The statement that this always CAN be done is called 'Ado's theorem'. One way to get an answer is to look for proofs of this theorem online. This blog post by Terrence Tao looks useful (but I didn't read it all): https://terrytao.wordpress.com/2011/05/10/ados-theorem/.
EDITTED IN: A good start (see also the blogpost) is to first try and find the center of the Lie algebra, so all linear combinations of your operators that have the property of commuting with all the operators. If your center is only the zero operator, things are very simple. And if it is one dimensional things are not that bad either.
FURTHER EDIT AFTER READING COMMENT: There is one thing you can always do but only gives the desired result when the center is zero (because it will always represent operators in the center with the zero-matrix whether it is justified or not) and that is taking the adjoint representation. Here is an algorithm:
The vector space the matrices act on is just the Lie algebra itself (so it is ten dimensional). The vector $[1, 0, 0, 0, 0, 0, 0, 0, 0, 0]$ corresponds to the first operator on your list, the vector $[0, 1, 0, 0, 0, 0, 0, 0, 0, 0]$ to the second etc. (When I write lying vectors I mean standing ones. They are just more annoying to type.) I will call the operators in order $X_1, \ldots, X_{10}$.
Now to create a matrix for an operator $X$ in the Lie algebra we remember that the first column of any matrix $A$ is just the vector you get when letting $A$ act on the first basisvector $[1, 0, 0, 0, 0, 0, 0, 0, 0, 0]$, you get the second column of $A$ by seeing what happens when you let $A$ act on $[0, 1, 0, 0, 0, 0, 0, 0, 0, 0]$ etc.
Now what we want the matrix assigned to operator $X$ to do is send $[1, 0, 0, 0, 0, 0, 0, 0, 0, 0]$ to the vector corresponding to $[X, X_1]$, send $[0, 1, 0, 0, 0, 0, 0, 0, 0, 0]$ to the vector corresponding to $[X, X_2]$ etc.
More explicit: if you calculate that $[X, X_1] = 3 X_2 + 5 X_3 + 10 X_{10}$ the first column of the matrix you look for is $[0, 3, 5, 0, 0, 0, 0, 0, 0, 10]$. Etc.
At the end of the day every operator has then both a matrix AND a vector attached to it, but once you have all the matrices you can (if you want) forget the vectors. From the description it is not completely clear that the matrices associated to the operators satisfy the same commutation relations as the operators themselves, but you can check that by hand. (Or prove it in the general case. Terrence Tao states that this 'follows from the Jacobi Identity', which is the identity $[X, [Y, Z]] + [Y, [Z, X]] + [Z, [X, Y]] = 0$. The Jacobi identity is part of the definition of an abstract Lie algebra but more relevantly, it is automatically true for every Lie algebra where the elements are linear operators and the Lie bracket is the commutator bracket.)
