Basically, given a Lagrangian formulation of a theory, by expanding the EL equations
\begin{align}
0 &= \frac{d}{dt} \frac{\partial L}{\partial \dot{q}^a}(t,q^a,\dot{q}^a) - \frac{\partial L}{\partial q^a} \\
&= \frac{\partial}{ \partial t} \frac{\partial L}{\partial \dot{q}^a} + \frac{\partial^2 L}{\partial q^b \partial \dot{q}^a} \dot{q}^b + \frac{\partial^2 L}{\partial \dot{q}^b \partial \dot{q}^a} \ddot{q}^b - \frac{\partial L}{\partial q^a} \\
&= \left( \frac{\partial^2 L}{\partial \dot{q}^b \partial \dot{q}^a} \right) \ddot{q}^b + \left( \frac{\partial^2 L}{\partial q^b \partial \dot{…

Consider the Lagrangian
\begin{align}
L = \frac{1}{2}(\dot{x}^2 + \dot{y}^2).
\end{align}
We know the equations of motion giving the time evolution of the $x$ and $y$ coordinate space variables are
\begin{align}
\ddot{x} &= 0, \\
\ddot{y} &= 0,
\end{align}
and we can solve these explicitly as
\begin{align}
x &= x_0 + v_0^x t, \\
y &= y_0 + v_0^y t.
\end{align}
Thus, the equations of motion coupled with boundary conditions which fix the integration constants completely determine the motion of the system. As you see above $\det M_{ba} \neq 0$.

We now re-analyse this system from a Hamiltonian perspective. The canonical momenta are
\begin{align}
p_x &= \frac{\partial L}{\partial \dot{x}} = \dot{x}, \\
p_y &= \frac{\partial L}{\partial \dot{y}} = \dot{y},
\end{align}
and the canonical Hamiltonian is
\begin{align}
H_c &= p_x \dot{x} + p_y \dot{y} - L \\
&= p_x p_x + p_y p_y - \frac{1}{2}(\dot{x}^2 + \dot{y}^2) \\
&= p_x^2 + p_y^2 - \frac{1}{2}(p_x^2 + p_y^2) \\
&= \frac{1}{2}(p_x^2 + p_y^2).
\end{align}
The possibility of constructing $H$ clearly relies on being able to solve for $\dot{x}$ and $\dot{y}$ as functions of $p_x$ and $p_y$.

We can now compute the time evolution from the Poisson brackets, which satisfy $\{x,p_x \} = 1$,
\begin{align}
\dot{x} &= \{x,H_c \} = \{x,\frac{1}{2}(p_x^2 + p_y^2) \} = p_x, \\
\dot{y} &= \{y,H_c \} = \{y,\frac{1}{2}(p_x^2 + p_y^2) \} = p_y, \\
\dot{p}_x &= \{p_x,H_c \} = 0 \ \ \ \leftrightarrow \ \ \ \ddot{x} = 0 \\
\dot{p}_y &= \{p_y,H_c \} = 0 \ \ \ \leftrightarrow \ \ \ \ddot{y} = 0.
\end{align}
This shows that if we can move freely between the Lagrangian and Hamiltonian descriptions, we can reproduce the equations of motion in either formalism. An important lesson from this is that t…

Consider the Lagrangian
\begin{align}
L = \frac{1}{2}(\dot{x} - y)^2
\end{align}
The equations of motion are
\begin{align}
\ddot{x} - \dot{y} &= 0, \\
\dot{x} - y &= 0.
\end{align}
This system has two degrees of freedom, $x$ and $y$, and two equations of motion, however the two equations of motion are not independent - the first equation is simply the time derivative of the second, in other words it is a consequence of the second equation of motion. But the second equation has no acceleration, this it is not an equation of motion, and since the equation with acceleration is simply a consequ…

If we consider the solution $x(t)$ as a Taylor expansion
\begin{align}
x(t) = \alpha + \beta t + \frac{\gamma}{2} t^2 + \phi(t)
\end{align}
then the constraint $\dot{x} = y$ implies
\begin{align}
y(y) = \beta + \gamma t + \dot{\phi}(t).
\end{align}
The boundary conditions on this system are
\begin{align}
\phi(0) &= \dot{\phi}(0) = \ddot{\phi}(0) = 0, \\
x(0) &= \alpha, \dot{x}(0) = \beta, \\
y(0) &= \beta, \dot{y}(0) = \gamma.
\end{align}
The function $\phi(t)$ is otherwise completely arbitrary, a particular choice of boundary conditions can fix $\alpha, \beta, \gamma$ but it can't fix $\ph…

Now let us consider the Hamiltonian formulation of this constrained system. The canonical momenta are
\begin{align}
p_x &= \frac{\partial L}{\partial \dot{x}} = \dot{x} - y, \\
p_y &= \frac{\partial L}{\partial \dot{y}} = 0.
\end{align}
Here we see that $\dot{x}$ can be inverted (in this case for $p_x$), however $\dot{y}$ cannot be inverted (not for $p_y$, nor in general for $p_x$). Another problem is that this system is already incompatible with the basic Poisson algebra for dynamical systems, which should satisfy $\{y,p_y \} = 1$ however here we see $\{y,p_y \} = 0$, again indicating that…

Looking back at the constrained system $L = \frac{1}{2}(\dot{x}-y)^2$ with it's canonical Hamiltonian $H = \frac{1}{2} p_x^2 + p_x y$ which leads to inconsistencies, we note the inconsistencies lie in the fact that $\{y,p_y\} = 0$ due to $p_y = 0$ when we should have $\{y,p_y \} = 1$. To make this system consistent we use Lagrange multipliers to form the constrained Hamiltonian
$$\tilde{H} = \frac{1}{2} p_x^2 + p_x y + \lambda p_y $$
and then impose the constraints 'weakly', i.e. $p_y = 0$ is to hold after applying the PB's, and even more the constraint $p_y = 0$ should not change with tim…