If a 1, b -1, c -1 and the system of equations, x=a(y+z), y=b(z+x), z=c(x+y) has a non-trivial solution, then.

Step-by-step solution

For system of homogeneous equation, if it has non-trivial solution, then =0 , so | array ccc 1 & -a & -a -b & 1 & -b -c & -c & 1 array |=0 | array ccc - 1 a & 1 & 1 1 & - 1 b & 1 1 & 1 & - 1 c array |=0 aligned & R₂ R₂-R₁ and R₃ R₃-R₁ & | array ccc - 1 a & 1 & 1 1+ 1 a & - 1 b -1 & 0 1+ 1 a & 0 & - 1 c -1 array |=0 & - 1 a ( 1 b +1 ) ( 1 c +1 )+ (1+ 1 a ) (1+ 1 c ) & . (1+a)(1+c) a c + (1+a)(1+b) a +1 ) (1+ 1 a )=0 & (1+b)(1+c) a b c & a 1+b + c 1+c = 1 1+a -1+1 & b 1+a + c 1+c =1 aligned