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 $(1),(2)$.


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.

  • $\begingroup$ Thanks, but the first eqn. has rhs = $-Bx_3$; and am not clear why sign is negative. I mean that the lhs having $-C$ as coefficient of $x_2$ is justified; but need reason for having rhs' coefficient as $-B$ instead of $B$. $\endgroup$
    – jiten
    Jul 2 '20 at 10:39
  • 1
    $\begingroup$ You can have it as $B$, then you will need to write $x_2=-B$. It's up to you $\endgroup$ Jul 2 '20 at 10:44

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