So I am trying to get a polynomial approximation for $\sin(x)$ using Pade approximation for $n = 2$ and $m = 3$ and the Maclaurin series $(\deg 5)$ for
$$\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!}$$
However, the matrix I am getting seems to be insolvable, but I'm not sure what I'm doing wrong since my professor says there should be a solution, infinitely many, but just to find one.
Using some MatLab code he provided, and working it myself by hand I get the matrix, \begin{bmatrix}0&0&0&-1&0\\1&0&0&0&-1\\0&1&0&0&0\\-1/6&0&1&0&0\\0&-1/6&0&0&0\end{bmatrix} and this multiplied by the $[q_1, q_2, q_3, p_1, p_2]$ variable matrix should equal \begin{bmatrix}-1\\0\\1/6\\0\\-1/120\end{bmatrix}
The main issue I see is that the $q_2$ term is defined to $2$ different values when solved by hand. Also that the determinant of the matrix is $0$.