Question

Solve the system of equations.

$\frac{2}{x}+\frac{3}{y}+\frac{10}{z}=4;\frac{4}{x}-\frac{6}{y}+\frac{5}{z}=1$ and $\frac{6}{x}+\frac{9}{y}-\frac{20}{z}=2$

Answer 1

Let $\frac{1}{x}=p,\frac{1}{y}=q$, and $\frac{1}{z}=r$, then the given equations become $2p+3q+10r=4,4p-6q+5r=1,6p+9q-20r=2$
The system can be written as AX=B where
$A=\left[\begin{array}{ccc}2& 3& 10\\ 4& -6& 5\\ 6& 9& -20\end{array}\right],X=\left[\begin{array}{c}p\\ q\\ r\end{array}\right],B=\left[\begin{array}{c}4\\ 1\\ 2\end{array}\right]$
Here, $|A|=\left[\begin{array}{ccc}2& 3& 10\\ 4& -6& 5\\ 6& 9& -20\end{array}\right]=2(120-45)-3(-80-30)+10(36+36)=150+330+720=1200$

Thus, A is non-singular. Therefore, its inverse exists.
Therefore, the above system is consistent and has a unique solution given by $X=A^{-1}B$
Cofactors of A are
$A_{11}=120-45=75,A_{12}=-(-80-30)=110,A_{13}=(36+36)=72A_{21}=-(-60-90)=150,A_{22}=(-40-60)=-100,A_{23}=-(18-18)=0A_{31}=15+60=75,A_{32}=-(10-40)=30,A_{33}=-12-12=-24$

$\therefore adj\left(A\right)={\left[\begin{array}{ccc}75& 110& 72\\ 150& -100& 0\\ 75& 30& -24\end{array}\right]}^T=\left[\begin{array}{ccc}75& 150& 75\\ 110& -100& 30\\ 72& 0& -24\end{array}\right]$
Now, $A^{-1}=\frac{1}{|A|}\left(adjA\right)=\frac{1}{1200}\left[\begin{array}{ccc}75& 150& 75\\ 110& -100& 30\\ 72& 0& -24\end{array}\right]$

$\therefore x=A^{-1}B\Rightarrow \left[\begin{array}{c}p\\ q\\ r\end{array}\right]=\frac{1}{1200}\left[\begin{array}{ccc}75& 150& 75\\ 110& -100& 30\\ 72& 0& -24\end{array}\right]\left[\begin{array}{c}4\\ 1\\ 2\end{array}\right]$

$=\frac{1}{1200}\left[\begin{array}{ccc}300& +150& +150\\ 440& -100& +60\\ 288& +0& -48\end{array}\right]=\frac{1}{1200}\left[\begin{array}{c}600\\ 400\\ 240\end{array}\right]=\left[\begin{array}{c}\frac{1}{2}\\ \frac{1}{3}\\ \frac{1}{5}\end{array}\right]$

$\Rightarrow p=\frac{1}{2},q=\frac{1}{3},r=\frac{1}{5}\Rightarrow \frac{1}{x}=\frac{1}{2},\frac{1}{y}=\frac{1}{3},\frac{1}{z}=\frac{1}{5}\Rightarrow x=2,y=3$ and $z=5$.