If $a \neq 1, b \neq-1, c \neq-1$ and the system of equations, $x=a(y+z), y=b(z+x), z=c(x+y)$ has a…

If $a \neq 1, b \neq-1, c \neq-1$ and the system of equations, $x=a(y+z), y=b(z+x), z=c(x+y)$ has a non-trivial solution, then.
  1. $\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}=0$
  2. $\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}=1$
  3. $\frac{a b c}{(a+1)(b+1)(c+1)}=1$
  4. $\frac{a+b+c}{(a+1)(b+1)(c+1)}=2$

Solution

For system of homogeneous equation, if it has non-trivial solution, then $\Delta=0$, so $ \left|\begin{array}{ccc} 1 & -a & -a \\ -b & 1 & -b \\ -c & -c & 1 \end{array}\right|=0 \Rightarrow\left|\begin{array}{ccc} -\frac{1}{a} & 1 & 1 \\ 1 & -\frac{1}{b} & 1 \\ 1 & 1 & -\frac{1}{c} \end{array}\right|=0 $ $ \begin{aligned} & R_2 \rightarrow R_2-R_1 \text { and } R_3 \rightarrow R_3-R_1 \\ & \Rightarrow\left|\begin{array}{ccc} -\frac{1}{a} & 1 & 1 \\ 1+\frac{1}{a} & -\frac{1}{b}-1 & 0 \\ 1+\frac{1}{a} & 0 & -\frac{1}{c}-1 \end{array}\right|=0 \\ & \Rightarrow-\frac{1}{a}\left(\frac{1}{b}+1\right)\left(\frac{1}{c}+1\right)+\left(1+\frac{1}{a}\right)\left(1+\frac{1}{c}\right) \\ & \left.\Rightarrow \frac{(1+a)(1+c)}{a c}+\frac{(1+a)(1+b)}{a}+1\right)\left(1+\frac{1}{a}\right)=0 \\ & \Rightarrow \quad \frac{(1+b)(1+c)}{a b c} \\ & \Rightarrow \quad \frac{a}{1+b}+\frac{c}{1+c}=\frac{1}{1+a}-1+1 \\ & \Rightarrow \frac{b}{1+a}+\frac{c}{1+c}=1 \end{aligned} $

Asked in: AP EAMCET 2018 (23 Apr Shift 2)

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