Given that the system of equations `x=cy+bz ,y=az+cx , z=bx +ay` has nonzero solutions and and at least one of the a,b,c is a proper fraction.
abc is

Options

A. `gt-1`
B. `gt1`
C. `lt2`
D. `lt3`

Answer

A

Explanation

The system of equation
`-x+cy+bz=0`
`cx-y+az=0`
`bx+ay-z=0`
has a nonzero solution if
`Delta = |{:(-1,,c,,b),(c,,-1,,a),(b,,a,,-1):}|=0`
then Clearly the system has infinitely many solutions .From (1) and (2) we have
`(x)/(ac+b) =(y)/(bc+a)=(z)/(1-c^(2)) `
`" or " (x^(2))/((1-a^(2))(1-c^(2)))=(y^(2))/((1-b^(2))(1-c^(2)a)) =(z^(2))/((1-c^(2))^(2))`[From (4)]
`" or " (x^(2))/(1-a^(2))=(y^(2))/(1-b^(2)) =(z^(2))/(1-c^(2))`
from (5) we see that `1-a^(2),1-b^(2),1-c^(2)` are all positive or all negative .Given that one of a,b,c is proper fraction so
`1-a^(2) gt ,1-b^(2) gt 0,1-c^(2) gt 0` which gives `a^(2) +b^(2)+c^(2) lt 3`
using (4) and (6) we get
`1lt 3+2 abc`
`" or " abc gt -1`