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