Solve by determinants: $\frac{1}{x}-\frac{2}{y}+\frac{1}{z}=3$
$\begin{aligned}
& \frac{2}{x}+\frac{2}{y}-\frac{3}{z}=9 \
& \frac{2}{x}-\frac{3}{y}+\frac{1}{z}=-9
\end{aligned}$

Question

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The Answer of Solve by determinants: $\frac{1}{x}-\frac{2}{y}+\frac{1}{z}=3$
$\begin{aligned}
& \frac{2}{x}+\frac{2}{y}-\frac{3}{z}=9 \
& \frac{2}{x}-\frac{3}{y}+\frac{1}{z}=-9
\end{aligned}$

Solve by Determinants 1x−2y+1z=3\frac{1}{x}-\frac{2}{y}+\frac{1}{z}=3x1​−y2​+z1​=3 2x+2y−3z=92x−3y+1z=−9\begin{aligned}& \frac{2}{x}+\frac{2}{y}-\frac{3}{z}=9 \\& \frac{2}{x}-\frac{3}{y}+\frac{1}{z}=-9\end{aligned}​x2​+y2​−z3​=9x2​−y3​+z1​=−9​

Solve by Determinants $\frac{1}{x}-\frac{2}{y}+\frac{1}{z}=3$ $\begin{aligned}& \frac{2}{x}+\frac{2}{y}-\frac{3}{z}=9 \\& \frac{2}{x}-\frac{3}{y}+\frac{1}{z}=-9\end{aligned}$