<s>[INST] 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
 [/INST] 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 $\;\cfrac1x\;$
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