# Solving a linear system of reciprocals.

Solve for $$\begin{cases}\frac{1}{x} +\frac{1}{y}+\frac{1}{z}=0\\\frac{4}{x} +\frac{3}{y}+\frac{2}{z}=5\\\frac{3}{x} +\frac{2}{y}+\frac{4}{z}=-4\end{cases}$$

I turn the equations into $$\begin{cases}yz+xz+xy=0\\4yz+3xz+2xy=5xyz\\3yz+2xz+4xy=-4xyz\end{cases}$$

Not sure if I am doing fine

If the reciprocals are freaking you out, just let $$t=\frac1x,u=\frac1y,v=\frac1z,$$ so you have the system $$\begin{cases}t +u+v=0\\4t +3u+2v=5\\3t +2u+4v=-4\end{cases}$$ Once you've solved this, as long as none of $$t,u,v$$ is $$0,$$ you can simply let $$x=\frac1t,y=\frac1u,z=\frac1v.$$ If one or more of $$t,u,v$$ is $$0,$$ then the system has no solution.

It's much easier to solve the linear system for the reciprocals then take the reciprocal to get the result.

how about this, let: $$X=\frac1x,\,Y=\frac1y,\,Z=\frac1z$$ and it is much easier to solve the following: $$\begin{pmatrix} 1&1&1\\ 4&3&2\\ 3&2&4 \end{pmatrix} \begin{pmatrix} X\\ Y\\ Z \end{pmatrix}= \begin{pmatrix} 0\\5\\-4 \end{pmatrix}$$

Observe that the system's matrix of the reciprocal of the unknowns is:

$$A=\begin{pmatrix}1&1&1&0\\ 4&3&2&5\\ 3&2&4&-4\end{pmatrix}\stackrel{R_2-4R_1,\,R_3-3R_1}\longrightarrow\begin{pmatrix}1&1&1&0\\ 0&-1&-2&5\\ 0&-1&1&-4\end{pmatrix}\stackrel{R_3-R_2}\longrightarrow$$

$$\begin{pmatrix}1&1&1&0\\ 0&-1&-2&5\\ 0&0&3&-9\end{pmatrix}$$

Try to finish the exercise now, taking into account that the third column represents $$\;\cfrac 1z\;$$ , the second $$\;\cfrac1y\;$$ and the first one $$\;\cfrac1x\;$$

• There's an error at the 1st step: the third row should be $\;0\;-1\; \color{red}{+}1\;-4$. – Bernard Apr 28 at 15:57
• @Bernard Good catch, edited. Thanks. – DonAntonio Apr 28 at 18:43