Let us denote the coefficient matrix of (\[equ:32\]) as ${\bf{P}}=\left(P_{ij}\right)_{3 \times 3}$. If the polynomial system (\[equ:31\]) has a nontrivial solution, there will exist a nontrivial solution for the homogeneous linear system (\[equ:32\]). According to the algebraic theory, (\[equ:32\]) has a nontrivial solution if and only if the determinant of $\bf{P}$ is zero, , $ \det \left( {\bf{P}} \right) = 0$. Define $F_4 = \det \left( {\bf{P}} \right)$. $F_4$ has the same form as $F_i$ in (\[equ:31\]). Similarly, we can construct two homogeneous linear systems for $\left[s_0, s_1^2, s_2\right]^T$ and $\left[s_0, s_1, s_2^2\right]^T$. This will yield two homogeneous equations $F_5=0$ and $F_6=0$, respectively. Stacking $F_i$ ($i=1,\cdots, 6$) together, we have a homogeneous linear system as $$\label{equ:33}
{\bf{C}}\left(s_3\right){\bf{h} }= {\bf{0}}_{6 \times 1}$$ where ${\bf{C}}\left(s_3\right)$ is a $6 \times 6$ matrix with polynomials in $s_3$ as elements, and ${\bf{h}} = [s_0^2, s_1^2, s_2^2, s_0s_1,  s_0s_2, s_1s_2]^T$. If (\[equ:33\]) has a nontrivial solution, if and only if $\det \left({\bf{C}}\left(s_3\right)\right) = 0$. This is an eighth order polynomial equation in $s_3$. We can get $s_3$ by solving this equation. Comparing (\[equ:30\]) and (\[equ:31\]), we can find that if $[s_1, s_2]$ is a solution of (\[equ:30\]), $[s_1, s_2, 1]$ is a solution of (\[equ:31\]), and 