In this video (final result at 8:54), the professor explains that the most general fundamental matrix of a system of linear ordinary differential equations is given by $XC$, where $X$ is any fundamental matrix of the system (the matrix whose columns are a linear independent set of solutions), and $C$ is a matrix of constants. He makes it sound as if the entries of $C$ for the $2 \times 2$ system are four arbitrary constants which bare no specific relation to each other, except that the determinant of $C$ must not be $0$.
Is this understanding correct? If so, why is it the case that generalizing the fundamental matrix of a $2 \times 2$ system requires four arbitrary constants, when the system itself only has room for two arbitrary constants?
Ok, I watched about a minute of the video, and I think that the brief answer to your question is that while only two constants are needed to generate all solutions, there is nothing preventing you from having four constants and repeating some of the solutions.
Let me give a simple example. Suppose you have a fundamental matrix $$ X = \begin{pmatrix} x_{11} & x_{12} \\ x_{21} & x_{22} \end{pmatrix} $$
I can write another fundamental matrix by exchanging the two columns: $$ Y = \begin{pmatrix} x_{12} & x_{11} \\ x_{22} & x_{21} \end{pmatrix} $$
This is not a new set of solutions, because it's the exact same set that you had before. But it is a new fundamental matrix because it is represented slightly differently. The basic problem is that there is some sense in which unique solutions are a "subset" of fundamental matrices (but I am not going to bother to define this rigorously).
In our particular case, you can write $$ Y = X \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} $$ which agrees with the lecturer's assertion that all fundamental matrices can be written in the form $XC$ for some constant matrix $C$.
The only remaining question is why $C$ must have a nonzero determinant. Clearly, for $XC$ to be a fundamental matrix, its columns have to be linearly independent and therefore it has to have a nonzero determinant. But $\det(XC)=\det(X)\det(C)$, so $C$ must also have a nonzero determinant. (Basically, if $C$ had a zero determinant, then you would end up with a fundamental matrix that has either zero or one independent solutions in it.)