# Help to prove

• Jan 24th 2006, 12:30 PM
DenMac21
Help to prove
I have to prove if a,b,c <> 0 and
$\displaystyle \frac{{ay - bx}}{c} = \frac{{cx - az}}{b} = \frac{{bz - cy}}{a}$ then is $\displaystyle \frac{x}{a} = \frac{y}{b} = \frac{z}{c}$

Help someone?
• Jan 24th 2006, 10:14 PM
rgep
Write the common value $\displaystyle \frac{{ay - bx}}{c} = \frac{{cx - az}}{b} = \frac{{bz - cy}}{a} = t$ and treat it as a system of three linear equations in four unknowns x,y,z,t. Note that the rank of the system is three, since the determinant of the 3-by-3 minor corresponding to x,y,z is 2abc: hence there is a one-dimensional family of solutions. Since x=a, y=b, z=c, t=0 is clearly a solution, all solutions are multiples of this.
• Jan 25th 2006, 01:57 AM
DenMac21
Quote:

Originally Posted by rgep
Write the common value $\displaystyle \frac{{ay - bx}}{c} = \frac{{cx - az}}{b} = \frac{{bz - cy}}{a} = t$ and treat it as a system of three linear equations in four unknowns x,y,z,t. Note that the rank of the system is three, since the determinant of the 3-by-3 minor corresponding to x,y,z is 2abc: hence there is a one-dimensional family of solutions. Since x=a, y=b, z=c, t=0 is clearly a solution, all solutions are multiples of this.

Can you explain me solution little bit more?
I didn't quite understand it.