=[q_1\,,q_2]^T$
$\Rightarrow$
$$\vec{v}_1=\frac{\partial \vec{R}_1}{\partial \vec{q}}\dot{q}$$
$$\vec{v}_2=\frac{\partial \vec{R}_2}{\partial \vec{q}}\dot{q}$$
the kinetic energy is:
$$T=\frac{1}{2}\,m\left(\vec{v}_1\cdot\vec{v}_1+
\vec{v}_2\cdot\vec{v}_2\right)
={\rho}^{2}{{\dot{q}}_{{1}}}^{2}m+{\rho}^{2}m{\dot{q}}_{{1}}{\dot{q}}_{{2}
}\cos \left( q_{{1}}-q_{{2}} \right) +\frac{1}{2}\,{\rho}^{2}m{{\dot{q}}_{{2}}}
^{2}
$$
$$M=\left[ \begin {array}{cc} {\frac {\partial ^{2}}{\partial {x}^{2}}}T
 \left( x,y \right) &{\frac {\partial ^{2}}{\partial y\partial x}}T
 \left( x,y \right) \\ {\frac {\partial ^{2}}{
\partial y\partial x}}T \left( x,y \right) &{\frac {\partial ^{2}}{
\partial {y}^{2}}}T \left( x,y \right) \end {array} \right] 
$$
with $x=\dot{q}_1$ and $y=\dot{q}_2$ you obtain
$$M=\left[ \begin {array}{cc} 2\,{\rho}^{2}m&{\rho}^{2}m\cos \left( q_{{1
}}-q_{{2}} \right) \\ {\rho}^{2}m\cos \left( q_{{1}}
-q_{{2}} \right) &{\rho}^{2}m\end {array} \right] 
$$
thus M is symmetric and not constant.
I hope it is now clear?
A: Your second question is answered by Lagrangian Mechanics.
The equations of motion $M \ddot{x} + K x = 0$ are derived from the general equation 
$$ \frac{{\rm d}}{{\rm d}t} \left( \frac{\partial T}{\partial \dot{x}} \right) + \frac{\partial V}{\partial x} = 0 $$
where $T = \tfrac{1}{2} \dot{x}^\top M \dot x$ and $V = \tfrac{1}{2} x^\top K x$.
Specifically
  $$ \frac{\partial V}{\partial x}  = K x $$
and
$$ \frac{\partial T}{\partial \dot x} = M \dot x $$
Your first question is unclear to me. I challenge you to find a real physical system that yields a non-symmetric mass matrix, nor a stiffness matrix.
The point being by the linked post is that if it where the case of non-symmetry, then those components would be simply ignored by the equations of motion and would bear no significance in the solution. You just have something there that is going to be multiplied by zero at some point, so why have it?
