Can't express variables from the system of equations

Question

It a homework task

Can it be transformed to the system, where each expression of N doesn't depend on another N(like $N1=$some stuff that doesn't contain any N):

$$N_3=\dfrac{-pB_{13}N_1}{-A_{31}-pB_{32}-A_{32}}$$ $$N_2=\dfrac{(pB_{32}+A_{32})N_3}{A_{21}}$$ $$N_1=\dfrac{A_{31}N_3+A_{21}N_2}{pB_{13}}$$

Consider $p$, $B_{32}$, $B_{13}$, $A_{32}$, $A_{21}$, $A_{31}$, (shortly all B and A and p) as a constant

Now I clarify, that I want to transform the system above to system

$$N_3=stuff\space without\space any\space N$$ $$N_2=stuff\space without\space any\space N$$ $$N_1=stuff\space without\space any\space N$$

I don't really need a complete solution, just answer is it possible or, because my math skills are not pro, and I am getting $N_1+N_1(some\space stuff)=0$

Is it possible?

Answer 1

If we plug the second into the third, then the first into the third we eliminate $N_2$ and $N_3$

$$N_1=\dfrac{A_{31}N_3+A_{21}N_2}{pB_{13}}\\ N_1=\dfrac{A_{31}N_3+(pB_{32}+A_{32})N_3}{pB_{13}}\\ N_1=\dfrac{A_{31}+(pB_{32}+A_{32})}{pB_{13}}N_3\\ N_1=\dfrac{A_{31}+(pB_{32}+A_{32})}{pB_{13}}\dfrac{-pB_{13}N_1}{-A_{31}-pB_{32}-A_{32}}$$ Now note that the $N_1$s divide out so your system is singular. If what is multiplying $N_1$ on the right is equal to $1$ or $0$ you can choose $N_1$ at will. Otherwise the equations are inconsistent.

Answer 2

We need to assume $p B_{1 3} \neq 0$ otherwise the last equation is meaningless. This last equation is then a consequence of the two first equations. Hence it can be removed. $N_1$ can take any value, the first equation gives $N_3$ and the second equation then gives $N_2$.

You cannot transform the system the way you want because this system doesn't have a unique solution. There is an arbitrary parameter in the solution. You can write it for example as \begin{equation} \begin{array}{} N_1 = k\cr N_2 = x k\cr N_3 = y k \end{array} \end{equation} where $k$ is arbitrary. If you need only one solution, choose all the $N_i=0$