Defining a smooth unique envelope function. Given $a(t), b(t)$ and $c(t)$ defined on $I \to \mathbb{R}$, we want to define the smooth function, i.e. the envelope, $\gamma(t)$, of the family of lines $a(t)x + b(t)y =c(t)$. 
To find $\gamma(t)$ we solve the system 
$$ a(t)x + b(t) y = c(t) $$
$$ a'(t)x + b'(t)y = c'(t)$$
If $\begin{vmatrix} a(t) & b(t) \\
a'(t) & b'(t) \end{vmatrix}\neq 0$ then $\gamma(t)$ is unique. 
I am now asked to show that the non-singularity and smoothness of $\gamma(t)$ relies on 
$$ \begin{vmatrix} a(t) & b(t) & c(t) \\
a'(t) & b'(t) & c'(t) \\
a''(t) & b''(t) & c''(t) \end{vmatrix} \neq 0$$
I have absolutely no idea how the determinant of this matrix is related to the above system of equations. Especially since we could differentiate infinitely since $a(t), b(t)$ and $c(t)$ are smooth. 
The above matrix looks a lot like the wronskian, so the nonvanishing of the determinant tells us that the three smooth functions are linearly independent. But why does this matter? 
 A: Let $\gamma(t) = (x(t),y(t))$. By differentiating both sides of the equality $a'(t)x(t) + b'(t)y(t) = c'(t)$, we see that
$$a''(t)x(t) + b''(t)y(t) + a'(t)x'(t) + b'(t)y'(t) = c''(t).$$
Thus, if $\gamma$ is singular at $t = t_0 \in I$, we have that
$$a''(t_0)x(t_0) + b''(t_0)y(t_0) = c''(t_0).$$
Combining this with the two equations defining $\gamma$, we obtain the system
\begin{align}
a(t_0)x(t_0) + b(t_0)y(t_0) - c(t_0) &= 0 \\
a'(t_0)x(t_0) + b'(t_0)y(t_0) - c'(t_0) &= 0 \\
a''(t_0)x(t_0) + b''(t_0)y(t_0) - c''(t_0) &= 0
\end{align}
which can be written in matrix form as
$$\begin{pmatrix}
a(t_0) & b(t_0) & c(t_0) \\
a'(t_0) & b'(t_0) & c'(t_0) \\
a''(t_0) & b''(t_0) & c''(t_0)
\end{pmatrix}\begin{pmatrix} x(t_0) \\ y(t_0) \\ -1 \end{pmatrix}
= \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$$
This shows that
$$\begin{vmatrix}
a(t) & b(t) & c(t) \\
a'(t) & b'(t) & c'(t) \\
a''(t) & b''(t) & c''(t)
\end{vmatrix} = 0$$
whenever $\gamma'(t) = 0$. In particular, if the above determinant is nonzero for all $t \in I$, then $\gamma$ is non-singular.
