Question

Solve by matrix inversion method.

$\frac{4}{x}-\frac{6}{y}+\frac{5}{z}=1$

$\frac{6}{x}+\frac{9}{y}-\frac{20}{z}=1$

$\frac{2}{x}+\frac{3}{y}-\frac{10}{z}=4$

Answer 1

Above system of equation can be written in matrix form as:

$AX=B$

$X=A^{-1}B$

$A=\left[\begin{array}{ccc}2& 3& 10\\ 4& -6& 5\\ 6& 9& -20\end{array}\right],X=\left[\begin{array}{c}u\\ v\\ w\end{array}\right],B=\left[\begin{array}{c}4\\ 1\\ 2\end{array}\right]$, where $u=\frac{1}{x},v=\frac{1}{y},w=\frac{1}{z}$

$\left|A\right|=\left|\begin{array}{ccc}2& 3& 10\\ 4& -6& 5\\ 6& 9& -20\end{array}\right|=2(120-45)-3(-80-30)+10(36+36)=150+330+720=1200\ne 0$

For adj$A:$ $A_{11}=120-45=75$

$A_{12}=-(-60-90)=150$

$A_{13}=15+60=75$

$A_{12}=-(-80-30)=110$

$A_{22}=-40-60=-100$

$A_{32}=-(10-40)=30$

$A_{13}=36+36=72$

$A_{23}=-(18-18)=0$

$A_{33}=-12-12=-24$

$adj.A={\left[\begin{array}{ccc}75& 110& 72\\ 150& -100& 0\\ 75& 30& -24\end{array}\right]}^T=\left[\begin{array}{ccc}75& 150& 75\\ 110& -100& 30\\ 72& 0& -24\end{array}\right]$

$A^{-1}=\frac{1}{\left|A\right|}.adj.A=\frac{1}{1200}\left[\begin{array}{ccc}75& 150& 75\\ 110& -100& 30\\ 72& 0& -24\end{array}\right]$

$\left[\begin{array}{c}u\\ v\\ w\end{array}\right]=\frac{1}{1200}\left[\begin{array}{ccc}75& 150& 75\\ 110& -100& 30\\ 72& 0& -24\end{array}\right]\left[\begin{array}{c}4\\ 1\\ 2\end{array}\right]$

$\left[\begin{array}{c}u\\ v\\ w\end{array}\right]=\frac{1}{1200}\left[\begin{array}{c}300+150+300\\ 440-100+60\\ 288+0-48\end{array}\right]$

$\left[\begin{array}{c}u\\ v\\ w\end{array}\right]=\frac{1}{1200}\left[\begin{array}{c}600\\ 400\\ 240\end{array}\right]$

$\left[\begin{array}{c}u\\ v\\ w\end{array}\right]=\left[\begin{array}{c}\frac{1}{2}\\ \frac{1}{3}\\ \frac{1}{5}\end{array}\right]$

$x=2,y=3,z=5$