It is given on pg. #106, 107 in the book by: Thomas Banchoff, John Wermer; titled: Linear Algebra Through Geometry, second edn..
Consider a system of two equations in three unknowns:
$$a_1x_1 + a_2x_2 + a_3x_3 = 0$$
$$b_1x_1 + b_2x_2 + b_3x_3 = 0$$
We set $A = \begin{bmatrix} a_1 \\ a_2 \\ a_3 \end{bmatrix}, B = \begin{bmatrix} b_1\\ b_2 \\ b_3 \end{bmatrix}$ in $\mathbb{R}^3$. A solution vector $X = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}$ gives $A.X=0, B.X=0$.
We may find such an $X$ by multiplying the first equation by $b_1$ and the
second by $a_1$ and subtracting
$$a_1b_1x_1 + a_2b_1x_2 + a_3b_1x_3 = 0,$$
$$a_1b_1x_1 + a_1b_2x_2 + a_1b_3x_3 = 0,$$
$$(1): \ (a_2b_1 – a_1b_2)x_2 + (a_3b_1 – a_1b_3)x_3 = 0.$$
Similarly, we may multiply the first equation by $b_2$ and the second by $a_2$
and subtract to get
$$(2):\ (a_1b_2 – a_2b_1)x_1 + (a_3b_2 – a_2b_3)x_3 = 0.$$
We can obtain a solution to the system $(1), (2)$ by choosing
$$x_1 = (a_2b_3-a_3b_2), x_2=(a_3b_1-a_1b_3), x_3 =(a_1b_2-a_2b_1)$$
I am confused over how the author made such a choice for the coefficients of $x_1, x_2, x_3$ so as to derive from $(1),(2)$.
Best Answer
The derivation of (1) (or (2)) follows the logic that we want to eliminate $x_1$ (or $x_2$).
The system (1), (2) itself can seen as: $$ -Cx_2=-Bx_3,\\ Cx_1=Ax_3. $$ Now it is almost obvious, that if $x_3=C$, $x_2=B$ and $x_1=A$, then the system will be satisfied.