Solve the system:$\,\small\begin{eqnarray}a_1P+b_1Q+c_1R&=&0\\ a_2P+b_2Q+c_2R&=&0\\ a_3P+b_3Q+c_3R&=&0 \end{eqnarray}$ where $\small PQR\neq 0$

Question

Consider the system of diophantine equations: \begin{eqnarray} a_1P+b_1Q+c_1R&=&0\\ a_2P+b_2Q+c_2R&=&0\\ a_3P+b_3Q+c_3R&=&0 \end{eqnarray} where $PQR\neq 0$. What is the easiest method to solve it?

What I have done: I calculated instead the ratios: $\dfrac{P}{R} $ and $\dfrac{Q}{R} $. I obtained for each 3 equivalent ratios. There it gets pretty cumbersome and I got lost. I wonder if there is easier method?

Answer 1

Correct me if I'm wrong, but there is no unique solution. The system can be written as $$\left(\begin{array}{ccc} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \\ \end{array}\right)\left(\begin{array}{c}P\\Q\\R\end{array}\right)=\left(\begin{array}{c}0\\0\\0\end{array}\right).$$

• If $\left(\begin{array}{ccc} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \\ \end{array}\right)$ is non-singular, then the above system has the unique solution $\left(\begin{array}{c}P\\Q\\R\end{array}\right)=\left(\begin{array}{c}0\\0\\0\end{array}\right)$. However, this is in contradiction with $PQR\not=0$. Hence, there is no solution.
• If $\left(\begin{array}{ccc} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \\ \end{array}\right)$ is singular, then the system has infinitely many solutions.

Answer 2

As Eric S. said, if the matrix is non-singular, then there is no nonzero solution. But if it is singular, then the column vectors are linearly dependent over $\mathbb{Q}$, which means, by definition, that we can write them as a linear combination of each other with nonzero rational coefficients. And this can be done with integers, as we're free to clear denominators.

This will be a solution with $PQR\ne 0$ as long as none or all of the columns are multiples of each other. If columns $a$ and $b$ are multiples of each other, then the images is going to be either 2 or 1 dimensional over $\mathbb{Q}$. If it is two dimensional, then the kernel is exactly the subspace with $R=0$ so there are no solutions with $PQR\ne 0$. If the image is one dimensional, then every column is a multiple of every other column and the kernel is two dimensional and so we can make $PQR\ne 0$.

Answer 3

Write it as the system $Ax=b$, i.e. $$\pmatrix{ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \\ }\pmatrix{P\\Q\\R}=\pmatrix{0\\0\\0}$$

Since you say that $PQR\ne 0$ you say that there is a dependency between the equations, i.e. this system can be transformed to an equivalent linear system with at most two equation, i.e. 1 equation or 2 equations. This would mean that the system has at least $1$ degree of freedom, perhaps even $2$.

If the system has one degree of freedom the solution will depend on a free choice of one of the unknowns, e.g. $P=f(R)$ and $Q=g(R)$, which would mean that given $R$ you can adjust your $P,Q$ such that it solves the system.

For two degrees of freedom your system is equivalent to one equation only and therefore your solution will be of the form of $P=f(Q,R)$ where $Q,R$ you can choose as you wish.