Question

Solve the system of the following 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

Let $\frac{1}{x}=p,\frac{1}{y}=q,\frac{1}{z}=r$.

Then the given system of equations is as follows:
2p+q+10r=4
4p-6q+5r=1
6p+9q-20r=2

This system can be written in the form of AX= B, where
A=$\begin{vmatrix} 2 &3 &10 \\\ 4&-6 &5 \\\ 6&9 &-20 \end{vmatrix}$, X=$\begin{vmatrix} p\\\ q\\\ r \end{vmatrix}$and B=$\begin{vmatrix} 4\\\ 1\\\ 2 \end{vmatrix}$

Now,
|A|=2(120-45)-3(-80-30)+10(36+36)
=150+330+720
=1200

Thus, A is non-singular. Therefore, its inverse exists.

Now,
A11=75,A12=110,A13=72
A21=150,A22=−100,A23=0
A31=75,A32=30,A33=−24

∴A-1=$\frac{1}{|A|}$adj A
=\frac{1}{1200}$$\begin{vmatrix} 75 &150 &75 \\\ 110&-100 &30 \\\ 72&0 &-24 \end{vmatrix}

Now,
X=A-1B
=\frac{1}{1200}$$\begin{vmatrix} 600\\\ 400\\\ 240 \end{vmatrix}=$\begin{vmatrix} \frac{1}{2}\\\ \frac{1}{3}\\\ \frac{1}{5} \end{vmatrix}$
∴p=$\frac{1}{2}$,q=$\frac{1}{3}$ and r=$\frac{1}{5}$

Hence,x=2,y=3 and z=5