# Finding a solution to system of bilinear equations

I've got a system of equations

$$\begin{array}{rl} x_1 y_1 &= p_1\\ x_1 y_2 &= 1-p_1\\ x_2 y_2 &= p_2\\ x_2 y_3 &= 1-p_2\\ x_3 y_3 &= p_3\\ x_3 y_1 &= 1-p_3 \end{array}$$

where $p_i$ are constants. Is there any way to determine the number of solutions to this system without solving for the variables explicitly? Specifically, I want to know the solutions residing in the open interval $x_i, y_i \in (0,1)$.

• What about $x_1 y_3$? Commented Aug 17, 2016 at 11:58
• No constraints on that. Commented Aug 17, 2016 at 12:01
• Take the cologarithms. This turns the problem in an easy linear system of equations with a positivity constraint. You can write the explicit solution by inversion of the matrix.
– user65203
Commented Aug 17, 2016 at 12:19

$$\begin{bmatrix} x_1\\ x_2\\ x_3\end{bmatrix} \begin{bmatrix} y_1\\ y_2\\ y_3\end{bmatrix}^T = \begin{bmatrix} p_1 & 1-p_1 & *\\ * & p_2 & 1-p_2\\ 1-p_3 & * & p_3\end{bmatrix}$$

Hence, we have the following matrix completion problem

Find $t := (t_1, t_2, t_3)$ such that $$\begin{bmatrix} p_1 & 1-p_1 & t_1\\ t_2 & p_2 & 1-p_2\\ 1-p_3 & t_3 & p_3\end{bmatrix}$$ is a rank-$1$ matrix.

If the matrix is rank-$1$, then the 2nd and 3rd columns are multiples of the 1st column

$$\begin{array}{rl} 1-p_1 &= \alpha \, p_1\\ p_2 &= \alpha \, t_2\\ t_3 &= \alpha (1 - p_3)\\\\ t_1 &= \beta \, p_1\\ 1-p_2 &= \beta \, t_2\\ p_3 &= \beta (1-p_3)\end{array}$$

Hence,

$$\alpha = \dfrac{1-p_1}{p_1}\qquad \qquad t_2 = \dfrac{p_1 p_2}{1-p_1}\qquad \qquad t_3 = \dfrac{(1-p_1) (1 - p_3)}{p_1}$$

$$t_1 = \dfrac{p_1 p_3}{1-p_3}\qquad \qquad t_2 = \dfrac{(1-p_2) (1 - p_3)}{p_3}\qquad \qquad \beta = \frac{p_3}{1 - p_3}$$

In order to have a solution, we must have

$$\dfrac{p_1 p_2}{1-p_1} = \dfrac{(1-p_2) (1 - p_3)}{p_3}$$

or, in a much nicer form,

$$p_1 p_2 p_3 = (1 - p_1) (1-p_2) (1 - p_3)$$

Completing and factoring the matrix, we find one solution

$$\begin{bmatrix} p_1 & 1-p_1 & \dfrac{p_1 p_3}{1-p_3}\\ \dfrac{p_1 p_2}{1-p_1} & p_2 & 1-p_2\\ 1-p_3 & \dfrac{p_2 p_3}{1 - p_2} & p_3\end{bmatrix} = \begin{bmatrix} p_1\\ \dfrac{p_1 p_2}{1-p_1}\\ \dfrac{p_1 p_3}{1-p_3}\end{bmatrix} \begin{bmatrix} 1\\ \alpha\\ \beta\end{bmatrix}^T$$

The solvability of the system depends on the constants $p_i$. For example, for $(p_1,p_2,p_3)=(2,2,2)$ there is no solution. Using Buchberger's algorithm, or elementary elimination we see that the system has solutions if and only if the parameters satisfy the equation $$2p_1p_2p_3 - p_1p_2 - p_1p_3 - p_2p_3 + p_1 + p_2 + p_3 - 1=0,$$ i.e., if and only if either $(p_2,p_3)=(1,0)$, or if $$p_1=\frac{(p_2-1)(p_3-1)}{2p_2p_3 - p_2 - p_3 + 1}.$$ In the first case all solutions are given by $$x_1=\frac{ - p_1 + 1}{y_2}, \; x_2=\frac{1}{y_2},\; x_3=\frac{ - p_1 + 1}{p_1y_2},\; y_1=\frac{ - p_1y_2}{p_1 - 1},\; y_3=0.$$ Since $y_2\neq 0$ is arbitrary we obtain infinitely many solutions. A similar "formula" for the solutions can be obtained for the second case, e.g., $$x_1=\frac{p_3( - p_2 + 1)}{y_3(2p_2p_3 - p_2 - p_3 + 1)}, \cdots ,\cdots$$

• Thanks! So if $0<p_i <1$, there are zero solutions, unless the constraint you mentioned is satisfied, for which there are infinitely many? Commented Aug 17, 2016 at 12:08
• Yes, there are solutions if and only if $2p_1p_2p_3 - p_1p_2 - p_1p_3 - p_2p_3 + p_1 + p_2 + p_3 - 1=0,$ in which case there are infinitely many. Commented Aug 17, 2016 at 12:11