The "stationary potential energy" condition for static equilibrium in mechanical systems

I've often read that, for a mechanical system which can be described by $n$ generalized coordinates $q_1,...,q_n$, a point $\mathbf{Q}=(Q_1,...,Q_n)$ is a point of equilibrium if and only if the potential energy $U$ is stationary at that point, i.e. iff $\frac{\partial U}{\partial q_i}(\mathbf{Q})=0$ for all $i$. I've only seen a proof for the unidimensional case $n=1$, that uses the Lagrangian of the system to show that if the system is in $q=Q$, with $\dot q=0$, then $\ddot{q}=0 \iff \frac{\partial U}{\partial q}(Q)=0$ (in this case it is also easy to obtain the stable equilibrium condition $\partial ^2 U/\partial q^2>0$).

Following this example I was trying to prove the statement for the general case, however probably some steps aren't correct for the general case. The Lagrangian is (using Einstein's notation for summations $a_ib_i\dot{=}\sum _i a_ib_i$): $$L (\mathbf{q},\mathbf{\dot q})= T(\mathbf{q},\mathbf{\dot q})-U(\mathbf{q})=\frac{1}{2}\dot {q_i} A_{ij} \dot {q_j}-U(\mathbf{q}),$$where I've expressed the kinetical energy as a quadratic form of the generalized velocities and $A_{ij}$ are functions of $\mathbf{q}$ (as I said, I don't know if this is generally possible). Now we have:$$\frac{d}{dt}\frac{\partial \cal L}{\partial \dot{q_i}}=\frac{d}{dt}[A_{ij}\dot q_j]=\dot{q_k}\frac{\partial A_{ij}}{\partial q_k}\dot{q_j}+A_{ij}\ddot{q_j},$$and $$\frac{\partial \cal{L}}{\partial{q_i}}=\frac{1}{2}\dot{q_k}\frac{\partial A_{kj}}{\partial q_i}\dot{q_j}-\frac{\partial U}{\partial q_i}.$$So, if all velocities $\dot{q_j}=0$, the equations of motion are:$$A_{ij}\ddot{q_j}=-\frac{\partial U}{\partial q_i},\qquad i=1,...,n.$$Now, I see that if the system is in equilibrium in $\mathbf{Q}$, then, for all $j$, $\ddot {q_j}=0$ and it is required that $\partial U / \partial q_j = 0$. What about the vice-versa? Does $\partial U / \partial q_j = 0 \ \forall j \implies \ddot{q_j} = 0 \ \forall j$? I see that this is equivalent to require $\det A\neq0$, where $A$'s elements are $A_{ij}$, so is there a reason why it must be so?Thank you for your time.