Question

Given that the system of equations $x=cy+bz,y=az+cx,z=bx+ay$ has nonzero system of solution such that:

• A. $x:y:z\equiv (1-2a^2):(1-2b^2):(1-2c^2)$
• B. $x:y:z\equiv \frac{1}{1-2a^2}:\frac{1}{1-2b^2}:\frac{1}{1-2c^2}$
• C. $x:y:z\equiv \frac{a}{1-a^2}:\frac{b}{1-b^2}:\frac{c}{1-c^2}$
• D. $x:y:z\equiv \sqrt{1-a^2}:\sqrt{1-b^2}:\sqrt{1-c^2}$

Answer 1

Answer: (D)

$x:y:z\equiv \sqrt{1-a^2}:\sqrt{1-b^2}:\sqrt{1-c^2}$

Given that the system of equations $x=cy+bz,y=az+cx,z=bx+ay$ has nonzero solutions.
Let $x=cy+bz$ ----(1)
$y=az+cx$ ------(2)
$z=bx+ay$ ------(3)
substituting equation (1) in (2) gives
$y=az+c(cy+bz)$
$\Rightarrow y(1-c^2)=(a+bc)z$
$\Rightarrow \frac{y}{z}=\frac{a+bc}{1-c^2}$ ---(4)
substituting equation (1) in (3) gives
$z=b(cy+bz)+ay$
$\Rightarrow z(1-b^2)=(a+bc)y$
$\Rightarrow \frac{y}{z}=\frac{1-b^2}{a+bc}$ ----(5)
from (4) and (5)
$\frac{y^2}{z^2}=\frac{a+bc}{1-c^2}\frac{1-b^2}{a+bc}=\frac{1-b^2}{1-c^2}$
$\Rightarrow \frac{y}{z}=\frac{\sqrt{1-b^2}}{\sqrt{1-c^2}}$
Similarly $\frac{x}{y}=\frac{\sqrt{1-a^2}}{\sqrt{1-b^2}}$
$\therefore x:y:z=\sqrt{1-a^2}:\sqrt{1-b^2}:\sqrt{1-c^2}$
Hence, option D.