Homogenous Linear Equations in the form of Determinants In Arfken's "Mathematical Methods for Physicists"
How did he arrive to:
$x_1/x_3 = \frac{(a_2b_3-a_3b_2)}{(a_1b_2-a_2b_1)}$
Starting from:- 
$$
a_1x_1+a_2x_2+a_3x_3=0 ; \\
b_1x_1+b_2x_2+b_3x_3=0 ; \\
c_1x_1+c_2x_2+c_3x_3=0
$$
Given that C is a linear combination of A and B.
 A: Since $C$ is linearly dependent on $A$ and $B$, we need not consider it, it does not add any information.
If you look at the system
$$\begin{align}
a_1 x_1 + a_3x_3 = -a_2x_2 \\
b_1 x_1 + b_3x_3 = -b_2x_2
\end{align}$$
you can see, by Cramer's rule, that
$$x_1 = \frac{\begin{vmatrix} -a_2x_2 & a_3 \\ -b_2x_2 & b_3 \\ \end{vmatrix}}{\begin{vmatrix} a_1 & a_3 \\ b_1 & b_3 \end{vmatrix}}
= -x_2 \frac{\begin{vmatrix} a_2 & a_3 \\ b_2 & b_3 \\ \end{vmatrix}}{\begin{vmatrix} a_1 & a_3 \\ b_1 & b_3 \end{vmatrix}}
$$
and also that
$$x_3 = \frac{\begin{vmatrix} a_1 & -a_2x_2 \\ b_1 & -b_2x_2 \end{vmatrix}}{\begin{vmatrix} a_1 & a_3 \\ b_1 & b_3 \end{vmatrix}}
= -x_2 \frac{\begin{vmatrix} a_1 & a_2 \\ b_1 & b_2 \end{vmatrix}}{\begin{vmatrix} a_1 & a_3 \\ b_1 & b_3 \end{vmatrix}}
$$
and so you will have:
$$\frac{x_1}{x_3} = -x_2 \frac{\begin{vmatrix} a_2 & a_3 \\ b_2 & b_3 \\ \end{vmatrix}}{\begin{vmatrix} a_1 & a_3 \\ b_1 & b_3 \end{vmatrix}} \Bigg/ -x_2 \frac{\begin{vmatrix} a_1 & a_2 \\ b_1 & b_2 \end{vmatrix}}{\begin{vmatrix} a_1 & a_3 \\ b_1 & b_3 \end{vmatrix}} =
\frac{\begin{vmatrix} a_2 & a_3 \\ b_2 & b_3 \\ \end{vmatrix}}{\begin{vmatrix} a_1 & a_2 \\ b_1 & b_2 \end{vmatrix}}
= \frac{a_2b_3 - a_3b_2}{a_1b_2 - a_2b_1}
$$
