Question

Using elementary transformation on matrix, solve the system of equations.
$\frac{2}{x}+\frac{3}{y}+\frac{10}{z}=4$
$\frac{4}{x}-\frac{6}{y}+\frac{5}{z}=1$
$\frac{6}{x}+\frac{9}{y}-\frac{20}{z}=2$

Answer 1

Given equations
$\frac{2}{x}+\frac{3}{y}+\frac{10}{z}=4$
$\frac{4}{x}-\frac{6}{y}+\frac{5}{z}=1$
$\frac{6}{x}+\frac{9}{y}-\frac{20}{z}=2$
Let $\frac{1}{x}=a,\frac{1}{x}=b$ and $\frac{1}{z}=c$, then$2a+3b+10c=4$
$4a-6b+5c=1$
$6a+9b-20c=2$
Here , $\Delta =\left|\begin{array}{ccc}2& 3& 10\\ 4& -6& 5\\ 6& 9& -20\end{array}\right|$
$=2(120-45)-3(-80-30)+10(36+36)$$=150+330+720$
$=1200$
${\Delta }_1=\left|\begin{array}{ccc}4& 3& 10\\ 1& -6& 5\\ 2& 9& -20\end{array}\right|$
$=4(120-45)-3(-20-10)+10(9+12)$
$=300+90+210$
$=600$
${\Delta }_2=\left|\begin{array}{ccc}2& 4& 10\\ 4& 1& 5\\ 6& 2& -20\end{array}\right|$$=2(-20-10)-4(-80-30)+10(8-6)$$=-60+440+20$$=400$
and ${\Delta }_3=\left|\begin{array}{ccc}2& 3& 4\\ 4& -6& 1\\ 6& 9& 2\end{array}\right|$
$=2(-12-9)-3(8-6)+4(36+36)$
$=-42-6+288$
$=240$
Using Carame's rule.
$x=\frac{{\Delta }_1}{\Delta }=\frac{600}{1200}=\frac{1}{2}$
$y=\frac{{\Delta }_2}{\Delta }=\frac{400}{1200}=\frac{1}{3}$
$z=\frac{{\Delta }_3}{\Delta }=\frac{240}{1200}=\frac{1}{5}$
$\therefore a=\frac{1}{2}\to \frac{1}{x}=\frac{1}{2}\Rightarrow x=2$
$b=\frac{1}{3}\to \frac{1}{y}=\frac{1}{3}\Rightarrow y=3$
$c=\frac{1}{5}\to \frac{1}{z}=\frac{1}{5}\Rightarrow z=5$
Hence, $x=2,y=3,z=5$