No, the matrix consists of the coeficients, not the functions.
f1(x)= 1 so f1'(x)= 0(1)+ 0(sin(x))+ 0(cos(x)): <0, 0, 0>
f2(x)= sin(x) so f2'(x)= 0(1)+ 0(sin(x))+ 1(cos(x)): <0, 0, 1>
f3(x)= cos(x) so f2'(x)= 0(1)+ (-1)(sin(x))+ 0(cos(x)): <0, -1, 0>
The matrix is