How to solve a system of $4$ equations?

Question

I would like to solve the following system of equations in $ \mathbb{C} $ : $$ \begin{cases} x+2y+3z+4t+3xy-2zt = 1 \\ 4x+y+2z+3t+2xy-4zt = 3 \\ 3x+4y+z+2t-xy+zt = -1 \\ 2x+3y+4z+t+4xy+2zt = 2 \end{cases} $$ I'm not able to solve it unfortunatly. The idea is to transform this system in another system of the form : $$ \begin{cases} a_{11}(x+y)+a_{12}(z+t)+a_{13} \ xy+a_{14} \ zt = b_{1} \\ a_{21}(x+y)+a_{22}(z+t)+a_{23} \ xy+a_{24}\ zt = b_{2} \\ a_{31}(x+y)+a_{32}(z+t)+a_{33} \ xy+a_{34} \ zt = b_{1} \\ a_{41}(x+y)+a_{42}(z+t)+a_{43} \ xy+a_{44} \ zt = b_{4} \end{cases} $$ such that : $ A = (a_{ij})_{1 \leq i,j \leq 4} $ is an invertible matrix, which is easy to solve in this case, right ? But, i don't know how to do it. Could you help me please ?

Thank you.

Answer 1

Using resultants or Buchberger's algorithm we obtain first two linear equations, which we can substitute: $$ x= \frac{132t^2 + 167t - 6}{8(31t - 1)},\; y= \frac{-144t^2 -125t -2 }{8(31t - 1)} $$ Then we immediately obtain $$ z=\frac{116t + 11}{4(31t - 1)}. $$ This yields, for the remaining equation, then $$ 19008t^4 + 37076t^3 + 18775t^2 - 2844t + 68=0. $$ This has $4$ solutions. $2$ of them are real.

Answer 2

You cannot do the substitution you want since the $4\times 4$ matrix with the coefficients of $x$, $y$, $z$, $t$ has rank $4$. What you can do is reduce to an equivalent system in $xy$ and $zt$, as follows:

HINT:

Write $xy = p$, $zt = q$, and think of them as parameters. Solve the linear system in $x$, $y$, $z$, $t$ and get $$x = 1/40 (2 p + 23 q + 25) \\y = 1/40 (22 p - 47 q - 35) \\ z = 1/40 (-58 p - 7 q + 35) \\ t = 1/40 (2 p + 43 q - 5)$$

So we get $$p = 1/1600(2p+23 q +25)(22 p - 47 q -35)\\ q = 1/1600(- 58 p - 7 q + 35)(2 p + 43 q -5)$$ hence an equivalent system of equations in $p$, $q$. That can be solved in principle.